Maharashtra Board Class 12 Physics Chapter 13 AC Circuits Exercise Solutions

Question. Choose the correct option.

Question. If the RMS current in a 50 Hz AC circuit is 5A, the value of the current \( \frac{1}{300} \) seconds after its value becomes zero is
(a) \( 5\sqrt{2} \) A
(b) \( 5\sqrt{\frac{3}{2}} \) A
(c) \( \frac{5}{6} \) A
(d) \( \frac{5}{\sqrt{2}} \) A
Answer: (b) \( 5\sqrt{\frac{3}{2}} \) A
In simple words: When AC current starts from zero and goes through its cycle, at a specific time it reaches this particular value based on the sine wave pattern.

📝 Teacher's Note: Use the formula \( i = i_0 \sin(2\pi ft) \) where \( i_0 = I_{rms}\sqrt{2} \). Substitute the given time and frequency to find the instantaneous current.

🎯 Exam Tip: Always convert RMS values to peak values first using \( i_0 = I_{rms}\sqrt{2} \), then use the time equation to find instantaneous values.

Question. A resistor of 500 Ω and an inductance of 0.5 H are in series with an AC source which is given by V = 100\( \sqrt{2} \) sin (1000t). The power factor of the combination
(a) \( \frac{1}{\sqrt{2}} \)
(b) \( \frac{1}{\sqrt{3}} \)
(c) 0.5
(d) 0.6
Answer: (a) \( \frac{1}{\sqrt{2}} \)
In simple words: Power factor tells us how much of the total power is actually useful - here it's about 70% efficiency.

📝 Teacher's Note: Calculate \( X_L = \omega L \) first, then find impedance \( Z = \sqrt{R^2 + X_L^2} \), and power factor = \( \frac{R}{Z} \).

🎯 Exam Tip: Power factor = \( \cos \phi = \frac{R}{Z} \) - remember this key formula for all AC circuit problems.

Question. In a circuit L, C & R are connected in series with an alternating voltage of frequency f. the current leads the voltage by 45°. The value of C is
(a) \( \frac{1}{\pi f(2\pi fL - R)} \)
(b) \( \frac{1}{2\pi f(2\pi fL - R)} \)
(c) \( \frac{1}{\pi f(2\pi fL + R)} \)
(d) \( \frac{1}{2\pi f(2\pi fL + R)} \)
Answer: (b) \( \frac{1}{2\pi f(2\pi fL - R)} \)
In simple words: When current leads voltage by 45°, the capacitive reactance must be greater than inductive reactance by a specific amount.

📝 Teacher's Note: For current leading by 45°, \( \tan 45° = 1 = \frac{X_C - X_L}{R} \). This gives the condition to solve for C.

🎯 Exam Tip: When current leads voltage, the circuit is capacitive. Use \( \tan \phi = \frac{X_C - X_L}{R} \) for the phase relationship.

Question. In an AC circuit, e and i are given by e = 150 sin (150t) V and i = 150 sin (150t + \( \frac{\pi}{3} \)) A. the power dissipated in the circuit is
(a) 106W
(b) 150W
(c) 5625W
(d) Zero
Answer: (c) 5625W
In simple words: Even though current and voltage are not in phase, there's still significant power consumption because the phase difference isn't 90°.

📝 Teacher's Note: Use \( P = V_{rms} I_{rms} \cos \phi \) where \( \phi \) is the phase difference between voltage and current. Here \( \phi = 60° \).

🎯 Exam Tip: Convert peak values to RMS by dividing by \( \sqrt{2} \), then use \( P = V_{rms} I_{rms} \cos \phi \).

Question. In a series LCR circuit the phase difference between the voltage and the current is 45°. Then the power factor will be
(a) 0.607
(b) 0.707
(c) 0.808
(d) 1
Answer: (b) 0.707
In simple words: Power factor of 0.707 means about 71% of the power is useful, which happens when the phase difference is 45°.

📝 Teacher's Note: Power factor = \( \cos \phi \). When \( \phi = 45° \), \( \cos 45° = \frac{1}{\sqrt{2}} = 0.707 \).

🎯 Exam Tip: Remember \( \cos 45° = 0.707 \) - this is a frequently tested value in AC circuits.

Answer in brief.

Question. An electric lamp is connected in series with a capacitor and an AC source is glowing with a certain brightness. How does the brightness of the lamp change on increasing the capacitance?
Answer: Impedance, \( Z = \sqrt{R^2 + \frac{1}{\omega^2 C^2}} \), where R is the resistance of the lamp, ω is the angular frequency of AC and C is the capacitance of the capacitor connected in series with the AC source and the lamp. When C is increased, \( \frac{1}{\omega C} \) decreases. Hence, Z decreases.
Power factor, \( \cos \Phi = \frac{R}{Z} \)
As Z decreases, the power factor increases.
Now, the average power over one cycle,
\( P_{av} = V_{rms} I_{rms} \cos \Phi \)
\( = V_{rms} \left( \frac{V_{rms}}{Z} \right) \cos \Phi \)
\( = \frac{V_{rms}^2}{Z} \cos \phi \)
\( \therefore P_{av} \) increases as Z decreases and \( \cos \Phi \) increases.
As the current through the lamp \( \left( \frac{V_{rms}}{Z} \right) \) increases, the brightness of the lamp will increase when C is increased.
In simple words: Increasing capacitance reduces the total resistance to current flow, so more current flows through the lamp making it brighter.

📝 Teacher's Note: Emphasize that capacitive reactance \( X_C = \frac{1}{\omega C} \) decreases as C increases, leading to lower total impedance and higher current.

🎯 Exam Tip: State that lower impedance → higher current → brighter lamp. Always explain the cause-effect relationship step by step.

Question. The total impedance of a circuit decreases when a capacitor is added in series with L and R. Explain why?
Answer: For an LR circuit, the impedance,
\( Z_{LR} = \sqrt{R^2 + X_L^2} \), where \( X_L \) is the reactance of the inductor.
When a capacitor of capacitance C is added in series with L and R, the impedance,
\( Z_{LCR} = \sqrt{R^2 + (X_L - X_C)^2} \) because in the case of an inductor the current lags behind the voltage by a phase angle of \( \frac{\pi}{2} \) rad while in the case of a capacitor the current leads the voltage by a phase angle of \( \frac{\pi}{2} \) rad. The decrease in net reactance decreases the total impedance (\( Z_{LCR} < Z_{LR} \)).
In simple words: The capacitor and inductor have opposite effects on current - they partially cancel each other out, reducing the total opposition to current flow.

📝 Teacher's Note: Draw phasor diagrams showing how \( X_L \) and \( X_C \) point in opposite directions, resulting in net reactance \( |X_L - X_C| \).

🎯 Exam Tip: Always mention that \( X_L \) and \( X_C \) are 180° out of phase, so the net reactance is their difference, not sum.

Question. For very high frequency AC supply, a capacitor behaves like a pure conductor. Why?
Answer: The reactance of a capacitor is \( X_C = \frac{1}{2\pi fC} \), where f is the frequency of the AC supply and C is the capacitance of the capacitor. For very high frequency, f, \( X_C \) is very small. Hence, for very high frequency AC supply, a capacitor behaves like a pure conductor.
In simple words: At very high frequencies, the capacitor's opposition to current becomes almost zero, so current flows through it easily like through a wire.

📝 Teacher's Note: Use numerical examples: if f = 1 MHz and C = 1 μF, then \( X_C = \frac{1}{2\pi \times 10^6 \times 10^{-6}} = 0.16 \) Ω, which is very small.

🎯 Exam Tip: State that as f → ∞, \( X_C \) → 0, making the capacitor act like a short circuit.

Question. What is wattless current?
Answer: The current that does not lead to energy consumption, hence zero power consumption, is called wattless current. In the case of a purely inductive circuit or a purely capacitive circuit, average power consumed over a complete cycle is zero and hence the corresponding alternating current in the circuit is called wattless current. [Note: In this case, the power factor is zero.]
In simple words: Wattless current is like a current that flows back and forth but doesn't actually consume any electrical energy - it's purely reactive.

📝 Teacher's Note: Explain that in pure L or C circuits, energy is stored and returned each half cycle, resulting in zero net energy consumption.

🎯 Exam Tip: Mention that wattless current occurs when voltage and current are 90° out of phase, making cos φ = 0.

Question. What is the natural frequency of LC circuit? What is the reactance of this circuit at this frequency
Answer: The natural frequency of LC circuit is \( \frac{1}{2\pi\sqrt{LC}} \), where L is the inductance and C is the capacitance. The reactance of this circuit at this frequency is
\( \frac{1}{2\pi C} - \frac{1}{2\pi fL} = \frac{1}{2\pi C} - \frac{1}{2\pi \cdot \frac{1}{2\pi\sqrt{LC}} \cdot L} = \frac{1}{2\pi C} - \frac{1}{2\pi L} \cdot \frac{2\pi\sqrt{LC}}{1} = \frac{1}{2\pi\sqrt{LC}} \cdot \frac{1}{\sqrt{\frac{C}{L}}} - \frac{1}{\sqrt{\frac{C}{L}}} = \frac{1}{\sqrt{\frac{C}{L}}} - \frac{1}{\sqrt{\frac{C}{L}}} = 0. \)
In simple words: At the natural frequency, the inductive and capacitive reactances exactly cancel each other out, making the total reactance zero.

📝 Teacher's Note: Emphasize that at resonance, \( X_L = X_C \), so the net reactance is zero and the circuit behaves like a pure resistor.

🎯 Exam Tip: Remember the resonance condition: \( \omega L = \frac{1}{\omega C} \), which gives \( \omega = \frac{1}{\sqrt{LC}} \).

Question 3. In a series LR circuit \( X_L = R \) and power factor of the circuit is \( P_1 \). When capacitor with capacitance C such that \( X_L = X_C \) is put in series, the power factor becomes \( P_2 \). Calculate \( P_1 / P_2 \).
Answer: For a series LR circuit, power factor,
cos \( \phi = \frac{R}{\sqrt{R^2 + X_L^2}} \)
If \( X_L = R \), power factor \( P_1 = \frac{R}{\sqrt{2R^2}} = \frac{1}{\sqrt{2}} \)
For a series LCR circuit, power factor,
cos \( \phi = \frac{R}{\sqrt{R^2 + (X_L - X_C)^2}} \)
If \( X_L = X_C \), power factor \( P_2 = \frac{R}{R} = 1 \)
\( \therefore \frac{P_1}{P_2} = \frac{1}{\sqrt{2}} \)

📝 Teacher's Note: Emphasize that when \( X_L = X_C \), the circuit behaves like a pure resistive circuit. Use phasor diagrams to show how reactances cancel out.

🎯 Exam Tip: Remember the key formula: power factor = R/Z. When reactances are equal and opposite, they cancel out, making Z = R.

Question 4. When an AC source is connected to an ideal inductor show that the average power supplied by the source over a complete cycle is zero.
Answer: In an AC circuit containing only an ideal inductor, the current i lags behind the emf e by a phase angle of \( \frac{\pi}{2} \) rad. Here, for e = \( e_0 \) sin \( \omega t \), we have, i = \( i_0 \) sin(\( \omega t \) - \( \frac{\pi}{2} \))
Instantaneous power, P = ei
= (\( e_0 \) sin \( \omega t \)) [\( i_0 \) (sin \( \omega t \) cos \( \frac{\pi}{2} \) - cos \( \omega t \) sin \( \frac{\pi}{2} \))]
= - \( e_0 i_0 \) sin \( \omega t \) cos \( \omega t \) as cos \( \frac{\pi}{2} \) = 0 and sin \( \frac{\pi}{2} \) = 1.
Average power over one cycle,
\( P_{av} = \frac{\text{work done in one cycle}}{\text{time for one cycle}} \)
\( = \frac{\int_0^T P dt}{T} \)
\( = \frac{\int_0^T [e_0 i_0 \cos \phi \sin^2 \omega t \pm e_0 i_0 \sin \phi \sin \omega t \cos \omega t] dt}{T} \)
\( = \frac{e_0 i_0}{T} [\cos \phi \int_0^T \sin^2 \omega t dt \pm \sin \phi \int_0^T \sin \omega t \cos \omega t dt] \)
Now, \( \int_0^T \sin^2 \omega t dt = \int_0^T \frac{1 - \cos 2\omega t}{2} dt \)
\( = \int_0^T \frac{1}{2} dt - \int_0^T \frac{\cos 2\omega t}{2} dt = \frac{T}{2} - \frac{1}{2} \left( \frac{\sin 2\omega t}{2\omega} \right)_0^T \)
\( = \frac{T}{2} - \frac{1}{4\omega} (\sin 2\omega T - \sin 0) \)
\( = \frac{T}{2} - \frac{1}{4\omega} \sin 2 \left( \frac{2\pi}{T} \right) T - 0 \)
\( = \frac{T}{2} - \frac{1}{4\omega} [0 - 0] = \frac{T}{2} \)
Also, \( \int_0^T \sin \omega t \cos \omega t dt = \frac{1}{2} \int_0^T \sin 2\omega t dt = \frac{1}{2} \left[ \frac{-\cos 2\omega t}{2\omega} \right]_0^T \)
\( = - \frac{1}{4\omega} \cos 2 \left( \frac{2\pi}{T} \right) T - \cos 0 = - \frac{1}{4\omega} [1 - 1] = 0 \)
Hence, \( P_{av} = \frac{e_0 i_0}{T} \cos \phi \times \frac{T}{2} = \frac{e_0 i_0}{2} \cos \phi \)
\( = \frac{e_0}{\sqrt{2}} \frac{i_0}{\sqrt{2}} \cos \phi \)
= \( e_{rms} i_{rms} \) cos \( \Phi = e_{rms} i_{rms} \left( \frac{R}{Z} \right) \), where the impedance Z = \( \sqrt{R^2 + (X_L - X_C)^2} \).
\( \therefore P_{av} \) = 0, i.e., the circuit does not dissipate power.

📝 Teacher's Note: Draw waveform graphs showing how emf and current are 90° out of phase, resulting in alternating positive and negative power cycles that cancel out over one complete cycle.

🎯 Exam Tip: Remember that for pure reactive components (L or C), average power is zero because current and voltage are 90° out of phase.

Question 5. Prove that an ideal capacitor in an AC circuit does not dissipate power
Answer: In an AC circuit containing only an ideal capacitor, the current i leads the emf e by a phase angle of \( \frac{\pi}{2} \) rad. Here, for e = \( e_0 \) sin \( \omega t \), we have, i = \( i_0 \) sin(\( \omega t \) + \( \frac{\pi}{2} \))
Instantaneous power, P = ei
= (\( e_0 \) sin \( \omega t \)) [\( i_0 \) (sin \( \omega t \) cos \( \frac{\pi}{2} \) + cos \( \omega t \) sin \( \frac{\pi}{2} \))]
= \( e_0 i_0 \) sin \( \omega t \) cos \( \omega t \) as cos \( \frac{\pi}{2} \) = 0 and sin \( \frac{\pi}{2} \) = 1.
Average power over one cycle, \( P_{av} \)
\( = \frac{\text{work done in one cycle}}{\text{time for one cycle}} \)
\( = \frac{\int_0^T P dt}{T} \)
\( = \frac{\int_0^T [e_0 i_0 \cos \phi \sin^2 \omega t \pm e_0 i_0 \sin \phi \sin \omega t \cos \omega t] dt}{T} \)
\( = \frac{e_0 i_0}{T} [\cos \phi \int_0^T \sin^2 \omega t dt \pm \sin \phi \int_0^T \sin \omega t \cos \omega t dt] \)
Now, \( \int_0^T \sin^2 \omega t dt = \int_0^T \frac{1 - \cos 2\omega t}{2} dt \)
\( = \int_0^T \frac{1}{2} dt - \int_0^T \frac{\cos 2\omega t}{2} dt = \frac{T}{2} - \frac{1}{2} \left( \frac{\sin 2\omega t}{2\omega} \right)_0^T \)
\( = \frac{T}{2} - \frac{1}{4\omega} (\sin 2\omega T - \sin 0) \)
\( = \frac{T}{2} - \frac{1}{4\omega} \sin 2 \left( \frac{2\pi}{T} \right) T - 0 \)
\( = \frac{T}{2} - \frac{1}{4\omega} [0 - 0] = \frac{T}{2} \)
Also, \( \int_0^T \sin \omega t \cos \omega t dt = \frac{1}{2} \int_0^T \sin 2\omega t dt = \frac{1}{2} \left[ \frac{-\cos 2\omega t}{2\omega} \right]_0^T \)
\( = - \frac{1}{4\omega} \cos 2 \left( \frac{2\pi}{T} \right) T - \cos 0 = - \frac{1}{4\omega} [1 - 1] = 0 \)
Hence, \( P_{av} = \frac{e_0 i_0}{T} \cos \phi \times \frac{T}{2} = \frac{e_0 i_0}{2} \cos \phi \)
\( = \frac{e_0}{\sqrt{2}} \frac{i_0}{\sqrt{2}} \cos \phi \)
= \( e_{rms} i_{rms} \) cos \( \Phi = e_{rms} i_{rms} \left( \frac{R}{Z} \right) \), where the impedance Z = \( \sqrt{R^2 + (X_L - X_C)^2} \).
\( \therefore P_{av} \) = 0, i.e., the circuit does not dissipate power.

📝 Teacher's Note: Compare this with the inductor case - both are reactive components where current and voltage are 90° out of phase, but in opposite directions. The mathematical proof is identical.

🎯 Exam Tip: For any pure reactive component, the phase difference is ±90°, making cos φ = 0, hence zero power dissipation.

Question 6. (a) An emf e = \( e_0 \) sin \( \omega t \) applied to a series L – C – R circuit derives a current I = \( I_0 \) sin\( \omega t \) in the circuit. Deduce the expression for the average power dissipated in the circuit. (b) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain.
Answer: (a) Instantaneous power,
P = ei
= (\( e_0 \) sin \( \omega t \)) [\( i_0 \) (sin \( \omega t \) ± \( \Phi \))]
= \( e_0 i_0 \) sin \( \omega t \)(sin \( \omega t \) cos \( \Phi \) ± cos \( \omega t \) sin \( \Phi \))
= \( e_0 i_0 \) sin² \( \omega t \) cos \( \Phi \) ± \( e_0 i_0 \) sin \( \Phi \) sin \( \omega t \) cos \( \omega t \)
Average power over one cycle,
\( P_{av} = \frac{\text{work done in one cycle}}{\text{time for one cycle}} \)
\( = \frac{\int_0^T P dt}{T} \)
\( = \frac{\int_0^T [e_0 i_0 \cos \phi \sin^2 \omega t \pm e_0 i_0 \sin \phi \sin \omega t \cos \omega t] dt}{T} \)
\( = \frac{e_0 i_0}{T} [\cos \phi \int_0^T \sin^2 \omega t dt \pm \sin \phi \int_0^T \sin \omega t \cos \omega t dt] \)
Now, \( \int_0^T \sin^2 \omega t dt = \int_0^T \frac{1 - \cos 2\omega t}{2} dt \)
\( = \int_0^T \frac{1}{2} dt - \int_0^T \frac{\cos 2\omega t}{2} dt = \frac{T}{2} - \frac{1}{2} \left( \frac{\sin 2\omega t}{2\omega} \right)_0^T \)
\( = \frac{e_0 i_0}{T} \cos \phi \times \frac{T}{2} = \frac{e_0 i_0}{2} \cos \phi \)
\( = \frac{e_0}{\sqrt{2}} \frac{i_0}{\sqrt{2}} \cos \phi \)
= \( e_{rms} i_{rms} \) cos \( \Phi = e_{rms} i_{rms} \left( \frac{R}{Z} \right) \), where the impedance Z = \( \sqrt{R^2 + (X_L - X_C)^2} \).
(b) \( P_{av} = e_{rms} i_{rms} \) cos \( \Phi \)
The factor cos \( \Phi \) is called as power factor. For circuits used for transporting electric power, a low power factor means the power available on transportation is much less than \( e_{rms} i_{rms} \). It means there is significant loss of power during transportation.

📝 Teacher's Note: Emphasize that power factor represents the efficiency of power transfer. Draw a power triangle showing active, reactive, and apparent power components.

🎯 Exam Tip: Low power factor means more current is needed for the same useful power, leading to higher I²R losses in transmission lines.

Question 7. A device Y is connected across an AC source of emf e = \( e_0 \) sin\( \omega t \). The current through Y is given as i = \( i_0 \) sin(\( \omega t \) + π/2) (a) Identify the device Y and write the expression for its reactance. (b) Draw graphs showing variation of emf and current with time over one cycle of AC for Y. (c) How does the reactance of the device Y vary with the frequency of the AC ? Show graphically (d) Draw the phasor diagram for the device Y.
Answer: (a) The device Y is a capacitor. Its reactance is \( X_c = \frac{1}{\omega C} \), where \( \omega \) is the angular frequency of the applied emf and C is the capacitance of the capacitor.
(b) [Graph shown in original document]
(c) \( X_C = \frac{1}{\omega C} = \frac{1}{2\pi fC} \). Thus \( X_C \propto \frac{1}{f} \), where f is the frequency of AC. Suppose C = \( \left( \frac{1000}{2\pi} \right) \) pF
For f= 100 Hz, \( X_C \) = 1 × 10⁷ Ω = 10MΩ;
for f = 200 Hz, \( X_C \) = 5 MΩ;
for f = 300 Hz, \( X_C = \frac{10}{3} \) MΩ;
for f = 400 Hz, \( X_C \) = 2.5 MΩ
for f = 500 Hz, \( X_C \) = 2 MΩ and so on
(d) The phasor representing the peak emf (\( e_0 \)) makes an angle (\( \omega t \)) in an anticlockwise direction with respect to the horizontal axis. As the current leads the voltage by 90°, the phasor representing the peak current (\( i_0 \)) is turned 90° anticlockwise with respect to the phasor representing emf \( e_0 \). The projections of these phasors on the vertical axis give instantaneous values of e and i.

📝 Teacher's Note: Use practical examples like a motor starter circuit to show how capacitors are used to improve power factor. Demonstrate the phase relationship with oscilloscope traces if available.

🎯 Exam Tip: Remember that current leads voltage in a capacitor (ICE - "current leads in capacitors and inductors lag"), and reactance is inversely proportional to frequency.

Question. For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain.
Answer: \( P_{av} = e_{rms} i_{rms} \cos \Phi \)
The factor \( \cos \Phi \) is called as power factor. For circuits used for transporting electric power, a low power factor means the power available on transportation is much less than \( e_{rms} i_{rms} \). It means there is significant loss of power during transportation.

📝 Teacher's Note: Use the analogy of a water pump — low power factor is like a pump that moves a lot of water but doesn't deliver much useful flow due to leaks and inefficiencies.

🎯 Exam Tip: Always mention both the mathematical relationship and the practical implication for power transmission to score full marks.

Question. A device Y is connected across an AC source of emf e = e₀ sinωt. The current through Y is given as i = i₀ sin(ωt + π/2)
(a) Identify the device Y and write the expression for its reactance.
(b) Draw graphs showing variation of emf and current with time over one cycle of AC for Y.
(c) How does the reactance of the device Y vary with the frequency of the AC? Show graphically
(d) Draw the phasor diagram for the device Y.
Answer:
(a) The device Y is a capacitor. Its reactance is \( X_c = \frac{1}{\omega C} \),
where ω is the angular frequency of the applied emf and C is the capacitance of the capacitor.

(b)

(c) \( X_C = \frac{1}{\omega C} = \frac{1}{2\pi fC} \). Thus \( X_C \propto \frac{1}{f} \), where f is the frequency of AC. Suppose C = \( \left( \frac{1000}{2\pi} \right) \) pF
For f= 100 Hz, \( X_C = 1 \times 10^7 \) Ω = 10MΩ;
for f = 200 Hz, \( X_C = 5 \) MΩ;
for f = 300 Hz, \( X_C = \frac{10}{3} \) MΩ;
for f = 400 Hz, \( X_C = 2.5 \) MΩ
for f = 500 Hz, \( X_C = 2 \) MΩ and so on

(d)
The phasor representing the peak emf (e₀) makes an angle (ωt) in an anticlockwise direction with respect to the horizontal axis. As the current leads the voltage by 90°, the phasor representing the peak current (i₀) is turned 90° anticlockwise with respect to the phasor representing emf e₀. The projections of these phasors on the vertical axis give instantaneous values of e and i.

📝 Teacher's Note: Emphasize that in a capacitor, current "rushes ahead" to charge the plates before voltage builds up — this helps students remember the phase relationship.

🎯 Exam Tip: Always draw phasor diagrams with proper labels and arrows showing direction of rotation to get full marks.

Question. Derive an expression for the impedance of an LCR circuit connected to an AC power supply.
Answer:
Figure shows an inductor of inductance L, capacitor of capacitance C, resistor of resistance R, key K and source (power supply) of alternating emf (e) connected to form a closed series circuit.

We assume the inductor, capacitor and resistor to be ideal. As these are connected in series, at any instant, they carry the same current i = i₀ sin ωt. The voltage across the resistor, eᵣ = Ri, is in phase with the current. The voltage across the inductor, eₗ = XₗI, leads the current by \( \frac{\pi}{2} \) rad and that across the capacitor, eᶜ = XᶜI, lags behind the current by \( \frac{\pi}{2} \) rad. This is shown in the phasor diagram.

\( Z = \frac{e_0}{i_0} = \sqrt{R^2 + (X_L - X_C)^2} \)

is the effective resistance of the circuit. It is called the impedance.

📝 Teacher's Note: Use vector addition analogy — resistance is horizontal, reactances are vertical (opposite directions), and impedance is the resultant vector.

🎯 Exam Tip: Always include the phasor diagram and derive the formula using Pythagoras theorem for maximum marks.

Question. Compare resistance and reactance.
Answer:
(1) Resistance is opposition to flow of charges (current) and appears in a DC circuit as well as in an AC circuit.
The term reactance appears only in an AC circuit. It occurs when an inductor and/or a capacitor is used.

(2) In a purely resistive circuit, current and voltage are always in phase.
When reactance is not zero, there is nonzero phase difference between current and voltage.

(3) Resistance does not depend on the frequency of AC.
Reactance depends on the frequency of AC. In case of an inductor, reactance increases linearly with frequency. In case of a capacitor, reactance decreases as frequency of AC increases; it is inversely proportional to frequency.

(4) Resistance gives rise to production of Joule heat in a component.
In a circuit with pure reactance, there is no production of heat.

📝 Teacher's Note: Use the analogy of a river — resistance is like rocks that always oppose flow, while reactance is like a dam that only opposes changing flow.

🎯 Exam Tip: Present the comparison in a clear point-by-point format and mention both inductors and capacitors for reactance.

Question. Show that in an AC circuit containing a pure inductor, the voltage is ahead of current by π/2 in phase.
Answer:
Figure 13.8 shows an AC source, generating a voltage e = e₀ sin ωt, connected to a key K and a pure inductor of inductance L to form a closed circuit.

On closing the key K, an emf is induced in the inductor as the magnetic flux linked with it changes with time. This emf opposes the applied emf and according to the laws of electromagnetic induction by Faraday and Lenz, we have,

e' = -L \( \frac{di}{dt} \) ................. (1)
where e' is the induced emf and i is the current through the inductor. To maintain the current; e and e' must be equal in magnitude and opposite in direction.

According to Kirchhoff's voltage law, as the resistance of the inductor is assumed to be zero, we have, e = -e' = L \( \frac{di}{dt} \) ... (2)

\( \frac{di}{dt} = \frac{e}{L} = \frac{e_0 \sin \omega t}{L} \)

\( \int di = \int \frac{e_0 \sin \omega t}{L} dt \)

\( i = -\frac{e_0}{\omega L} \cos \omega t + C \)

where C is the constant of integration. C must be time independent and have the dimension of current. As e oscillates about zero, i also oscillates about zero and hence there cannot be any time independent component of current.
∴ C = 0. ∴ i = -\( \frac{e_0}{\omega L} \) cos ωt = -\( \frac{e_0}{\omega L} \) sin(\( \frac{\pi}{2} \) - ωt)
∴ i = \( \frac{e_0}{\omega L} \) sin(ωt - \( \frac{\pi}{2} \)) ............. (3)
as sin (-θ) = - sin θ
From Eq. (3), i_peak = i₀ = \( \frac{e_0}{\omega L} \)
∴ i = i₀ sin(ωt - \( \frac{\pi}{2} \)) ................ (4)
Comparison of this equation with e = e₀ sin ωt shows that e leads i by \( \frac{\pi}{2} \) rad, i.e., the voltage is ahead of current by \( \frac{\pi}{2} \) rad in phase.

📝 Teacher's Note: Emphasize the physical meaning — an inductor "resists" change in current, so current takes time to catch up with voltage changes.

🎯 Exam Tip: Always start with Faraday's law and show all mathematical steps clearly to demonstrate the π/2 phase difference.

Question. An AC source generating a voltage e = e₀ sinωt is connected to a capacitor of capacitance C. Find the expression for the current i flowing through it. Plot a graph of e and i versus ωt.
Answer:
Figure 13.12 shows an AC source, generating a voltage e = e₀ sin ωt, connected to a capacitor of capacitance C. The plates of the capacitor get charged due to the applied voltage. As the alternating voltage is reversed in each half cycle, the capacitor is alternately charged and discharged. If q is the charge on the capacitor, the corresponding potential difference across the plates of the capacitor is V = \( \frac{q}{C} \)
∴ q = CV. q and V are functions of time, with V = e = e₀ sin ωt. The instantaneous current in the circuit is i = \( \frac{dq}{dt} = \frac{d}{dt} \)(CV) = C \( \frac{dv}{dt} \) = C \( \frac{d}{dt} \)(e₀ sin ωt) = ωC e₀ cos ωt
∴ i = \( \frac{e_0}{(1/\omega C)} \) sin(ωt + \( \frac{\pi}{2} \)) = i₀ sin(ωt + \( \frac{\pi}{2} \))
where i₀ = \( \frac{e_0}{(1/\omega C)} \) is the peak value of the current.
Table gives the values of e and i for different values of cot and Fig shows graphs of e and i versus ωt. i leads e by phase angle of \( \frac{\pi}{2} \) rad.

📝 Teacher's Note: Help students understand that in a capacitor, current flows to charge/discharge the plates, so current must flow before voltage can build up.

🎯 Exam Tip: Include both the mathematical derivation and the graph to show the π/2 phase lead of current over voltage.

Question. If the effective current in a 50 cycle AC circuit is 5 A, what is the peak value of current? What is the current \( \frac{1}{600} \) sec.after if was zero?
Answer:
Data : f = 50 Hz, \( I_{rms} \) = 5 A, t = \( \frac{1}{600} \) s
The peak value of the current,
\( I_0 = I_{rms} \sqrt{2} = (5)(1.414) = 7.07 A \)
\( i = I_0 \sin (2\pi ft) \)
\( = 7.07 \sin [2\pi(50) \times \frac{1}{600}] \)
\( = 7.07 \sin \left(\frac{\pi}{6}\right) = (7.07)(0.5) \)
\( = 3.535 A \)
This is the required current.

📝 Teacher's Note: Use the relation between RMS and peak values to first find peak current, then apply the instantaneous current formula. Emphasize that we must convert time and frequency to radians for the sine function.

🎯 Exam Tip: Always convert the time-frequency product to radians before evaluating sine functions. Check that \( 2\pi ft = \frac{\pi}{6} \) gives \( \sin(\frac{\pi}{6}) = 0.5 \).

Question. A light bulb is rated 100W for 220 V AC supply of 50 Hz. Calculate (a) resistance of the bulb. (b) the rms current through the bulb.
Answer:
Data: Power (\( V_{rms} I_{rms} \)) = 100 W, \( V_{rms} \) = 220V, f = 50 Hz
The rms current through the bulb,
\( I_{rms} = \frac{\text{power}}{V_{rms}} = \frac{100}{220} = 0.4545 A \)
The resistance of the bulb,
\( R = \frac{V_{rms}}{I_{rms}} = \frac{220}{(100/220)} = (22) \times (22) = 484 \Omega \)

📝 Teacher's Note: Remind students that for resistive circuits, power = VI, and Ohm's law applies directly. The frequency doesn't affect resistance calculations for pure resistors.

🎯 Exam Tip: Use \( P = \frac{V^2}{R} \) or \( P = VI \) consistently. Double-check units: power in watts, voltage in volts, current in amperes, resistance in ohms.

Question. A 15.0 µF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what will happen to the capacitive reactance and the current.
Answer:
Data : C = 15 µF = 15 × 10⁻⁶ F, \( V_{rms} \) = 220V, f = 50 Hz,
The capacitive reactance = \( \frac{1}{2\pi fC} \)
\( = \frac{1}{2(3.142)(50)(15 \times 10^{-6})} = \frac{100 \times 100}{(3.142)(15)} \)
\( = 212.2\Omega \)
\( I_{rms} = \frac{V_{rms}}{\text{capacitive reactance}} = \frac{220}{212.2} \)
\( = 1.037 A \)
\( I_{peak} = I_{rms} \sqrt{2} = (1.037)(1.414) = 1.466 A \)
If the frequency is doubled, the capacitive reactance will be halved and the current will be doubled.

📝 Teacher's Note: Emphasize the inverse relationship between frequency and capacitive reactance: \( X_C = \frac{1}{2\pi fC} \). As frequency increases, capacitive reactance decreases, allowing more current to flow.

🎯 Exam Tip: Remember that capacitors "like" high frequencies (low reactance) while inductors "like" low frequencies. State the frequency effect clearly when asked.

Question. An AC circuit consists of only an inductor of inductance 2 H. If the current is represented by a sine wave of amplitude 0.25 A and frequency 60 Hz, calculate the effective potential difference across the inductor (\( \pi = 3.142 \))
Answer:
Data : L = 2H, \( I_0 \) = 0.25 A, f = 60 Hz, \( \pi = 3.142 \)
\( \omega L = 2\pi fL = 2(3.142)(60)(2) = 754.1 \Omega \)
The effective potential difference across the inductor = \( \omega L I_{rms} = \omega L \frac{I_0}{\sqrt{2}} \)
\( = \frac{(754.1)(0.25)}{1.414} = 133.3 V \)

📝 Teacher's Note: Students often confuse peak and RMS values. Always convert peak current to RMS using \( I_{rms} = \frac{I_0}{\sqrt{2}} \) before calculating voltage across reactive components.

🎯 Exam Tip: Write down the inductive reactance \( X_L = \omega L = 2\pi fL \) first, then use \( V_{rms} = X_L I_{rms} \). Don't forget the \( \sqrt{2} \) conversion factor.

Question. Alternating emf of e = 220 sin 100 πt is applied to a circuit containing an inductance of \( \frac{1}{\pi} \) henry. Write an equation for instantaneous current through the circuit. What will be the reading of the AC galvanometer connected in the circuit?
Answer:
Data: e = 220 sin 100 πt, L = \( \frac{1}{\pi} \)H
Comparing e = 220 sin 100 πt with e = \( e_0 \sin \omega t \), we get
\( \omega = 100 \pi \)
\( \omega L = (100 \pi) \times \frac{1}{\pi} = 100 \Omega \)
The instantaneous current through the circuit
\( i = \frac{e_0}{\omega L} \sin(100 \pi t - \frac{\pi}{2}) \)
\( = \frac{220}{100} \sin (100 \pi t - \frac{\pi}{2}) = 2.2 \sin (100 \pi t - \frac{\pi}{2}) \) in ampere [assuming that e is in volt.]
\( I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{2.2}{1.414} = 1.556 A \) is the reading of the AC galvanometer connected in the circuit.

📝 Teacher's Note: In a pure inductor, current lags voltage by 90°. The amplitude of current is \( \frac{e_0}{\omega L} \), and an AC galvanometer reads RMS values, not peak values.

🎯 Exam Tip: For inductors, current always lags by \( \frac{\pi}{2} \). AC instruments measure RMS values, so divide peak current by \( \sqrt{2} \) for the galvanometer reading.

Question. A 25 µF capacitor, a 0.10 H inductor and a 25Ω resistor are connected in series with an AC source whose emf is given by e = 310 sin 314 t (volt). What is the frequency, reactance, impedance, current and phase angle of the circuit?
Answer:
Data: C = 25 µF = 25 × 10⁻⁶ F, L = 0.10H, R = 25 Ω, e = 310 sin (314 t) [volt]
Comparing e = 310 sin (314 t) with e = \( e_0 \sin (2\pi ft) \), we get,
the frequency of the alternating emf as
\( f = \frac{314}{2\pi} = \frac{314}{2(3.14)} = 50 Hz \)
Reactance = \( |\omega L - \frac{1}{\omega C}| = |2\pi fL - \frac{1}{2\pi fC}| \)
\( = |2(3.14)(50)(0.10) - \frac{1}{2(3.14)(50)(25 \times 10^{-6})}| \)
\( = |31.4 - \frac{100 \times 10^2}{(3.14)(25)}| = |31.4 - 127.4| \)
\( = 96\Omega \)
\( Z^2 = R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2 = (25)^2 + (96)^2 = 9841 \Omega^2 \)
Impedance, \( Z = \sqrt{9841} \Omega = 99.2\Omega \)
Peak current, \( I_0 = \frac{e_0}{Z} = \frac{310}{99.2} A \)
\( I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{310}{(1.414)(99.2)} A = 2.21 A \)
\( \cos \Phi = \frac{R}{Z} = \frac{25}{99.2} = 0.2520 \)
The phase angle, \( \Phi = \cos^{-1}(0.2520) = 75.40° = 1.316 \) rad

📝 Teacher's Note: This is a comprehensive LCR circuit problem. Calculate \( X_L \) and \( X_C \) separately first, then find net reactance \( |X_L - X_C| \). Use the impedance triangle to find Z and phase angle.

🎯 Exam Tip: Always check whether \( X_L > X_C \) or \( X_C > X_L \) to determine if the circuit is inductive or capacitive. The net reactance determines the phase relationship.

Question. A capacitor of 100 µF, a coil of resistance 50Ω and an inductance 0.5 H are connected in series with a 110 V-50Hz source. Calculate the rms value of current in the circuit.
Answer:
Data : C = 100 µF = 100 × 10⁻⁶ F = 10⁻⁴ F, R = 50 Ω, L = 0.5H, f = 50 Hz, \( V_{rms} \) = 110 V
\( \omega L = 2\pi fL = 2 (3.142)(50)(0.5) = 157.1 \Omega \)
\( \frac{1}{\omega C} = \frac{1}{2\pi fC} = \frac{1}{2(3.142)(50)(10^{-4})} = \frac{100 \times 100}{3.142} = 31.83\Omega \)
\( Z^2 = R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2 = (50)^2 + (157.1 - 31.83)^2 \)
2500 + 15700 = 18200 Ω²
Impedance, \( Z = \sqrt{18200} \Omega = 134.9 \Omega \)
The rms value of the current in the circuit,
\( I_{rms} = \frac{V_{rms}}{Z} = \frac{110}{134.9} A \)
= 0.8154 A

📝 Teacher's Note: Calculate inductive and capacitive reactances separately, then find the net reactance. Since \( X_L > X_C \), the circuit is inductive overall.

🎯 Exam Tip: Organize calculations step by step: find \( X_L \), find \( X_C \), calculate net reactance, find impedance Z, then apply Ohm's law for AC circuits.

Question. Find the capacity of a capacitor which when put in series with a 10Ω resistor makes the power factor equal to 0.5. Assume an 80V-100Hz AC supply.
Answer:
Data : R = 10 Ω, power factor = 0.5, f = 100 Hz
Power factor = \( \frac{1}{2\pi fCR} \)
\( 0.5 = \frac{1}{2(3.142)(100)C(10)} \)
\( C = \frac{1}{3.142 \times 10^3} = \frac{10 \times 10^{-4}}{3.142} \)
\( = 3.182 \times 10^{-4} F \)
This is the capacity of the capacitor.

📝 Teacher's Note: For an RC circuit, power factor = \( \cos\phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_C^2}} \). This can be simplified when we know the power factor value.

🎯 Exam Tip: Use the relationship: if power factor = 0.5, then \( \frac{R}{Z} = 0.5 \), which means \( Z = 2R \). This helps find the reactance quickly.

Question. Find the time required for a 50 Hz alternating current to change its value from zero to the rms value.
Answer:
Data : f = 50 Hz, \( i = \frac{i_0}{\sqrt{2}} \), \( \frac{i}{i_0} = \frac{1}{\sqrt{2}} \)
\( i = i_0 \sin\omega t \)
\( \sin\omega t = \frac{i}{i_0} = \frac{1}{\sqrt{2}} \)
\( \omega t = \frac{\pi}{4} \) rad
\( 2\pi ft = \frac{\pi}{4} \)
\( t = \frac{1}{8f} = \frac{1}{8(50)} = \frac{1}{400} \)
\( = \frac{1000 \times 10^{-3}}{400} = 2.5 \times 10^{-3} s \)
This is the required time.

📝 Teacher's Note: The RMS value equals \( \frac{\text{peak}}{\sqrt{2}} \), so \( \sin\omega t = \frac{1}{\sqrt{2}} \) gives \( \omega t = \frac{\pi}{4} \). This occurs at 1/8 of the period.

🎯 Exam Tip: Remember that \( \sin 45° = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \). The time for 45° phase = \( \frac{T}{8} = \frac{1}{8f} \).

Question. Calculate the value of capacitance in picofarad, which will make 101.4 micro henry inductance to oscillate with frequency of one megahertz.
Answer:
Data : \( f_r = 10^6 Hz \), L = 101.4 × 10⁻⁶ H
\( f_r = \frac{1}{2\pi\sqrt{LC}} \)
\( f_r^2 = \frac{1}{4\pi^2 LC} \)
\( C = \frac{1}{4\pi^2 f_r^2 L} = \frac{1}{4(3.142)^2(10^6)^2(101.4 \times 10^{-6})} \)
\( = \frac{10000 \times 10^{-10}}{4(3.142)^2(101.4)} = 2.497 \times 10^{-10} F \)
\( = 249.7 \times 10^{-12} F = 249.7 \) picofarad
This is the value of the capacity.

📝 Teacher's Note: For LC oscillations, resonant frequency \( f_r = \frac{1}{2\pi\sqrt{LC}} \). This is the natural frequency at which the circuit oscillates when energy transfers between electric and magnetic fields.

🎯 Exam Tip: Remember the units: 1 picofarad = 10⁻¹² F. Rearrange the resonant frequency formula to solve for the unknown component (C in this case).

Question. A 10 µF capacitor is charged to a 25 volt of potential. The battery is disconnected and a pure 100 m H coil is connected across the capacitor so that LC oscillations are set up. Calculate the maximum current in the coil.
Answer:
Data: C = 10 µF = 10 × 10⁻⁶ F = 10⁻⁵ F, L = 100mH = 100 × 10⁻³ H = 10⁻¹ H, V = 25V
For reference, see the solved example (8) above.
\( \frac{1}{2}CV^2 = \frac{1}{2}Li^2 \)
\( i^2 = \frac{C}{L}V^2 \)
\( i = \sqrt{\frac{C}{L}}V = \sqrt{\frac{10^{-5}}{10^{-1}}} \times 25 \)
\( i = 25 \times 10^{-2} A = 0.25 A \)
This is the maximum current in the coil.

📝 Teacher's Note: In LC oscillations, energy conservation applies: initial electric energy in the capacitor converts completely to magnetic energy in the inductor at maximum current. Use \( \frac{1}{2}CV^2 = \frac{1}{2}Li_{max}^2 \).

🎯 Exam Tip: Maximum current occurs when all the electric energy is converted to magnetic energy. Set the two energy expressions equal and solve for current.

Question. A 100 µF capacitor is charged with a 50 V source supply. Then source supply is removed and the capacitor is connected across an inductance, as a result of which 5A current flows through the inductance. Calculate the value of the inductance.
Answer:
Data: C = 100 µF = 100 × 10⁻⁶ F = 10⁻⁴ F, V = 50V, i = 5A
The energy stored in the electric field in the capacitor
\( = \frac{1}{2}CV^2 \)
The energy stored in the magnetic field in the inductor = \( \frac{1}{2}Li^2 \)
Here, \( \frac{1}{2}CV^2 = \frac{1}{2}Li^2 \)
\( L = C\frac{V^2}{i^2} \)
\( L = C\left(\frac{V}{i}\right)^2 = 10^{-4} \times \left(\frac{50}{5}\right)^2 \)
\( = 10^{-4} \times 10^2 \)
\( = 10^{-2} H \)
This is the value of the inductance.

📝 Teacher's Note: This is another energy conservation problem in LC circuits. When the capacitor discharges through the inductor, all electric energy converts to magnetic energy at the moment of maximum current.

🎯 Exam Tip: Use energy conservation: \( U_C = U_L \) when all energy transfers from capacitor to inductor. This gives \( \frac{1}{2}CV^2 = \frac{1}{2}Li^2 \), so \( L = C\left(\frac{V}{i}\right)^2 \).

Question 23. A 100 µF capacitor is charged with a 50 V source supply. Then source supply is removed and the capacitor is connected across an inductance, as a result of which 5A current flows through the inductance. Calculate the value of the inductance.
Answer:
Data: C = 100 µF = 100 × 10⁻⁶ F = 10⁻⁴ F,
V = 50V, i = 5A
The energy stored in the electric field in the capacitor
= \( \frac{1}{2} \)CV²
The energy stored in the magnetic field in the inductor = \( \frac{1}{2} \)Li²
Here, \( \frac{1}{2} \)CV² = \( \frac{1}{2} \)Li²
\( \implies \) L = C \( \frac{V^2}{i^2} \)
\( \implies \) L = C \( \left(\frac{V}{i}\right)^2 \) = 10⁻⁴ \( \left(\frac{50}{5}\right)^2 \) = 10⁻⁴ × 10²
= 10⁻²H
This is the value of the inductance.
In simple words: The energy stored in the capacitor transfers to the inductor, so we equate both energies and solve for inductance.

📝 Teacher's Note: Emphasize energy conservation in LC oscillations. Show students that electrical energy becomes magnetic energy and vice versa, like a pendulum converting kinetic to potential energy.

🎯 Exam Tip: Always write the energy conservation equation first: ½CV² = ½Li². This immediately gives you the relationship between all variables.