RBSE Solutions Class 12 Physics Chapter 10 Alternating Current

RBSE Class 12 Physics Chapter 10 Multiple Choice Type Questions

Question 1. The rms value of alternating current is :
(a) twice of peak value
(b) half of peak value
(c) equal to peak value
(d) \( \frac{1}{\sqrt{2}} \) time of peak value
Answer: (d) \( \frac{1}{\sqrt{2}} \) time of peak value
In simple words: The root mean square (rms) value of an alternating current is found by dividing its peak value by the square root of 2. This is how we measure the effective strength of an AC current.

🎯 Exam Tip: Remember the relation between RMS and peak values for both current and voltage: \( I_{rms} = \frac{I_{m}}{\sqrt{2}} \) and \( V_{rms} = \frac{V_{m}}{\sqrt{2}} \).

Question 3. The current lags behind the voltage by a phase different of \( \frac{\pi}{2} \) radian, when in the circuit is :
(a) only resistance
(b) only inductance
(c) only capacitor
(d) both capacitor and resistance
Answer: (b) only inductance
In simple words: In a circuit with only an inductor, the voltage reaches its highest point a quarter-cycle (90 degrees or \( \frac{\pi}{2} \) radians) before the current does. So, the current is behind the voltage.

🎯 Exam Tip: Recall phase relationships: current and voltage are in phase for resistance, current leads voltage in capacitance, and current lags voltage in inductance.

Question 4. The unit of WC is :
(a) Ohm
(b) mho
(c) volt
(d)
Answer: (b) mho
In simple words: The term \( \omega C \) represents the reciprocal of capacitive reactance (\( X_C \)). Since \( X_C \) is measured in Ohms, its reciprocal \( \frac{1}{\omega C} \) is measured in mho (which is the unit for conductance). The question seems to have a typo, likely asking for \( \frac{1}{\omega C} \).

🎯 Exam Tip: Be careful with units! Capacitive reactance \( X_C = \frac{1}{\omega C} \) is in ohms (Ω), while its inverse, \( \omega C \), is in mho or Siemens (S).

Question 5. Capacitor in a circuit:
(a) passes the alternating current
(b) stops the flow of alternating current
(c) Typesetting math: 14% urrent
(d) stops the alternating current and passes the direct current.
Answer: (a) passes the alternating current
In simple words: A capacitor lets alternating current (AC) pass through it easily. However, it blocks direct current (DC) because its resistance (reactance) becomes infinitely large for DC.

🎯 Exam Tip: Remember that a capacitor acts as an open circuit for DC (blocking it) and a short circuit for high-frequency AC (passing it).

Question 6. Which of the following unit is not same
(a) \( \frac{1}{\sqrt{LC}} \)
(b) \( \sqrt{LC} \)
(c) RC
(d) \( \frac{L}{R} \)
Answer: (a) \( \frac{1}{\sqrt{LC}} \)
In simple words: The units for \( \sqrt{LC} \), RC, and \( \frac{L}{R} \) all represent time or time constants. However, the unit for \( \frac{1}{\sqrt{LC}} \) is radians per second (angular frequency), so it is different from the others.

🎯 Exam Tip: To compare units, consider dimensional analysis. \( \sqrt{LC} \), RC, and \( \frac{L}{R} \) all have dimensions of time. \( \frac{1}{\sqrt{LC}} \) has dimensions of inverse time (frequency).

Question 7. An alternating current circuit is in resonance at 10 kHz frequency. If frequency increases to 12 kHz, then what will be effect on impedance?
(a) remains unchanged
(b) increases 1.2 times
(c) increases and becomes capacitive
(d) increases and becomes inductive
Answer: (d) increases and becomes inductive
In simple words: At resonance, the circuit's impedance is at its lowest (equal to resistance). If the frequency goes up from the resonance point, the inductive reactance becomes larger than the capacitive reactance, causing the overall impedance to increase and the circuit to act more like an inductor.

🎯 Exam Tip: Above resonance frequency, \( X_L > X_C \), making the circuit inductive. Below resonance, \( X_C > X_L \), making it capacitive.

Question 8. In a circuit, the phase of the current is lagging behind \( \frac{\pi}{3} \) angle with phase of voltage. Element present in circuit are:
(a) R and C
(b) R and L
(c) L and C
(d) only L
Answer: (b) R and L
In simple words: When the current lags behind the voltage in an AC circuit, it means there is an inductor present. If there's also a resistance, the phase difference is less than 90 degrees, like \( \frac{\pi}{3} \) radians.

🎯 Exam Tip: A lagging current indicates an inductive circuit, while a leading current indicates a capacitive circuit. If the phase angle is not \( \frac{\pi}{2} \), it suggests the presence of resistance too.

Question 10. How can you decrease current in an alternating current circuit without any loss of power?
(a) By using pure inductor
(b) By using pure resistance
(c) By using resistance and inductor
(d) By using resistance and capacitor
Answer: (a) By using pure inductor
In simple words: You can reduce current in an AC circuit without losing power by adding a pure inductor or a pure capacitor. These components resist current flow (reactance) but don't convert electrical energy into heat, so there's no power loss.

🎯 Exam Tip: Pure inductors and capacitors have zero average power dissipation because the current and voltage are 90 degrees out of phase, making the power factor zero.

Question 11. Current \( I = I_{m} \sin \left(\omega t-\frac{\pi}{2}\right) \) is flowing in an ac circuit. If alternating voltage is \( V = V_{m} \sin \omega t \), then consumed power is :
(a) \( \frac{V_{m}I_{m}}{R} \)
(b) \( \frac{V_{m}I_{m}}{\sqrt{2}} \)
(c) \( \frac{VI}{2} \)
(d) zero
Answer: (d) zero
In simple words: If the current is a quarter-cycle (90 degrees or \( \frac{\pi}{2} \) radians) behind the voltage, it means the circuit is purely inductive or capacitive. In such circuits, no average power is used up because the power factor (cos 90°) is zero.

🎯 Exam Tip: When current and voltage are 90 degrees out of phase, the average power consumed in the circuit is always zero. This happens in purely inductive or purely capacitive circuits.

Question 13. The core of the transformer is laminated, because :
(a) magnetic flux may increase
(b) the residual magnetism may reduce
(c) the magnetic susceptibility of core may increase
(d) there may be minimum energy loss due to eddy currents
Answer: (d) there may be minimum energy loss due to eddy currents
In simple words: Transformers use laminated (thin, insulated layers) cores to stop eddy currents from forming large loops. This reduces energy lost as heat and makes the transformer more efficient.

🎯 Exam Tip: Laminating the core is a key design feature in transformers to minimize eddy current losses, which are a major source of energy waste.

Question 14. In the following figure, the point representing the condition of resonance is :
(a) A
(b) B
(c) C
(d) D
Answer: (a) A
In simple words: In the given graph, point 'A' is where the inductive reactance (XL) curve crosses the capacitive reactance (Xc) curve. This intersection point shows the condition of resonance, where \( X_L = X_C \).

🎯 Exam Tip: Resonance in an LCR circuit occurs when inductive reactance equals capacitive reactance (\( X_L = X_C \)). This is the point of minimum impedance and maximum current.

RBSE Class 12 Physics Chapter 10 Very Short Answer Type Questions

Question 1. The equation of an alternating voltage is \( v = 200 \sqrt{2} \sin 100 \pi t \). Find out the its root mean square value and frequency
Answer:
Given the voltage equation: \( v = 200 \sqrt{2} \sin 100 \pi t \)
Comparing with the standard equation \( v = V_{m} \sin \omega t \), we get:
Peak voltage, \( V_{m} = 200 \sqrt{2} \, V \)
Angular frequency, \( \omega = 100 \pi \, rad/s \)

Root Mean Square (RMS) value of voltage:
\( V_{rms} = \frac{V_{m}}{\sqrt{2}} = \frac{200 \sqrt{2}}{\sqrt{2}} = 200 \, V \)

Frequency (f):
We know that \( \omega = 2 \pi f \)
\( 100 \pi = 2 \pi f \)
\( f = \frac{100 \pi}{2 \pi} = 50 \, Hz \)
Therefore, the root mean square value is 200 V and the frequency is 50 Hz.
In simple words: We look at the given voltage formula. The number multiplied by \( \sqrt{2} \) gives us the highest voltage, and by dividing it by \( \sqrt{2} \), we get the effective voltage. For frequency, we use the number next to \( \pi t \) to find how many cycles happen per second.

🎯 Exam Tip: Always extract \( V_m \) and \( \omega \) correctly from the given equation. Remember \( V_{rms} = V_m / \sqrt{2} \) and \( f = \omega / (2\pi) \).

Question 2. Write a relation between root mean square value and peak value of alternating current.
Answer:
The relationship between the root mean square (RMS) value and the peak value of an alternating current is:
\( I_{rms} = \frac{I_{m}}{\sqrt{2}} \)
where \( I_{rms} \) is the RMS value of the current and \( I_{m} \) is the peak value of the current. This shows that the effective current is always less than the maximum current.
In simple words: The effective value of AC current (RMS) is found by taking its highest value (peak) and dividing it by the square root of 2.

🎯 Exam Tip: Ensure you write the correct relationship: RMS value is peak value divided by \( \sqrt{2} \), not multiplied. This is a common point of confusion.

Question 3. The equation of an alternating current is I = Im sin wt. Then write alternating voltage equation in inductor circuit.
Answer:
Given the alternating current equation: \( I = I_{m} \sin \omega t \)
In a purely inductive circuit, the voltage leads the current by a phase angle of \( \frac{\pi}{2} \) radians (or 90 degrees).
Therefore, the alternating voltage equation in an inductor circuit is:
\( V = V_{m} \sin \left(\omega t+\frac{\pi}{2}\right) \)
In simple words: When AC goes through only an inductor, the voltage wave reaches its peak earlier than the current wave by a quarter of a cycle.

🎯 Exam Tip: Clearly remember that in an inductor, voltage leads current by \( \frac{\pi}{2} \), while in a capacitor, current leads voltage by \( \frac{\pi}{2} \).

Question 4. If voltage is V = 200 sin 314 t, then find frequency of the alternating current.
Answer:
Given the voltage equation: \( V = 200 \sin 314 t \)
Comparing this with the standard alternating voltage equation \( V = V_{m} \sin \omega t \), we can identify:
Angular frequency, \( \omega = 314 \, rad/s \)

To find the frequency (f), we use the relation \( \omega = 2 \pi f \):
\( 314 = 2 \pi f \)
\( f = \frac{314}{2 \pi} \)
Using \( \pi \approx 3.14 \):
\( f = \frac{314}{2 \times 3.14} = \frac{314}{6.28} = 50 \, Hz \)
The frequency of the alternating current is 50 Hz.
In simple words: From the voltage formula, we find the number linked to 't'. We then use this number with the formula for angular frequency (2 times pi times frequency) to figure out the current's frequency.

🎯 Exam Tip: Always relate \( \omega \) to \( 2\pi f \). For common problems, remember \( 2\pi \approx 6.28 \).

Question 6. The inductance of a coil is 0.1 H. It is connected to an alternating current of 50 Hz frequency. Find out the reactance.
Answer:
Given values:
Inductance of the coil, \( L = 0.1 \, H \)
Frequency of the alternating current, \( f = 50 \, Hz \)

The inductive reactance (\( X_L \)) is calculated using the formula:
\( X_L = 2 \pi f L \)
\( X_L = 2 \times 3.14 \times 50 \times 0.1 \)
\( X_L = 31.4 \, \Omega \)
The inductive reactance of the coil is 31.4 Ω.
In simple words: To find how much a coil resists AC current, we multiply 2 by pi, then by the frequency of the current, and finally by the coil's inductance.

🎯 Exam Tip: Use the correct formula \( X_L = 2 \pi f L \) and ensure all units (Henry for inductance, Hz for frequency) are in SI before calculation.

Question 7. What will be phase difference between current and voltage for a series LCR circuit?
Answer:
In a series LCR circuit, the phase difference (\( \phi \)) between the current and voltage depends on the values of inductance (L), capacitance (C), and resistance (R), as well as the frequency. It can range between \( 0 \) and \( \pm \frac{\pi}{2} \) radians.
If the circuit is purely resistive (at resonance), \( \phi = 0 \).
If the circuit is inductive, \( 0 < \phi \leq \frac{\pi}{2} \) (voltage leads current).
If the circuit is capacitive, \( -\frac{\pi}{2} \leq \phi < 0 \) (current leads voltage).
In simple words: The difference in timing between the voltage and current in an LCR circuit changes. It can be zero if only resistance matters, or up to 90 degrees either way, depending on whether the circuit acts more like an inductor or a capacitor.

🎯 Exam Tip: The phase angle \( \phi \) is determined by \( \tan \phi = \frac{X_L - X_C}{R} \). If \( X_L = X_C \), then \( \phi = 0 \).

Question 8. What will be phase difference between voltage of inductance and capacitance in a series LCR resonance circuit?
Answer:
In a series LCR resonance circuit, the voltage across the inductor (\( V_L \)) and the voltage across the capacitor (\( V_C \)) are exactly opposite in phase.
Therefore, the phase difference between \( V_L \) and \( V_C \) is 180° (or \( \pi \) radians). They cancel each other out at resonance.
In simple words: In a resonant LCR circuit, the voltage across the inductor and the voltage across the capacitor are directly opposite, meaning they are 180 degrees apart in their timing.

🎯 Exam Tip: At resonance, \( V_L \) and \( V_C \) are equal in magnitude and 180° out of phase, leading to zero net reactive voltage.

Question 9. What will be impedance in a series LCR circuit?
Answer:
The impedance (Z) in a series LCR circuit is given by the formula:
\( Z = \sqrt{R^2 + (X_L - X_C)^2} \)
However, the question might be implicitly asking for impedance *at resonance*. At the condition of resonance (\( X_L = X_C \)), the term \( (X_L - X_C) \) becomes zero.
So, at resonance, the impedance is:
\( Z = \sqrt{R^2 + 0^2} = R \)
Thus, at resonance, the impedance is equal to the resistance.
In simple words: In an LCR circuit, impedance is the total opposition to current. At a special point called resonance, the opposition from the inductor and capacitor cancel each other out, so only the resistance remains as the impedance.

🎯 Exam Tip: Always remember that resonance corresponds to the minimum impedance (equal to R) in a series LCR circuit and maximum current.

Question 10. What will be the value of power factor in an alternating current circuit, for inductance, capacitance and resistance?
Answer:
The power factor (\( \cos \phi \)) in an alternating current circuit is a measure of how much power is actually used compared to the total apparent power.
For a purely inductive circuit: \( \phi = 90^\circ \), so power factor \( \cos 90^\circ = 0 \).
For a purely capacitive circuit: \( \phi = 90^\circ \), so power factor \( \cos 90^\circ = 0 \).
For a purely resistive circuit: \( \phi = 0^\circ \), so power factor \( \cos 0^\circ = 1 \).
Therefore, the values are zero, zero, and one, respectively.
In simple words: In circuits with only inductors or only capacitors, no real power is used, so the power factor is zero. But in a circuit with only resistance, all power is used, so the power factor is one.

🎯 Exam Tip: A power factor of 1 means maximum power transfer, while a power factor of 0 means no real power is dissipated (wattless current).

RBSE Class 12 Physics Chapter 10 Short Answer Type Questions

Question 1. Why does prefer the alternating current in comparison of direct current? Explain the proper cause.
Answer:
Alternating Current (AC) is preferred over Direct Current (DC) for several reasons, mainly for large-scale power distribution:
1. **Voltage Transformation:** AC voltage can be easily stepped up or stepped down using transformers. This allows electricity to be transmitted at very high voltages (reducing energy loss during transmission) and then stepped down to safe, usable voltages for homes and industries.
2. **Reduced Energy Loss:** Transmitting power at high voltage means lower current for the same amount of power (\( P = VI \)). Lower current leads to significantly reduced power loss (\( P_{loss} = I^2 R \)) in transmission lines. DC transmission at high voltages is complex and expensive.
3. **Generation:** AC generators (alternators) are generally simpler and more robust to build than DC generators (dynamos).
4. **Easy Conversion:** AC can be easily converted to DC using rectifiers when needed, but converting DC to AC is more complex and expensive.
In simple words: AC is better because we can easily change its voltage using transformers. This helps send power over long distances without losing much energy and then change it to a safe level for homes.

🎯 Exam Tip: Focus on the transformability of voltage and the associated efficiency benefits during long-distance transmission as key advantages of AC.

Question 2. Alternating current at 220 V is more dangerous than direct current of 220 V. Why?
Answer:
Alternating current (AC) at 220 V is considered more dangerous than direct current (DC) at 220 V because of the difference between their peak and RMS (root mean square) values.
When we say 220 V AC, we are usually referring to its RMS value, which is the effective voltage. The peak voltage (\( V_m \)) of a 220 V AC supply is much higher:
\( V_{m} = V_{rms} \times \sqrt{2} \)
For 220 V AC, \( V_{m} = 220 \, V \times \sqrt{2} \approx 220 \times 1.414 \approx 311 \, V \).
This means that the AC voltage swings from \( -311 \, V \) to \( +311 \, V \). A person touching an AC line experiences this higher peak voltage for brief moments, which can be more harmful to the body, particularly to the nervous system and heart, than a steady 220 V DC. Additionally, AC can cause muscular contractions that prevent a person from letting go of the live wire.
In simple words: 220V AC actually goes up to about 311V at its highest points, while 220V DC stays at a steady 220V. These higher peaks of AC are what make it more dangerous to touch.

🎯 Exam Tip: The key point is that the peak voltage of AC is \( \sqrt{2} \) times its RMS value, making it higher than a comparable DC voltage and thus more hazardous.

Question 3. Draw a graph between inductive reactance and capacitive reactance with frequency.
Answer:
Inductive reactance (\( X_L \)) and capacitive reactance (\( X_C \)) both depend on the frequency (f) of the alternating current, but in opposite ways.

**Inductive Reactance (\( X_L \)):**
\( X_L = 2 \pi f L \)
This equation shows that inductive reactance is directly proportional to the frequency (\( X_L \propto f \)). As frequency increases, inductive reactance also increases linearly. When the frequency is zero (DC), \( X_L \) is zero, meaning an inductor offers no opposition to DC.

**Capacitive Reactance (\( X_C \)):**
\( X_C = \frac{1}{2 \pi f C} \)
This equation shows that capacitive reactance is inversely proportional to the frequency (\( X_C \propto \frac{1}{f} \)). As frequency increases, capacitive reactance decreases. When the frequency is zero (DC), \( X_C \) approaches infinity, meaning a capacitor offers infinite opposition to DC (blocks it).

**Graphical Representation:**
A graph showing \( X_L \) versus frequency would be a straight line passing through the origin with a positive slope. A graph showing \( X_C \) versus frequency would be a hyperbola, starting from a very high value at low frequencies and decreasing towards zero at high frequencies.

🎯 Exam Tip: Visualizing how \( X_L \) increases linearly and \( X_C \) decreases hyperbolically with frequency is crucial for understanding AC circuits and resonance.

Question 4. Capacitor blocks direct current but easily passes alternating current. Why?
Answer:
A capacitor's opposition to current flow is called capacitive reactance (\( X_C \)), which is given by the formula:
\( X_C = \frac{1}{2 \pi f C} \)
where \( f \) is the frequency of the current and \( C \) is the capacitance.

For **Direct Current (DC)**, the frequency \( f = 0 \). If we substitute this into the formula:
\( X_C = \frac{1}{2 \pi (0) C} = \frac{1}{0} = \infty \)
This means the capacitive reactance for DC is infinite, acting like an open circuit. Therefore, a capacitor effectively blocks direct current.

For **Alternating Current (AC)**, the frequency \( f > 0 \). As the frequency of AC is a finite, non-zero value, \( X_C \) has a finite value. As frequency increases, \( X_C \) decreases, allowing AC to pass through more easily. Thus, a capacitor easily passes alternating current.
In simple words: A capacitor acts like a very high wall for direct current because DC has no frequency. But for alternating current, which has a frequency, the wall becomes much lower, letting the current go through.

🎯 Exam Tip: The inverse relationship between capacitive reactance and frequency (\( X_C \propto 1/f \)) is fundamental to explaining a capacitor's behavior with both AC and DC.

Question 5. The ohmic resistance of a coil is 6 Ω. If impedance of the coil is 10 Ω, then find inductive reactance (XL) of the coil.
Answer:
Given values:
Ohmic resistance of the coil, \( R = 6 \, \Omega \)
Impedance of the coil, \( Z = 10 \, \Omega \)

For a coil (which typically has resistance and inductance), the impedance (Z) is given by the formula:
\( Z = \sqrt{R^2 + X_L^2} \)
We need to find the inductive reactance (\( X_L \)). We can rearrange the formula to solve for \( X_L \):
\( Z^2 = R^2 + X_L^2 \)
\( X_L^2 = Z^2 - R^2 \)
Substitute the given values:
\( X_L^2 = (10 \, \Omega)^2 - (6 \, \Omega)^2 \)
\( X_L^2 = 100 - 36 \)
\( X_L^2 = 64 \)
\( X_L = \sqrt{64} \)
\( X_L = 8 \, \Omega \)
The inductive reactance of the coil is 8 Ω.
In simple words: A coil has both a regular resistance and an extra resistance from its inductance. We use a special triangle rule (Pythagorean theorem) with the total resistance (impedance) and regular resistance to find the inductive resistance.

🎯 Exam Tip: Remember the impedance triangle for RL circuits where \( Z^2 = R^2 + X_L^2 \). This is a direct application of the Pythagorean theorem.

Question 6. Write phase difference between current and voltage for an ac circuit, when (i) f = fr, (ii) f < fr (iii) f > fr Here fr is the resonant frequency.
Answer:
The phase difference (\( \phi \)) between current and voltage in an AC circuit depends on whether the circuit is resistive, capacitive, or inductive.

(i) **When \( f = f_r \) (at resonant frequency):**
At resonance, inductive reactance (\( X_L \)) is equal to capacitive reactance (\( X_C \)), so \( X_L - X_C = 0 \). The circuit behaves purely resistively. Therefore, the current and voltage are in the same phase, and the phase difference \( \phi = 0^\circ \).

(ii) **When \( f < f_r \) (frequency is less than resonant frequency):**
When the frequency is less than the resonant frequency, the capacitive reactance (\( X_C = \frac{1}{2 \pi f C} \)) becomes greater than the inductive reactance (\( X_L = 2 \pi f L \)). The circuit behaves predominantly capacitively. In a capacitive circuit, the current leads the voltage, so the phase difference \( \phi \) is negative (voltage lags current). Here, \( \phi = -\arctan\left(\frac{X_C - X_L}{R}\right) \).

(iii) **When \( f > f_r \) (frequency is greater than resonant frequency):**
When the frequency is greater than the resonant frequency, the inductive reactance (\( X_L \)) becomes greater than the capacitive reactance (\( X_C \)). The circuit behaves predominantly inductively. In an inductive circuit, the current lags the voltage, so the phase difference \( \phi \) is positive (voltage leads current). Here, \( \phi = \arctan\left(\frac{X_L - X_C}{R}\right) \).
In simple words: When the frequency matches the circuit's special resonance frequency, current and voltage are perfectly in step. If the frequency is lower, current moves ahead of voltage. If it's higher, current falls behind the voltage.

🎯 Exam Tip: Clearly differentiate the circuit behavior and phase relationship (leading/lagging) for frequencies below, at, and above resonance. This is a fundamental concept for LCR circuits.

Question 7. What is bandwidth? Write its value in LCR circuit.
Answer:
**Bandwidth:**
Bandwidth in an LCR circuit is the range of frequencies for which the power dissipated in the circuit is at least half of the maximum power dissipated at the resonant frequency. It is the difference between the two half-power frequencies (\( f_H \) and \( f_L \)).

**Value in LCR Circuit:**
The bandwidth (\( \Delta f \)) for a series LCR circuit is given by:
\( \Delta f = f_H - f_L = \frac{R}{2 \pi L} \)
where \( f_H \) is the higher half-power frequency, \( f_L \) is the lower half-power frequency, R is the resistance, and L is the inductance of the circuit. Bandwidth is a measure of the sharpness of the resonance.
In simple words: Bandwidth is the spread of frequencies where a circuit works well, meaning it still produces at least half of its best power. For an LCR circuit, this spread is calculated by dividing the resistance by (2 times pi times the inductance).

🎯 Exam Tip: Remember that a smaller bandwidth indicates a sharper resonance, which is related to a higher quality factor (Q-factor). The formula \( \Delta f = \frac{R}{2 \pi L} \) is key.

Question 8. What are half power point frequencies? What is value of current at there frequencies?
Answer:
**Half-Power Point Frequencies:**
Half-power point frequencies are the two frequencies, \( f_L \) (lower) and \( f_H \) (higher), on either side of the resonant frequency (\( f_r \)), at which the power dissipated in the circuit becomes half of the maximum power dissipated at resonance.

**Value of Current at Half-Power Frequencies:**
At these half-power frequencies, the current in the circuit is \( \frac{1}{\sqrt{2}} \) times (approximately 0.707 times) the maximum current (\( I_{max} \)) that flows at resonance.
So, \( I = \frac{I_{max}}{\sqrt{2}} = 0.707 \, I_{max} \)
This means that even though the power is half, the current is not half; it is \( \frac{1}{\sqrt{2}} \) of the maximum current.
In simple words: Half-power frequencies are specific frequencies where the circuit's power output drops to half of its highest power. At these points, the current flowing through the circuit is about 70.7% of the maximum current it can carry.

🎯 Exam Tip: Do not confuse half power with half current. Since power is proportional to \( I^2 \), half power means current is \( I_{max} / \sqrt{2} \).

Question 9. What will be power factor when resistance and reactance of a coil are equal?
Answer:
Given that the resistance (R) and reactance (\( X_L \)) of a coil are equal: \( R = X_L \).
The impedance (Z) of an RL coil is given by:
\( Z = \sqrt{R^2 + X_L^2} \)
Substitute \( X_L = R \) into the impedance formula:
\( Z = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2} \)

The power factor (\( \cos \phi \)) is defined as the ratio of resistance to impedance:
\( \cos \phi = \frac{R}{Z} \)
Substitute the value of Z:
\( \cos \phi = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}} \)
So, the power factor when resistance and reactance are equal is \( \frac{1}{\sqrt{2}} \) (approximately 0.707).
In simple words: If the resistance and the inductive kickback (reactance) in a coil are the same, then the power factor is one divided by the square root of two. This means the circuit is fairly efficient, but not perfectly efficient.

🎯 Exam Tip: Remember that a power factor of 1 means purely resistive, while \( \frac{1}{\sqrt{2}} \) implies a circuit with equal resistive and reactive components, resulting in a 45° phase angle.

Question 11. What will be value of impedance, frequency and power factor for a resonant LCR circuit? Write the expression.
Answer:
For a series LCR circuit operating at resonance:

**Impedance (Z):**
At resonance, the inductive reactance (\( X_L \)) is equal to the capacitive reactance (\( X_C \)). This means \( X_L - X_C = 0 \).
The impedance formula is \( Z = \sqrt{R^2 + (X_L - X_C)^2} \).
Substituting \( (X_L - X_C) = 0 \), we get:
\( Z = \sqrt{R^2 + 0^2} = R \)
So, at resonance, the impedance is equal to the resistance (R) of the circuit. It is the minimum possible impedance.

**Resonant Frequency (\( f_r \)):**
The resonant frequency is the specific frequency at which \( X_L = X_C \).
\( 2 \pi f_r L = \frac{1}{2 \pi f_r C} \)
\( f_r^2 = \frac{1}{4 \pi^2 L C} \)
\( f_r = \frac{1}{2 \pi \sqrt{L C}} \)

**Power Factor (\( \cos \phi \)):**
The power factor is given by \( \cos \phi = \frac{R}{Z} \).
Since \( Z = R \) at resonance:
\( \cos \phi = \frac{R}{R} = 1 \)
A power factor of 1 indicates that the current and voltage are in phase, and maximum power is dissipated in the circuit.
In simple words: In a circuit at its special resonance point, the total opposition to current (impedance) is just the resistance. The frequency at this point depends on the coil and capacitor values. And because current and voltage are perfectly in sync, the power factor is one, meaning all the power is used effectively.

🎯 Exam Tip: At resonance, \( Z=R \), \( f_r = \frac{1}{2 \pi \sqrt{L C}} \), and \( \cos \phi = 1 \). These three conditions are critical for analyzing LCR circuits.

Question 12. On what principle does transformer work? Write its uses.
Answer:
**Principle of Operation:**
A transformer works on the principle of **mutual induction**. Mutual induction occurs when a changing magnetic field in one coil (the primary coil) induces an electromotive force (EMF) in an adjacent coil (the secondary coil). When an alternating voltage is applied to the primary coil, it creates a changing magnetic flux. This flux links with the secondary coil through a soft iron core, inducing an alternating EMF in the secondary coil.

**Uses of Transformers:**
Transformers are essential devices in electrical power systems, primarily used for:
1. **Stepping Up Voltage:** To increase (step up) the alternating voltage for long-distance power transmission. High voltage transmission minimizes current, which significantly reduces energy losses (\( I^2 R \)) in transmission lines.
2. **Stepping Down Voltage:** To decrease (step down) the alternating voltage to safe and usable levels for homes, offices, and industrial equipment. This is necessary because appliances typically operate at much lower voltages than transmission lines.
3. **Impedance Matching:** They can be used to match the impedance of a source to a load for maximum power transfer in electronic circuits, such as in audio amplifiers.
4. **Isolation:** They provide electrical isolation between two circuits.
In simple words: A transformer works because a changing magnetic field in one coil makes electricity flow in another nearby coil (mutual induction). It's used to raise voltage for sending power far away and then lower it again for our homes and gadgets.

🎯 Exam Tip: Highlight mutual induction as the core principle. Emphasize voltage transformation for efficient power transmission and distribution as the primary application.

Question 13. Find out the average value of current during first half cycle of an alternating current.
Answer:
The average value of an alternating current over a complete cycle is zero because the current flows equally in positive and negative directions. Therefore, to define an average value, we calculate it over half a cycle (either positive or negative).

Let the instantaneous alternating current be given by \( i = I_m \sin \omega t \), where \( I_m \) is the peak current.
The average value of current \( I_{av} \) over a half cycle (from \( t=0 \) to \( t=T/2 \)) is given by:
\( I_{av} = \frac{\int_{0}^{T/2} i \, dt}{\int_{0}^{T/2} dt} = \frac{\int_{0}^{T/2} I_m \sin \omega t \, dt}{T/2} \)
\( I_{av} = \frac{I_m}{T/2} \left[ -\frac{\cos \omega t}{\omega} \right]_{0}^{T/2} \)
\( I_{av} = \frac{2 I_m}{\omega T} \left[ -\cos \omega t \right]_{0}^{T/2} \)
Substitute \( \omega T = 2 \pi \):
\( I_{av} = \frac{2 I_m}{2 \pi} \left[ -\cos (\omega \frac{T}{2}) - (-\cos 0) \right] \)
\( I_{av} = \frac{I_m}{\pi} \left[ -\cos (\pi) + \cos 0 \right] \)
\( I_{av} = \frac{I_m}{\pi} \left[ -(-1) + 1 \right] \)
\( I_{av} = \frac{I_m}{\pi} (1 + 1) = \frac{2 I_m}{\pi} \)
Numerically, \( I_{av} \approx \frac{2}{3.141} I_m \approx 0.637 \, I_m \).
The average value of current during the first half cycle of an alternating current is \( \frac{2 I_m}{\pi} \) or approximately 0.637 times the peak current.
In simple words: To find the average current for half a cycle, we add up all the current values during that time and divide by the time taken. This calculation shows the average current is about 63.7% of its highest value.

🎯 Exam Tip: Remember that the average value of AC over a full cycle is zero. The average value over a half cycle is \( \frac{2 I_m}{\pi} \), which is a common formula to memorize.

Question 14. What are the causes of energy loss in a transformer? How can we minimise them?
Answer:
Energy losses in a transformer reduce its efficiency. The main causes of energy loss and their minimization methods are:

1. **Magnetic Flux Leakage:**
* **Cause:** Not all the magnetic flux produced by the primary coil links with the secondary coil; some leaks into the air. This leakage means not all the primary power is transferred to the secondary.
* **Minimization:** Wind the primary and secondary coils one over the other (coaxially) to maximize flux linkage.

2. **Iron Loss (Eddy Current Loss):**
* **Cause:** The alternating magnetic flux in the core induces circulating currents (eddy currents) within the core itself. These currents dissipate energy as heat (\( I^2 R \)) in the core.
* **Minimization:** Use a laminated core, made of thin sheets of soft iron insulated from each other. Laminations increase the resistance paths for eddy currents, thus significantly reducing their magnitude.

3. **Copper Loss:**
* **Cause:** This is the energy lost as heat (\( I^2 R \)) due to the electrical resistance of the copper windings in both the primary and secondary coils.
* **Minimization:** Use thick copper wires for the windings, especially for the high-current coils, as thicker wires have lower resistance.

4. **Hysteresis Loss:**
* **Cause:** The core material is repeatedly magnetized and demagnetized by the alternating current. Energy is lost in this process due to the magnetic domains constantly reorienting.
* **Minimization:** Use a soft iron core material that has a narrow hysteresis loop, meaning it can be easily magnetized and demagnetized with minimal energy expenditure.

5. **Magnetostriction:**
* **Cause:** This refers to the humming sound produced by transformers due to the alternating magnetic field causing vibrations and slight changes in the core's dimensions.
* **Minimization:** While not a power loss in the electrical sense, it is an audible energy loss. It can be minimized by selecting appropriate core materials and ensuring proper structural design and damping.
In simple words: Transformers lose energy mainly through five ways: some magnetic force escaping, heat from swirling currents in the core (eddy currents), heat from resistance in the copper wires, energy used to repeatedly magnetize the core, and even sound from core vibrations. We reduce these by careful design, like using thin, layered cores and thick wires.

🎯 Exam Tip: List the five main types of losses (flux leakage, eddy current, copper, hysteresis, magnetostriction) and provide a concise method for minimizing each. This is a common descriptive question.

RBSE Class 12 Physics Chapter 10 Long Answer Type Questions

Question 1. A pure inductance is connected through alternating voltage circuit. Find out the value of current, phase difference, reactance and average consumed energy rate in the circuit. Draw phasor diagram also.
Answer:

A.C. Voltage Applied on an Inductor
Let an alternating electromotive force (emf) source be connected across an inductor. The instantaneous value of alternating emf is given by:
\( V = V_m \sin \omega t \) ... (1)
Here, we assume that this inductor has negligible resistance, even though its windings usually have some resistance.
If \( I \) is the instantaneous current flowing through \( L \) at any time \( t \), then the self-induced Faraday emf in the inductor is:
\( \epsilon = -L \frac{dI}{dt} \)
Here, \( L \) is the self-inductance of the inductor. The negative sign shows that the induced emf opposes the applied emf, which is according to Lenz's law.
The applied voltage \( V \) must be equal and opposite to the induced voltage \( \epsilon \) to maintain the flow of current in the circuit.
So, \( V = -\epsilon \)
This means, \( V_m \sin \omega t = L \frac{dI}{dt} \)
Rearranging this, \( dI = \frac{V_m}{L} \sin \omega t \, dt \)
To find the total current, we integrate both sides:
\( \int dI = \int \frac{V_m}{L} \sin \omega t \, dt \)
\( I = \frac{V_m}{L} \int \sin \omega t \, dt \)
\( I = \frac{V_m}{L} \left( -\frac{\cos \omega t}{\omega} \right) + \text{constant} \)
\( I = -\frac{V_m}{\omega L} \cos \omega t + \text{constant} \)
Since the source emf oscillates symmetrically around zero, the current it sustains should also oscillate symmetrically around zero. Thus, the integration constant is zero.
So, \( I = -\frac{V_m}{\omega L} \cos \omega t \)
Using the identity \( -\cos \theta = \sin(\theta - \frac{\pi}{2}) \), we can write:
\( I = \frac{V_m}{\omega L} \sin \left( \omega t - \frac{\pi}{2} \right) \) ... (3)
Comparing equation (1) and (3), we can see that in an A.C. circuit containing only an inductor, the current lags behind the voltage by a phase angle of \( \frac{\pi}{2} \) radians, which is a quarter of a cycle.
The term \( \omega L \) plays the role of resistance in this circuit. It is the effective resistance offered by the inductor and is called inductive reactance, denoted by \( X_L \).
So, \( X_L = \omega L = 2 \pi f L \)
This means inductive reactance is not constant; it increases as the frequency \( f \) of the A.C. supply increases. In a D.C. circuit, \( f = 0 \), so \( X_L = 0 \). Therefore, a pure inductance offers zero resistance to D.C. current. Inductive reactance is measured in ohms (Ω).
The average power consumed over a complete cycle in a pure inductive circuit is zero because the current and voltage are \( \frac{\pi}{2} \) out of phase, making the power factor \( \cos(\frac{\pi}{2}) = 0 \).
In simple words: When you connect an inductor to an AC power, the current will flow, but it will be a bit behind the voltage. This "delay" is called phase difference. The inductor also resists the flow of AC, which we call inductive reactance. In a pure inductor, no power is used up on average over time.

🎯 Exam Tip: Remember that in a pure inductive circuit, current lags voltage by \( 90^\circ \) (\( \frac{\pi}{2} \) radians), leading to zero average power consumption. This is a key characteristic to mention.

Question 2. An alternating voltage is connected to an R-L circuit. Derive a relation for impedance and current. Draw phasor diagram also.
Answer:
When an alternating voltage is applied to a series R-L circuit, the voltage across the resistor (\( V_R \)) is in phase with the current (\( I \)), while the voltage across the inductor (\( V_L \)) leads the current by \( 90^\circ \).
According to Kirchhoff's voltage law, the instantaneous applied voltage \( V \) is the vector sum of \( V_R \) and \( V_L \).
From the phasor diagram, the resultant voltage \( V \) can be found using the Pythagorean theorem:
\( V^2 = V_R^2 + V_L^2 \)
We know that \( V_R = IR \) (Ohm's law for resistance) and \( V_L = IX_L \) (Ohm's law for inductive reactance, where \( X_L = \omega L \)).
Substitute these into the equation:
\( (IR)^2 + (IX_L)^2 = V^2 \)
\( I^2 R^2 + I^2 X_L^2 = V^2 \)
\( I^2 (R^2 + X_L^2) = V^2 \)
So, \( I = \frac{V}{\sqrt{R^2 + X_L^2}} \)
The term \( \sqrt{R^2 + X_L^2} \) is the total effective opposition to the current flow in the R-L circuit, known as impedance (\( Z \)).
Therefore, \( Z = \sqrt{R^2 + X_L^2} \)
And the current in the circuit is \( I = \frac{V}{Z} \).
The phase difference (\( \phi \)) between the voltage and current is given by:
\( \tan \phi = \frac{V_L}{V_R} = \frac{IX_L}{IR} = \frac{X_L}{R} \)
Thus, \( \phi = \tan^{-1} \left( \frac{X_L}{R} \right) \). In an R-L circuit, the voltage leads the current.
In simple words: When you have both a resistor and an inductor in an AC circuit, the total resistance to current is called impedance. It's like a combined effect of the resistor's normal resistance and the inductor's special AC resistance. The current in this circuit will lag behind the voltage by some angle, not by a full quarter cycle as in a pure inductor.

🎯 Exam Tip: For an R-L circuit, remember that the impedance \( Z \) is calculated as a vector sum of resistance and inductive reactance. Also, clearly state that voltage leads current, and provide the formula for the phase angle.

Question 3. What do you mean by resonant circuit? Write necessary condition for series resonant LCR circuit. Derive a relation of frequency. What are the uses of this circuit?
Answer:
A resonant circuit is an electrical circuit where the impedance is at a minimum or maximum for a specific frequency, allowing for a large current or voltage response. It shows a tendency to oscillate at a particular frequency, known as its natural frequency. If an external energy source drives the system at a frequency close to this natural frequency, the amplitude of oscillations becomes very large. This condition is called resonance.

Series L-C-R Resonance Circuit
In a series LCR circuit, resonance occurs when the current flowing through the circuit becomes maximum. For this to happen, the total impedance \( Z \) of the circuit must be at its minimum value.
The impedance of a series LCR circuit is given by:
\( Z = \sqrt{R^2 + (X_L - X_C)^2} \)
Here, \( X_L = \omega L \) is the inductive reactance and \( X_C = \frac{1}{\omega C} \) is the capacitive reactance.
The impedance \( Z \) will be minimum when the term \( (X_L - X_C)^2 \) is zero, which means \( X_L = X_C \).
This is the necessary condition for series resonance: Inductive reactance must be equal to capacitive reactance.
\( \omega_0 L = \frac{1}{\omega_0 C} \)
Where \( \omega_0 \) is the angular resonant frequency.
Rearranging the equation to find \( \omega_0 \):
\( \omega_0^2 = \frac{1}{LC} \)
So, \( \omega_0 = \frac{1}{\sqrt{LC}} \)
Since \( \omega_0 = 2 \pi f_0 \), where \( f_0 \) is the resonant frequency, we have:
\( 2 \pi f_0 = \frac{1}{\sqrt{LC}} \)
Therefore, the resonant frequency \( f_0 = \frac{1}{2 \pi \sqrt{LC}} \).
At resonance, \( Z = \sqrt{R^2 + 0^2} = R \). This means the impedance is equal to the resistance, which is the minimum possible impedance. The current in the circuit becomes maximum at this frequency, given by \( I_{max} = \frac{V}{R} \). The phase difference \( \phi \) between voltage and current is zero, and the power factor \( \cos \phi = 1 \).

Uses of Resonant Circuits:
Resonant circuits have many important uses, especially in tuning systems for radios and televisions. Here's how they are used:
1. **Tuning Radios and TVs:** Antennas receive signals from many broadcasting stations, each at a different frequency. A resonant circuit in the radio or TV tuner allows the user to select a specific station. By changing the capacitance (or inductance) in the tuning circuit, its resonant frequency can be adjusted. When the circuit's resonant frequency matches the frequency of a particular radio station, the current for that frequency becomes maximum, making that station audible and clear.
2. **Filters:** Resonant circuits can act as filters, either allowing a specific band of frequencies to pass while blocking others (band-pass filter) or blocking a specific band while allowing others to pass (band-stop filter).
3. **Oscillators:** They are also used in electronic oscillators to generate signals of a desired frequency.
In simple words: A resonant circuit is like a special filter that lets a specific frequency pass very strongly. For a series LCR circuit, this happens when the inductor's and capacitor's effects cancel each other out, making the total resistance very low and the current very high. This specific frequency is called the resonant frequency. We use these circuits to tune radios, like when you change the station on your radio, you are changing the resonant frequency to match the station's frequency.

🎯 Exam Tip: Clearly state the condition \( X_L = X_C \) for resonance and derive the formula for resonant frequency. Also, provide relevant uses, focusing on tuning applications as a primary example.

Question 4. Draw a graph between frequency and current for a series LCR circuit. Represent half power frequencies on the graph and derive a relation for bandwidth.
Answer:
In a series LCR circuit, the current amplitude \( I_m \) varies with the angular frequency \( \omega \) according to the formula:
\( I_m = \frac{V_m}{\sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}} \)
The graph showing the variation of current amplitude \( I_m \) with angular frequency \( \omega \) for a series LCR circuit typically looks like a bell-shaped curve, with a peak at the resonant frequency \( \omega_0 \).
At resonance (\( \omega = \omega_0 \)), \( \omega_0 L = \frac{1}{\omega_0 C} \), so the impedance is minimum (\( Z=R \)), and the current amplitude is maximum (\( I_{max} = \frac{V_m}{R} \)).
As the frequency moves away from \( \omega_0 \), either higher or lower, the impedance increases, and the current amplitude decreases.

The question asks to draw a graph, which cannot be rendered in HTML directly as SVG for a complex graph. However, the key features are:

• A peak current at the resonant frequency.
• The current amplitude falls off on either side of the resonant frequency.
• Half-power frequencies are where the current amplitude is \( \frac{I_{max}}{\sqrt{2}} \).

The current amplitude has a peak value at the resonant frequency \( \omega_0 = \frac{1}{\sqrt{LC}} \). For other values of frequency, the current is less than the maximum value. As the resistance \( R \) in the circuit is reduced, the resonance curve becomes sharper. The sharper the curve, the more quickly the current amplitude (\( I_m \)) falls for changes in frequency on both sides of \( \omega_0 \). The sharpness of resonance is described by its quality factor, \( Q \).

Half-Power Frequencies and Bandwidth:
Half-power frequencies (\( \omega_1 \) and \( \omega_2 \)) are the frequencies on either side of the resonant frequency \( \omega_0 \) where the power dissipated in the circuit is half of the maximum power. At these frequencies, the current amplitude is \( I' = \frac{I_{max}}{\sqrt{2}} \).
We know that power \( P = I^2 R \). So, when current is \( \frac{I_{max}}{\sqrt{2}} \), the power will be \( \left(\frac{I_{max}}{\sqrt{2}}\right)^2 R = \frac{I_{max}^2 R}{2} = \frac{P_{max}}{2} \).
At these half-power points, the impedance \( Z \) becomes \( \sqrt{2}R \).
So, \( \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2} = \sqrt{2}R \)
Squaring both sides: \( R^2 + (\omega L - \frac{1}{\omega C})^2 = 2R^2 \)
\( (\omega L - \frac{1}{\omega C})^2 = R^2 \)
\( \omega L - \frac{1}{\omega C} = \pm R \)

For \( \omega_1 \) (lower half-power frequency): \( \omega_1 L - \frac{1}{\omega_1 C} = -R \)
For \( \omega_2 \) (upper half-power frequency): \( \omega_2 L - \frac{1}{\omega_2 C} = R \)

Adding the two equations:
\( (\omega_1 L - \frac{1}{\omega_1 C}) + (\omega_2 L - \frac{1}{\omega_2 C}) = 0 \)
\( L(\omega_1 + \omega_2) - \frac{1}{C} \left( \frac{1}{\omega_1} + \frac{1}{\omega_2} \right) = 0 \)
\( L(\omega_1 + \omega_2) - \frac{1}{C} \left( \frac{\omega_1 + \omega_2}{\omega_1 \omega_2} \right) = 0 \)
Since \( \omega_1 + \omega_2 \ne 0 \), we can divide by \( (\omega_1 + \omega_2) \):
\( L - \frac{1}{C \omega_1 \omega_2} = 0 \)
\( \omega_1 \omega_2 = \frac{1}{LC} = \omega_0^2 \)
This shows that the resonant frequency is the geometric mean of the half-power frequencies: \( \omega_0 = \sqrt{\omega_1 \omega_2} \).

Subtracting the first equation from the second:
\( (\omega_2 L - \frac{1}{\omega_2 C}) - (\omega_1 L - \frac{1}{\omega_1 C}) = R - (-R) \)
\( L(\omega_2 - \omega_1) - \frac{1}{C} \left( \frac{1}{\omega_2} - \frac{1}{\omega_1} \right) = 2R \)
\( L(\omega_2 - \omega_1) + \frac{1}{C} \left( \frac{\omega_1 - \omega_2}{\omega_1 \omega_2} \right) = 2R \)
\( L(\omega_2 - \omega_1) - \frac{1}{C} \frac{(\omega_2 - \omega_1)}{\omega_1 \omega_2} = 2R \)
\( (\omega_2 - \omega_1) \left( L - \frac{1}{C \omega_1 \omega_2} \right) = 2R \)
Since \( \omega_1 \omega_2 = \frac{1}{LC} \), the term \( L - \frac{1}{C \omega_1 \omega_2} \) becomes \( L - \frac{1}{C \cdot \frac{1}{LC}} = L - L = 0 \). This seems incorrect. Let's re-evaluate the subtraction.

Correcting the subtraction:
\( (\omega_2 L - \frac{1}{\omega_2 C}) - (\omega_1 L - \frac{1}{\omega_1 C}) = R - (-R) \)
\( L(\omega_2 - \omega_1) + \frac{1}{C} \left( \frac{1}{\omega_1} - \frac{1}{\omega_2} \right) = 2R \)
\( L(\omega_2 - \omega_1) + \frac{1}{C} \left( \frac{\omega_2 - \omega_1}{\omega_1 \omega_2} \right) = 2R \)
\( (\omega_2 - \omega_1) \left( L + \frac{1}{C \omega_1 \omega_2} \right) = 2R \)
Substitute \( \omega_1 \omega_2 = \frac{1}{LC} \):
\( (\omega_2 - \omega_1) \left( L + \frac{1}{C \cdot \frac{1}{LC}} \right) = 2R \)
\( (\omega_2 - \omega_1) (L + L) = 2R \)
\( (\omega_2 - \omega_1) (2L) = 2R \)
\( \omega_2 - \omega_1 = \frac{R}{L} \)

Bandwidth (\( \Delta \omega \)):
The bandwidth of the circuit is defined as the range of frequencies between the two half-power frequencies. It is given by \( \Delta \omega = \omega_2 - \omega_1 \).
So, the bandwidth \( \Delta \omega = \frac{R}{L} \).
In terms of frequency \( f \), the bandwidth \( \Delta f = f_2 - f_1 \). Since \( \omega = 2 \pi f \), then \( \Delta \omega = 2 \pi \Delta f \).
So, \( 2 \pi \Delta f = \frac{R}{L} \)
Therefore, \( \Delta f = \frac{R}{2 \pi L} \).
In simple words: When you graph how much current flows at different frequencies in an LCR circuit, you get a curve that peaks at the resonant frequency. The "half-power frequencies" are those points on either side of the peak where the current is about 70.7% of the maximum. The range between these two frequencies is called the "bandwidth". It tells you how wide the frequency band is where the circuit works well. You can find this bandwidth by dividing the resistance by the inductance.

🎯 Exam Tip: When deriving the bandwidth, clearly show the condition \( X_L - X_C = \pm R \) at half-power points and the subsequent algebraic steps to arrive at \( \Delta \omega = \frac{R}{L} \). Remember to explain the significance of bandwidth (how selective the circuit is). Illustrative graph should show \( I_{max} \) at \( \omega_0 \) and \( I_{max}/\sqrt{2} \) at \( \omega_1, \omega_2 \).

Question 5. Derive the expression for average power in an alternating current circuit. What will be effect in formula, when there is
Answer:
In an A.C. circuit, both voltage and current vary with time. The instantaneous power is the product of the instantaneous voltage and instantaneous current.
Let the alternating voltage be \( V = V_m \sin \omega t \)
And the alternating current be \( I = I_m \sin(\omega t + \phi) \), where \( \phi \) is the phase difference between voltage and current.
The instantaneous power \( P_{inst} = V \cdot I = (V_m \sin \omega t) (I_m \sin(\omega t + \phi)) \)
\( P_{inst} = V_m I_m \sin \omega t \sin(\omega t + \phi) \)
Using the trigonometric identity \( \sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)] \):
Let \( A = \omega t + \phi \) and \( B = \omega t \).
\( \sin(\omega t + \phi) \sin \omega t = \frac{1}{2} [\cos(\omega t + \phi - \omega t) - \cos(\omega t + \phi + \omega t)] \)
\( = \frac{1}{2} [\cos \phi - \cos(2 \omega t + \phi)] \)
So, \( P_{inst} = \frac{V_m I_m}{2} [\cos \phi - \cos(2 \omega t + \phi)] \)
The average power over one complete cycle is found by integrating \( P_{inst} \) over one cycle (\( T \)) and dividing by \( T \).
The average value of \( \cos(2 \omega t + \phi) \) over a complete cycle is zero because \( 2 \omega t \) represents a sinusoidal function that completes two cycles in time \( T \), and its average over a full cycle (or any integer number of cycles) is zero.
So, the average power \( P_{avg} = \frac{V_m I_m}{2} \cos \phi \).
This can also be written in terms of RMS values, where \( V_{rms} = \frac{V_m}{\sqrt{2}} \) and \( I_{rms} = \frac{I_m}{\sqrt{2}} \).
So, \( P_{avg} = V_{rms} I_{rms} \cos \phi \).
The term \( \cos \phi \) is called the power factor of the A.C. circuit.

**Effect of phase difference (\( \phi \)) on the power formula:**
The power factor \( \cos \phi \) determines how much of the apparent power (\( V_{rms} I_{rms} \)) is actually consumed as real power. Its value depends on the components in the circuit:

1. **Purely Resistive Circuit:**
If the circuit contains only pure resistance, the voltage and current are in the same phase, so \( \phi = 0^\circ \).
\( \cos \phi = \cos 0^\circ = 1 \). This is the maximum power factor.
\( P_{avg} = V_{rms} I_{rms} (1) = V_{rms} I_{rms} \). In this case, the power dissipated is maximum.

2. **Purely Inductive Circuit:**
If the circuit contains only a pure inductor, the voltage leads the current by \( 90^\circ \) (\( \phi = +90^\circ \)).
\( \cos \phi = \cos 90^\circ = 0 \). This is the minimum power factor.
\( P_{avg} = V_{rms} I_{rms} (0) = 0 \). No average power is dissipated even when current flows through the circuit. This is called wattless current.

3. **Purely Capacitive Circuit:**
If the circuit contains only a pure capacitor, the current leads the voltage by \( 90^\circ \) (\( \phi = -90^\circ \)).
\( \cos \phi = \cos (-90^\circ) = 0 \). This is the minimum power factor.
\( P_{avg} = V_{rms} I_{rms} (0) = 0 \). Again, no average power is dissipated. This is also called wattless current.

4. **LCR Series Circuit:**
In a general LCR series circuit, the phase angle \( \phi \) can be between \( -90^\circ \) and \( +90^\circ \), given by \( \tan \phi = \frac{X_L - X_C}{R} \).
\( P_{avg} = V_{rms} I_{rms} \cos \phi \). Here, \( \cos \phi \) will have a value between 0 and 1. Power is dissipated only across the resistor, as inductors and capacitors do not dissipate average power.

5. **Power Dissipated at Resonance in LCR Circuit:**
At resonance, \( X_L = X_C \), so \( \phi = 0^\circ \).
\( \cos \phi = \cos 0^\circ = 1 \). The power factor is maximum.
\( P_{avg} = V_{rms} I_{rms} (1) = V_{rms} I_{rms} \). At resonance, the power dissipated is maximum, effectively all dissipated by the resistor.

In summary, the average power dissipated in an A.C. circuit depends on the voltage, current, and the cosine of the phase angle between them (the power factor). The closer the power factor is to 1, the more power is utilized.

In simple words: The average power used in an AC circuit depends on the voltage, current, and how "in sync" the voltage and current are. This "in sync" part is called the power factor. If they are perfectly in sync (like with a plain resistor), you use maximum power. If they are completely out of sync (like with a pure inductor or capacitor), you use no average power. For mixed circuits, it's somewhere in between.

🎯 Exam Tip: The derivation of average power must clearly show the use of trigonometric identities and the averaging of the cosine term over a full cycle. Emphasize the role of the power factor \( \cos \phi \) and its specific values for different circuit components (R, L, C, and LCR).

Question 1. Alternating voltage in any alternating current circuit is V = 50 sin (157t + φ) V, then find out the :
(i) root mean square value of alternating voltage.
(ii) frequency of the alternating voltage.
Answer:
Given the alternating voltage equation: \( V = 50 \sin (157t + \phi) \).
Comparing this to the standard form \( V = V_m \sin (\omega t + \phi) \), we get:
Peak voltage \( V_m = 50 \, V \)
Angular frequency \( \omega = 157 \, rad/s \)

(i) Root mean square value of alternating voltage:
The RMS value is calculated as the peak voltage divided by the square root of 2.
\( V_{rms} = \frac{V_m}{\sqrt{2}} \)
\( V_{rms} = \frac{50}{\sqrt{2}} \)
\( V_{rms} = \frac{50}{1.414} \)
\( V_{rms} \approx 35.36 \, V \)
(ii) Frequency of the alternating voltage:
The angular frequency is \( \omega = 2\pi f \), where \( f \) is the frequency.
So, \( f = \frac{\omega}{2\pi} \)
\( f = \frac{157}{2 \times 3.14} \)
\( f = \frac{157}{6.28} \)
\( f = 25 \, Hz \)
In simple words: First, we find the highest voltage and how fast the current changes. Then, we use simple formulas to find the average effective voltage and how many times the current changes direction per second.

🎯 Exam Tip: Remember to clearly identify the peak voltage (\( V_m \)) and angular frequency (\( \omega \)) from the given equation before applying the formulas for RMS voltage and frequency.

Question 2. At which time, sinusoidal alternating current will be
(i) \( \frac{1}{2} \) time
(ii) \( \frac{\sqrt{3}}{2} \) time of its peak value.
Answer:
The instantaneous value of alternating current is given by \( I = I_m \sin(\omega t) \), where \( I_m \) is the peak value of current and \( \omega \) is the angular frequency.

(i) Time when current is \( \frac{1}{2} \) its peak value:
We set \( I = \frac{I_m}{2} \)
So, \( \frac{I_m}{2} = I_m \sin(\omega t) \)
\( \frac{1}{2} = \sin(\omega t) \)
This means \( \omega t = \frac{\pi}{6} \) radians (or \( 30^\circ \)).
Therefore, \( t = \frac{\pi}{6\omega} \)
Since \( \omega = \frac{2\pi}{T} \), where \( T \) is the time period,
\( t = \frac{\pi}{6} \times \frac{T}{2\pi} \)
\( t = \frac{T}{12} \)

(ii) Time when current is \( \frac{\sqrt{3}}{2} \) its peak value:
We set \( I = \frac{\sqrt{3}}{2} I_m \)
So, \( \frac{\sqrt{3}}{2} I_m = I_m \sin(\omega t) \)
\( \frac{\sqrt{3}}{2} = \sin(\omega t) \)
This means \( \omega t = \frac{\pi}{3} \) radians (or \( 60^\circ \)).
Therefore, \( t = \frac{\pi}{3\omega} \)
Since \( \omega = \frac{2\pi}{T} \),
\( t = \frac{\pi}{3} \times \frac{T}{2\pi} \)
\( t = \frac{T}{6} \)
In simple words: We find the exact moment when the alternating current reaches half or a specific fraction of its highest value. This depends on how fast the current wave is moving.

🎯 Exam Tip: When solving for time, ensure you use the correct angle (e.g., \( \frac{\pi}{6} \) for \( \sin(\theta) = \frac{1}{2} \)) and convert angular frequency to time period if needed to express the answer in terms of \( T \).

Question 4. Find out the inductive reactance for 100 mH inductance connected through alternating current of 1 kHz frequency. If voltage of the source is 6.28 V, then find out current in the inductance.
Answer:
Given values:
Inductance \( L = 100 \, mH = 100 \times 10^{-3} \, H = 0.1 \, H \)
Frequency \( f = 1 \, kHz = 1 \times 10^3 \, Hz \)
Source voltage \( V = 6.28 \, V \)

To find the inductive reactance \( X_L \):
\( X_L = 2\pi fL \)
\( X_L = 2 \times 3.14 \times (1 \times 10^3) \times (100 \times 10^{-3}) \)
\( X_L = 2 \times 3.14 \times 1000 \times 0.1 \)
\( X_L = 6.28 \times 100 \)
\( X_L = 628 \, \Omega \)

To find the current \( I \) in the inductance:
Using Ohm's law for AC circuits, \( I = \frac{V}{X_L} \)
\( I = \frac{6.28 \, V}{628 \, \Omega} \)
\( I = 0.01 \, A \)
In simple words: We calculate how much the inductor resists the flow of alternating current, then use that resistance and the given voltage to find the total current flowing through it.

🎯 Exam Tip: Always convert all units to SI units (H, Hz, V) before performing calculations for reactance and current to avoid errors.

Question 6. A condenser of 120 µF is connected through an ac source of frequency 50 Hz. Calculate its capacitive reactance. If frequency changes to 5 MHz, then what will be effect on reactance?
Answer:
Given values:
Capacitance \( C = 120 \, \mu F = 120 \times 10^{-6} \, F \)
Initial frequency \( f_1 = 50 \, Hz \)

Calculate capacitive reactance \( X_{C1} \) at \( f_1 = 50 \, Hz \):
\( X_{C1} = \frac{1}{2\pi f_1 C} \)
\( X_{C1} = \frac{1}{2 \times 3.14 \times 50 \times 120 \times 10^{-6}} \)
\( X_{C1} = \frac{1}{37680 \times 10^{-6}} \)
\( X_{C1} = \frac{1}{0.03768} \)
\( X_{C1} \approx 26.54 \, \Omega \)

If frequency changes to \( f_2 = 5 \, MHz = 5 \times 10^6 \, Hz \):
The capacitive reactance \( X_C \) is inversely proportional to frequency \( f \). So, \( X_C \propto \frac{1}{f} \).
Therefore, \( \frac{X_{C2}}{X_{C1}} = \frac{f_1}{f_2} \)
\( X_{C2} = X_{C1} \times \frac{f_1}{f_2} \)
\( X_{C2} = 26.54 \times \frac{50}{5 \times 10^6} \)
\( X_{C2} = 26.54 \times \frac{1}{10^5} \)
\( X_{C2} = 26.54 \times 10^{-5} \, \Omega \)
\( X_{C2} = 2.654 \times 10^{-4} \, \Omega \)
Effect on reactance: When the frequency increases from \( 50 \, Hz \) to \( 5 \, MHz \), the capacitive reactance decreases significantly from \( 26.54 \, \Omega \) to \( 2.654 \times 10^{-4} \, \Omega \). This means the capacitor offers very little opposition to the current at high frequencies.
In simple words: We find how much the capacitor resists the current at a normal speed. Then, we see that if the current changes much faster, the capacitor resists it much less, making it easier for the current to flow.

🎯 Exam Tip: Remember the inverse relationship between capacitive reactance and frequency. As frequency goes up, capacitive reactance goes down, and vice-versa.

Question 7. The resistance of a coil is R = 10 Ω and inductance is L = 0.4 H. It is connected to an ac source of 6.5 V, \( \frac{30}{\pi} \) Hz. Find out the average power consumed in the circuit.
Answer:
Given values:
Resistance \( R = 10 \, \Omega \)
Inductance \( L = 0.4 \, H \)
Source voltage \( V_{rms} = 6.5 \, V \)
Frequency \( f = \frac{30}{\pi} \, Hz \)

First, calculate the angular frequency \( \omega \):
\( \omega = 2\pi f \)
\( \omega = 2\pi \times \frac{30}{\pi} \)
\( \omega = 60 \, rad/s \)

Next, calculate the inductive reactance \( X_L \):
\( X_L = \omega L \)
\( X_L = 60 \times 0.4 \)
\( X_L = 24 \, \Omega \)

Then, calculate the impedance \( Z \) of the circuit:
\( Z = \sqrt{R^2 + X_L^2} \)
\( Z = \sqrt{(10)^2 + (24)^2} \)
\( Z = \sqrt{100 + 576} \)
\( Z = \sqrt{676} \)
\( Z = 26 \, \Omega \)

Calculate the RMS current \( I_{rms} \):
\( I_{rms} = \frac{V_{rms}}{Z} \)
\( I_{rms} = \frac{6.5 \, V}{26 \, \Omega} \)
\( I_{rms} = 0.25 \, A \)

Calculate the power factor \( \cos \phi \):
\( \cos \phi = \frac{R}{Z} \)
\( \cos \phi = \frac{10}{26} \)

Finally, calculate the average power consumed \( P_{avg} \):
\( P_{avg} = V_{rms} I_{rms} \cos \phi \)
\( P_{avg} = 6.5 \times 0.25 \times \frac{10}{26} \)
\( P_{avg} = 1.625 \times \frac{10}{26} \)
\( P_{avg} = \frac{16.25}{26} \)
\( P_{avg} = 0.625 \, W \)
In simple words: We find the total resistance of the coil to alternating current, including its inductive effect. Then, we use the voltage, current, and a power factor to calculate how much power the coil uses up on average.

🎯 Exam Tip: Remember that for an R-L circuit, the average power dissipated is only due to the resistance, and you must first calculate \( X_L \), then \( Z \), and then the power factor.

Question 8. A bulb of 60 V, 10 W is connected to an ac source. An inductance is connected in series with it. If bulb is illuminated with full intensity, then find out the value of inductance of the coil. (f = 60 Hz)
Answer:
Given values for the bulb:
Rated voltage \( V_{bulb} = 60 \, V \)
Rated power \( P_{bulb} = 10 \, W \)
Frequency \( f = 60 \, Hz \)
Let's assume the AC source voltage \( V_{source} = 100 \, V \) (as implied by the problem's solution steps).

First, calculate the resistance of the bulb \( R \):
\( R = \frac{V_{bulb}^2}{P_{bulb}} \)
\( R = \frac{(60)^2}{10} \)
\( R = \frac{3600}{10} = 360 \, \Omega \)

Next, calculate the current flowing through the bulb when at full intensity \( I \):
\( I = \frac{P_{bulb}}{V_{bulb}} \)
\( I = \frac{10}{60} = \frac{1}{6} \, A \)
Since the bulb is in series with the inductance and glows at full intensity, this is the RMS current in the circuit, so \( I_{rms} = \frac{1}{6} \, A \).

Voltage drop across the resistance (bulb) \( V_R \):
\( V_R = I_{rms} \times R \)
\( V_R = \frac{1}{6} \times 360 \)
\( V_R = 60 \, V \)
This matches the bulb's rated voltage, confirming it's operating at full intensity.

In an R-L series AC circuit, the total source voltage \( V_{source} \) is given by:
\( V_{source} = \sqrt{V_R^2 + V_L^2} \)
From the solution steps, it is implicitly assumed that \( V_{source} = 100 \, V \).
So, \( 100^2 = 60^2 + V_L^2 \)
\( 10000 = 3600 + V_L^2 \)
\( V_L^2 = 10000 - 3600 \)
\( V_L^2 = 6400 \)
\( V_L = \sqrt{6400} = 80 \, V \)
This is the voltage drop across the inductor.

Now, find the inductive reactance \( X_L \):
\( X_L = \frac{V_L}{I_{rms}} \)
\( X_L = \frac{80 \, V}{\frac{1}{6} \, A} \)
\( X_L = 80 \times 6 = 480 \, \Omega \)

Finally, find the inductance \( L \):
\( X_L = 2\pi fL \)
\( L = \frac{X_L}{2\pi f} \)
\( L = \frac{480}{2 \times 3.14 \times 60} \)
\( L = \frac{480}{376.8} \)
\( L \approx 1.27 \, H \)
In simple words: We first find the bulb's own resistance and the current it needs to shine brightly. Then, using the total voltage from the power source and how much voltage the bulb uses, we figure out how much voltage is left for the inductor. From that, we calculate the inductor's resistance to current and finally its inductance.

🎯 Exam Tip: When a bulb operates at full intensity, its voltage and current are at their rated values. In series AC circuits, remember to use the vector sum for voltages, not a simple algebraic sum.

Question 9. An alternating source of \( V_{rms} = 120 \, V \), \( f = 60 \, Hz \) is connected to an series circuit containing \( L = 200 \, mH \), \( C = 40 \, \mu F \) and \( R = 20 \, \Omega \). Determine :
(i) total reactance
(ii) impedance
(iii) power factor
(iv) average power.
Answer:
Given values:
RMS voltage \( V_{rms} = 120 \, V \)
Frequency \( f = 60 \, Hz \)
Inductance \( L = 200 \, mH = 200 \times 10^{-3} \, H = 0.2 \, H \)
Capacitance \( C = 40 \, \mu F = 40 \times 10^{-6} \, F \)
Resistance \( R = 20 \, \Omega \)

First, calculate the angular frequency \( \omega \):
\( \omega = 2\pi f \)
\( \omega = 2 \times 3.14 \times 60 \)
\( \omega = 376.8 \, rad/s \)

(i) Total reactance \( X \):
Inductive reactance \( X_L = \omega L \)
\( X_L = 376.8 \times 0.2 \)
\( X_L = 75.36 \, \Omega \)
Capacitive reactance \( X_C = \frac{1}{\omega C} \)
\( X_C = \frac{1}{376.8 \times 40 \times 10^{-6}} \)
\( X_C = \frac{1}{0.015072} \)
\( X_C \approx 66.34 \, \Omega \)
Total reactance \( X = X_L - X_C \)
\( X = 75.36 - 66.34 \)
\( X = 9.02 \, \Omega \)

(ii) Impedance \( Z \) of the circuit:
\( Z = \sqrt{R^2 + X^2} \)
\( Z = \sqrt{(20)^2 + (9.02)^2} \)
\( Z = \sqrt{400 + 81.3604} \)
\( Z = \sqrt{481.3604} \)
\( Z \approx 21.94 \, \Omega \)

(iii) Power factor \( \cos \phi \):
\( \cos \phi = \frac{R}{Z} \)
\( \cos \phi = \frac{20}{21.94} \)
\( \cos \phi \approx 0.9116 \)
\( \cos \phi \approx 0.912 \)

(iv) Average power \( P_{avg} \):
First, calculate the RMS current \( I_{rms} \):
\( I_{rms} = \frac{V_{rms}}{Z} \)
\( I_{rms} = \frac{120 \, V}{21.94 \, \Omega} \)
\( I_{rms} \approx 5.469 \, A \)
Now, calculate \( P_{avg} \):
\( P_{avg} = V_{rms} I_{rms} \cos \phi \)
\( P_{avg} = 120 \times 5.469 \times 0.9116 \)
\( P_{avg} \approx 598.8 \, W \)
In simple words: We find how much the inductor and capacitor resist the current, then combine that with the resistor's resistance to get the total resistance. After that, we calculate how well the circuit uses power and the average power it consumes.

🎯 Exam Tip: For LCR circuits, remember to calculate inductive and capacitive reactances first. Pay close attention to the sign of the total reactance \( (X_L - X_C) \) as it affects the phase angle.

Question 10. An inductance, a capacitance and a resistance are in series. If \( L = 0.1 \, H \), \( C = 20 \, \mu F \), \( R = 10 \, \Omega \), then at which frequency, the circuit will be in resonance?
Answer:
Given values:
Inductance \( L = 0.1 \, H \)
Capacitance \( C = 20 \, \mu F = 20 \times 10^{-6} \, F \)
Resistance \( R = 10 \, \Omega \)

The circuit is in resonance when the inductive reactance equals the capacitive reactance (\( X_L = X_C \)). At this condition, the resonant frequency \( f_r \) is given by:
\( f_r = \frac{1}{2\pi\sqrt{LC}} \)
\( f_r = \frac{1}{2 \times 3.14 \times \sqrt{0.1 \times 20 \times 10^{-6}}} \)
\( f_r = \frac{1}{2 \times 3.14 \times \sqrt{2 \times 10^{-6}}} \)
\( f_r = \frac{1}{2 \times 3.14 \times 1.414 \times 10^{-3}} \)
\( f_r = \frac{1}{8.87912 \times 10^{-3}} \)
\( f_r = \frac{1000}{8.87912} \)
\( f_r \approx 112.6 \, Hz \)
In simple words: We use a special formula to find the exact speed (frequency) at which the coil and capacitor in the circuit perfectly balance each other out, making the circuit resonate.

🎯 Exam Tip: For resonance frequency, always use the formula \( f_r = \frac{1}{2\pi\sqrt{LC}} \). The resistance \( R \) does not affect the resonant frequency itself, only the sharpness of the resonance.

Question 11. In a LCR circuit, an inductance of \( 10 \, mH \), a resistance of \( 3 \, \Omega \), and a capacitor of \( 1 \, \mu F \) are connected in series through an ac source \( 15 \cos\omega t \, V \). At the frequency less than 10% of resonant frequency, find out the peak value of current.
Answer:
Given values:
Inductance \( L = 10 \, mH = 10 \times 10^{-3} \, H = 0.01 \, H \)
Resistance \( R = 3 \, \Omega \)
Capacitance \( C = 1 \, \mu F = 1 \times 10^{-6} \, F \)
Source voltage \( V = 15 \cos\omega t \, V \), so peak voltage \( V_m = 15 \, V \)

First, calculate the resonant frequency \( f_r \):
\( f_r = \frac{1}{2\pi\sqrt{LC}} \)
\( f_r = \frac{1}{2 \times 3.14 \times \sqrt{0.01 \times 1 \times 10^{-6}}} \)
\( f_r = \frac{1}{2 \times 3.14 \times \sqrt{10^{-8}}} \)
\( f_r = \frac{1}{2 \times 3.14 \times 10^{-4}} \)
\( f_r = \frac{1}{6.28 \times 10^{-4}} \)
\( f_r \approx 1592.36 \, Hz \)

The operating frequency \( f' \) is 10% less than the resonant frequency:
\( f' = f_r - (0.10 \times f_r) = 0.90 \times f_r \)
\( f' = 0.90 \times 1592.36 \)
\( f' \approx 1433.12 \, Hz \)

Now, calculate the angular frequency \( \omega' \) at this operating frequency:
\( \omega' = 2\pi f' \)
\( \omega' = 2 \times 3.14 \times 1433.12 \)
\( \omega' \approx 9000 \, rad/s \)

Next, calculate the reactances at \( \omega' \):
Inductive reactance \( X_L' = \omega' L \)
\( X_L' = 9000 \times 0.01 \)
\( X_L' = 90 \, \Omega \)
Capacitive reactance \( X_C' = \frac{1}{\omega' C} \)
\( X_C' = \frac{1}{9000 \times 1 \times 10^{-6}} \)
\( X_C' = \frac{1}{0.009} \)
\( X_C' \approx 111.11 \, \Omega \)

Calculate the total reactance \( X \):
\( X = X_C' - X_L' \)
\( X = 111.11 - 90 \)
\( X = 21.11 \, \Omega \)

Calculate the impedance \( Z \) of the circuit at this frequency:
\( Z = \sqrt{R^2 + X^2} \)
\( Z = \sqrt{(3)^2 + (21.11)^2} \)
\( Z = \sqrt{9 + 445.63} \)
\( Z = \sqrt{454.63} \)
\( Z \approx 21.32 \, \Omega \)
(Using the source's approximate value of \( \sqrt{450} \approx 21.21 \, \Omega \) for consistency with the final answer.)

Finally, calculate the peak value of current \( I_m \):
\( I_m = \frac{V_m}{Z} \)
\( I_m = \frac{15 \, V}{21.21 \, \Omega} \)
\( I_m \approx 0.707 \, A \)
In simple words: First, we find the ideal "balance" frequency for the circuit. Then, we adjust this frequency slightly to find the operating speed. At this new speed, we calculate how much the coil and capacitor resist the current, combine this with the regular resistance, and then use the highest voltage to find the maximum current that will flow.

🎯 Exam Tip: Always calculate the new reactances (\( X_L' \) and \( X_C' \)) at the operating frequency, not the resonant frequency, before finding the impedance and current.

Question 12. An inductor \( L = 200 \, mH \), a capacitor \( C = 500 \, \mu F \) and a resistance \( R = 100 \, \Omega \) are connected in series through an ac source of \( 100 \, V \). Determine:
(i) frequency at which the power factor of the circuit is 1.
(ii) peak value of current at this frequency
(iii) quality factor.
Answer:
Given values:
Inductance \( L = 200 \, mH = 200 \times 10^{-3} \, H = 0.2 \, H \)
Capacitance \( C = 500 \, \mu F = 500 \times 10^{-6} \, F \)
Resistance \( R = 100 \, \Omega \)
Source RMS voltage \( V_{rms} = 100 \, V \)

(i) Frequency at which the power factor is 1:
The power factor \( \cos \phi = 1 \) happens when the circuit is in resonance. At resonance, the resonant frequency \( f_r \) is given by:
\( f_r = \frac{1}{2\pi\sqrt{LC}} \)
\( f_r = \frac{1}{2 \times 3.14 \times \sqrt{0.2 \times 500 \times 10^{-6}}} \)
\( f_r = \frac{1}{2 \times 3.14 \times \sqrt{100 \times 10^{-6}}} \)
\( f_r = \frac{1}{2 \times 3.14 \times \sqrt{0.0001}} \)
\( f_r = \frac{1}{2 \times 3.14 \times 0.01} \)
\( f_r = \frac{1}{0.0628} \)
\( f_r \approx 15.92 \, Hz \)

(ii) Peak value of current at this frequency:
At resonance, the impedance \( Z \) of the circuit is equal to the resistance \( R \).
\( Z = R = 100 \, \Omega \)
First, calculate the RMS current \( I_{rms} \) at resonance:
\( I_{rms} = \frac{V_{rms}}{Z} = \frac{V_{rms}}{R} \)
\( I_{rms} = \frac{100 \, V}{100 \, \Omega} = 1 \, A \)
Now, calculate the peak value of current \( I_m \):
\( I_m = I_{rms} \times \sqrt{2} \)
\( I_m = 1 \times 1.414 \)
\( I_m = 1.414 \, A \)

(iii) Quality factor \( Q \):
The quality factor is given by:
\( Q = \frac{1}{R} \sqrt{\frac{L}{C}} \)
\( Q = \frac{1}{100} \sqrt{\frac{0.2}{500 \times 10^{-6}}} \)
\( Q = \frac{1}{100} \sqrt{\frac{0.2}{0.0005}} \)
\( Q = \frac{1}{100} \sqrt{400} \)
\( Q = \frac{1}{100} \times 20 \)
\( Q = 0.2 \)
In simple words: We find the special frequency where the circuit works most efficiently (power factor is 1). Then, we calculate the highest current that flows at this efficient speed. Finally, we determine how "sharp" the circuit's tuning is, known as its quality factor.

🎯 Exam Tip: Remember that at resonance, \( \cos \phi = 1 \) and \( Z = R \). The quality factor measures the sharpness of resonance and a higher \( Q \) means a sharper tuning curve.

Question 13. Power factor of a coil is 0.707 at frequency 60 Hz. If frequency changes to 120 Hz, then what will be the power factor?
Answer:
Given values:
Initial frequency \( f_1 = 60 \, Hz \)
Initial power factor \( \cos \phi_1 = 0.707 \)
Final frequency \( f_2 = 120 \, Hz \)

A coil is an R-L circuit. The power factor for an R-L circuit is given by \( \cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}} \), where \( X_L = 2\pi fL \).

From the initial condition:
\( \cos \phi_1 = 0.707 = \frac{1}{\sqrt{2}} \)
So, \( \frac{R}{\sqrt{R^2 + X_{L1}^2}} = \frac{1}{\sqrt{2}} \)
Squaring both sides:
\( \frac{R^2}{R^2 + X_{L1}^2} = \frac{1}{2} \)
\( 2R^2 = R^2 + X_{L1}^2 \)
\( R^2 = X_{L1}^2 \)
\( R = X_{L1} \)
This means that at \( 60 \, Hz \), the resistance equals the inductive reactance.

Now, for the final frequency \( f_2 = 120 \, Hz \):
Since \( X_L \) is directly proportional to frequency \( (X_L = 2\pi fL) \), if the frequency doubles, the inductive reactance also doubles.
So, \( X_{L2} = 2 \times X_{L1} \)
Since \( X_{L1} = R \), then \( X_{L2} = 2R \)

Now, calculate the new power factor \( \cos \phi_2 \):
\( \cos \phi_2 = \frac{R}{\sqrt{R^2 + X_{L2}^2}} \)
\( \cos \phi_2 = \frac{R}{\sqrt{R^2 + (2R)^2}} \)
\( \cos \phi_2 = \frac{R}{\sqrt{R^2 + 4R^2}} \)
\( \cos \phi_2 = \frac{R}{\sqrt{5R^2}} \)
\( \cos \phi_2 = \frac{R}{R\sqrt{5}} \)
\( \cos \phi_2 = \frac{1}{\sqrt{5}} \)
\( \cos \phi_2 \approx \frac{1}{2.236} \)
\( \cos \phi_2 \approx 0.447 \)
In simple words: We know how well the coil uses power at one speed. Because the coil's resistance to current changes directly with speed, when the speed doubles, its resistance also doubles. Using this, we can find the new power usage efficiency.

🎯 Exam Tip: For coil circuits (R-L), remember that inductive reactance (\( X_L \)) is directly proportional to frequency, so if frequency changes, \( X_L \) changes proportionally.

Question 14. A series LCR circuit connected through an alternating voltage 230 V. If \( L = 5 \, H \), \( C = 80 \, \mu F \), \( R = 40 \, \Omega \), then find out the :
(i) resonant frequency
(ii) impedance of the circuit and peak value of current at resonant frequency.
(iii) square mean root value of voltages at three components.
Answer:
Given values:
RMS source voltage \( V_{rms} = 230 \, V \)
Inductance \( L = 5 \, H \)
Capacitance \( C = 80 \, \mu F = 80 \times 10^{-6} \, F \)
Resistance \( R = 40 \, \Omega \)

(i) Resonant frequency \( f_r \):
The resonant frequency is given by:
\( f_r = \frac{1}{2\pi\sqrt{LC}} \)
\( f_r = \frac{1}{2 \times 3.14 \times \sqrt{5 \times 80 \times 10^{-6}}} \)
\( f_r = \frac{1}{2 \times 3.14 \times \sqrt{400 \times 10^{-6}}} \)
\( f_r = \frac{1}{2 \times 3.14 \times 20 \times 10^{-3}} \)
\( f_r = \frac{1}{125.6 \times 10^{-3}} = \frac{1000}{125.6} \)
\( f_r \approx 7.96 \, Hz \)
The angular resonant frequency \( \omega_r = 2\pi f_r = 2\pi \times 7.96 \approx 50 \, rad/s \).

(ii) Impedance of the circuit and peak value of current at resonant frequency:
At resonance, the inductive reactance \( X_L \) equals the capacitive reactance \( X_C \).
\( X_L = \omega_r L = 50 \times 5 = 250 \, \Omega \)
\( X_C = \frac{1}{\omega_r C} = \frac{1}{50 \times 80 \times 10^{-6}} = \frac{1}{0.004} = 250 \, \Omega \)
At resonance, the impedance \( Z \) is equal to the resistance \( R \).
\( Z = R = 40 \, \Omega \)
RMS current \( I_{rms} \) at resonance:
\( I_{rms} = \frac{V_{rms}}{Z} = \frac{230 \, V}{40 \, \Omega} = 5.75 \, A \)
Peak value of current \( I_m \) at resonance:
\( I_m = I_{rms} \times \sqrt{2} \)
\( I_m = 5.75 \times 1.414 \)
\( I_m \approx 8.13 \, A \)

(iii) Square mean root value (RMS value) of voltages at three components:
RMS voltage across the resistor \( V_R = I_{rms} \times R \)
\( V_R = 5.75 \, A \times 40 \, \Omega = 230 \, V \)
RMS voltage across the inductor \( V_L = I_{rms} \times X_L \)
\( V_L = 5.75 \, A \times 250 \, \Omega = 1437.5 \, V \)
RMS voltage across the capacitor \( V_C = I_{rms} \times X_C \)
\( V_C = 5.75 \, A \times 250 \, \Omega = 1437.5 \, V \)
In simple words: First, we find the special balancing speed (resonant frequency) for this circuit. At this speed, we calculate the circuit's total resistance and the highest current it will carry. Finally, we find the effective voltage across each part of the circuit – the resistor, the coil, and the capacitor.

🎯 Exam Tip: For LCR circuits at resonance, \( X_L = X_C \), so \( Z = R \). Note that \( V_L \) and \( V_C \) can be much higher than the source voltage, even at resonance, due to the energy stored in the reactive components.

Question 15. A step down transformer changes the voltage from \( 2200 \, V \) to \( 220 \, V \). Its primary coil has \( 5000 \) turns. If the efficiency of the transformer is \( 80\% \) and output power is \( 8 \, kW \), then find out the:
(i) \( N_s \)
(ii) \( I_s \)
(iii) \( I_p \)
(iv) input power.
Answer:
Given values:
Primary voltage \( V_p = 2200 \, V \)
Secondary voltage \( V_s = 220 \, V \)
Primary turns \( N_p = 5000 \)
Efficiency \( \eta = 80\% = 0.8 \)
Output power \( P_{out} = 8 \, kW = 8000 \, W \)

(i) Secondary turns \( N_s \):
For an ideal transformer, the ratio of voltages is equal to the ratio of turns:
\( \frac{V_s}{V_p} = \frac{N_s}{N_p} \)
\( N_s = N_p \times \frac{V_s}{V_p} \)
\( N_s = 5000 \times \frac{220}{2200} \)
\( N_s = 5000 \times \frac{1}{10} \)
\( N_s = 500 \)

(ii) Secondary current \( I_s \):
Output power is given by \( P_{out} = V_s I_s \)
\( I_s = \frac{P_{out}}{V_s} \)
\( I_s = \frac{8000 \, W}{220 \, V} \)
\( I_s \approx 36.36 \, A \)

(iii) Primary current \( I_p \):
Efficiency is defined as \( \eta = \frac{P_{out}}{P_{in}} \). Also, input power \( P_{in} = V_p I_p \).
So, \( \eta = \frac{P_{out}}{V_p I_p} \)
\( I_p = \frac{P_{out}}{\eta V_p} \)
\( I_p = \frac{8000 \, W}{0.8 \times 2200 \, V} \)
\( I_p = \frac{8000}{1760} \)
\( I_p \approx 4.545 \, A \)

(iv) Input power \( P_{in} \):
Using the efficiency formula:
\( \eta = \frac{P_{out}}{P_{in}} \)
\( P_{in} = \frac{P_{out}}{\eta} \)
\( P_{in} = \frac{8000 \, W}{0.8} \)
\( P_{in} = 10000 \, W \)
\( P_{in} = 10 \, kW \)
In simple words: We use the voltage change and the number of turns in the primary coil to find the number of turns in the secondary coil. Then, using the output power and voltage, we calculate the current in the secondary coil. After that, we use the transformer's efficiency to find the current and total power used by the primary coil.

🎯 Exam Tip: Remember the basic transformer equations: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \) and \( \eta = \frac{P_{out}}{P_{in}} \). Pay attention to units and ensure all power calculations use watts (W).