CBSE Class 12 Physics Electromagnetic Induction Assignment

 Important Questions for NCERT Class 12 Physics Electromagnetic Induction

Question. A coil of copper having 1000 turns is placed in a magnetic field (B = 4 × 10^–5) perpendicular to its axis. The cross sectional area of the coil is 0.05 m². If it turns through 180° in 0.01 second, then the e.m.f. induced in the coil will be ; 
(a) 4 V
(b) 0.04 V
(c) 0.4 V
(d) 0.2 V

Answer :  C

Question. The current in self inductance L = 40 mH is to be increased uniformly from 1 amp to 11 amp in 4 milliseconds. The e.m.f. induced in inductor during process is
(a) 100 volt
(b) 0.4 volt
(c) 4.0 volt
(d) 440 volt 

Answer :  A

Question. An inductor may store energy in
(a) its electric field
(b) its coils
(c) its magnetic field
(d) both in electric and magnetic fields

Answer :  C

Question. In a coil of self inductance of 5 henry, the rate of change of current is 2 ampere per second, the e.m.f. induced in the coil is : 
(a) 5 V
(b) – 5V
(c) – 10 V
(d) 10 V

Answer :  C

Question. According to Lenz’s law there is a conversion of:
(a) momentum
(b) collision 
(c) voltage
(d) energy

Answer :  D

Question. A 50 turn circular coil has a radius of 3 cm, it is kept in a magnetic field acting normal to the area of the coil. The magnetic field B is increased from 0.10 T to 0.35 T in 2 milli second, the average induced emf will be: 
(a) 177 V
(b) 1.77 V
(c) 0.177 V
(d) 17.7 V

Answer :  B

Question. A solenoid is 1.5 m long and its inner diameter is 4.0 cm. It has 3 layers of windings of 1000 turns each and carries a current of 2.0 amperes. The magnetic flux for a cross-section of the solenoid is nearly 
(a) 4.1´10-5 weber
(b) 5.2´10-5 weber
(c) 6.31´10-3 weber
(d) 2.5´10-7 weber 

Answer :  C

Question. The current in self inductance L = 40 mH is to be increased uniformly from 1 amp to 11 amp in 4 milliseconds. The e.m.f. induced in inductor during process is   
(a) 100 volt
(b) 0.4 volt
(c) 4.0 volt
(d) 440 volt 

Answer :  A

Important Questions for NCERT Class 12 Physics Alternating Induction

Question. In an A.C. circuit, the current flowing is I = 5 sin (100t – p/2) ampere and the potential difference is V = 200 sin (100t) volts. The power consumption is equal to
(a) 20 W
(b) 0 W
(c) 1000 W
(d) 40 W

Answer   B

Question. A transistor-oscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C in series produces oscillations of frequency f. If L is doubled and C is changed to 4C, the frequency will be
(a) f/2
(b) f/4
(c) 8f
(d) f /2 2

Answer   D

Question. A transformer having efficiency of 90% is working on 200 V and 3 kW power supply. If the current in the secondary coil is 6 A, the voltage across the secondary coil and the current in the primary coil respectively are
(a) 300 V, 15 A
(b) 450 V, 15 A
(c) 450 V, 13.5 A
(d) 600 V, 15 A 

Answer   B

Question. In an ideal parallel LC circuit, the capacitor is charged by connecting it to a D.C. source which is then disconnected. The current in the circuit 
(a) becomes zero instantaneously 
(b) grows monotonically
(c) decays monotonically
(d) oscillates instantaneously

Answer   D

Question. With the decrease of current in the primary coil from 2 amperes to zero value in 0.01s the emf generated in the secondary coil is 1000 volts.
The mutual inductance of the two coils is C
(a) 1.25 H
(b) 2.50 H
(c) 5.00 H
(d) 10.00 H

Answer   C

Question. An AC source of angular frequency w is fed across a resistor R and a capacitor C in series.
The current registered is I. If now the frequency of source is changed to w/3 (but maintaining the same voltage), the current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency w A
(a) √3/5
(b) √2/5
(c) √1/5
(d) √4/5

Answer   A

Important Questions for NCERT Class 12 Physics Electromagnetic Induction

Question. The flux linked with a coil at any instant 't' is given by f = 10t2 – 50t + 250. The induced emf at t = 3s is 
(a) –190 V
(b) –10 V 
(c) 10 V
(d) 190 V
Answer :  B

Question. In an AC generator, a coil with N turns, all of the same area A and total resistance R, rotates with frequency w in a magnetic field B. The maximum value of emf generated in the coil is 
(a) N.A.B.R.w
(b) N.A.B.
(c) N.A.B.R.
(d) N.A.B.w
Answer :  D

Question. A conducting ring of radius 1 metre is placed in an uniform magnetic field B of 0.01 tesla oscillating with frequency 100 Hz with its plane at right angle to B. What will be the induced electric field ? 
(a) pvolt /m
(b) 2volt /m
(c) 10 volt/m
(d) 62 volt/m
Answer :  B

Question. A wheel with ten metallic spokes each 0.50m long is rotated with a speed of 120 rev/min in a plane normal to the earth’ s magnetic field at the place.
If the magnitude of the field is 0.40 G, the induced emf between the axle and the rim of the wheel is equal to 
(a) 1.256 × 10 ^{– 3} V
(b) 6.28 × 10 ^{– 4} V
(c) 1.256 × 10 ^{– 4} V
(d) 6.28 × 10 ^{– 5} V
Answer :  D

Question. The magnetic flux through a circuit carrying a current of 2.0 A is 0.8 weber. If the current reduces to 1.5 A in 0.1 s, the induced emf be :
(a) 2.0 V
(b) 4.0 V
(c) 8.0 V
(d) none of the above
Answer :  A

Question. A charged particle moves through a magnetic field in a direction perpendicular to it. Then the 
(a) velocity remains unchanged 
(b) speed of the particle remains unchanged
(c) direction of the particle remains unchanged
(d) acceleration remains unchanged
Answer :  B

Question. In an inductor of self-inductance L = 2 mH, current changes with time according to relation i = t2e–t. At what time emf is zero? 
(a) 4s
(b) 3s
(c) 2s
(d) 1s
Answer :  C

Question. A cycle wheel of radius 0.5 m is rotated with constant angular velocity of 10 rad/s in a region of magnetic field of 0.1 T which is perpendicular to the plane of the wheel. The EMF generated between its centre and the rim is    
(a) 0.25 V
(b) 0.125 V
(c) 0.5 V
(d) zero
Answer :  B

Important Questions for NCERT Class 12 Physics Alternating Induction

Question. For the circuit shown in the fig., the current through the inductor is 0.9 A while the current through the condenser is 0.4 A. Then 
(a) current drawn from generator I = 1.13 A
(b) w = 1/(1.5 L C)
(c) I = 0.5 A
(d) I = 0.6 A
Answer   C

Question. The primary of a transformer when connected to a dc battery of 10 volt draws a current of 1 mA. The number of turns of the primary and secondary windings are 50 and 100 respectively. The voltage in the secondary and the current drawn by the circuit in the secondary are respectively
(a) 20 V and 2.0 mA
(b) 10 V and 0.5 mA
(c) Zero volt and therefore no current
(d) 20 V and 0.5 mA 
Answer   C

Question. A 220 volt input is supplied to a transformer. The output circuit draws a current of 2.0 ampere at 440 volts. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is
(a) 3.6 ampere
(b) 2.8 ampere
(c) 2.5 ampere
(d) 5.0 ampere 
Answer   D

Question. In an A.C. circuit, Irms and I0 are related as D
(a) Irms = pI0
(b) Irms = 2 I0
(c) Irms = I0/p
(d) Irms = I0 / 2 
Answer   D

Question. The primary and secondary coils of a transformer have 50 and 1500 turns respectively. If the magnetic flux f linked with the primary coil is given by f = f0 + 4t, where f is in webers, t is time in seconds and f0 is a constant, the output voltage across the secondary coil is 
(a) 120 volts
(b) 220 volts
(c) 30 volts
(d) 90 volts 
Answer   A

Question. An ideal coil of 10H is connected in series with a resistance of 5W and a battery of 5V. After 2 sec the connection is made, the current flowing in ampere in the circuit is A
(a) (1 – e^–1)
(b) (1 – e)
(c) e
(d) e^–1
Answer   A

ASSERTION REASON TYPE QUESTIONS

Directions: These questions consist of two statements, each printed as Assertion and Reason. While answering these questions, you are required to choose any one of the following four responses.
(a) If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
(b) If both Assertion and Reason are correct but Reason is not a correct explanation of the
Assertion. (c) If the Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are incorrect.

Question. Assertion: Induced emf will always occur whenever there is change in magnetic flux.
Reason : Current always induces whenever there is change in magnetic flux.
Answer. C

Question. Assertion: Faraday’s laws are consequence of conservation of energy.
Reason : Self inductance of a coil depends on current flowing through the coil.
Answer. C

Question. Assertion : Only a change in magnetic flux will maintain an induced current in the coil.
Reason : The presence of large magnetic flux through a coil maintain a current in the coil if the circuit is continuous.
Answer. C

Question. Assertion: Lenz’s law violates the principle of conservation of energy.
Reason: Induced emf always opposes the change in magnetic flux responsible for its production.
Answer. A

Question. Assertion : In the phenomenon of mutual induction, self-induction of each of the coils persists.
Reason : Self -induction arises when strength of current in same coil changes. In mutual induction, current is changing in both the individual coils.
Answer. B

Question. Assertion : Acceleration of a magnet falling through a long solenoid decreases.
Reason : The induced current produced in a circuit always flow in such direction that it opposes the change to the cause that produced it.
Answer. A

Question. Assertion : An emf can be induced by moving a conductor in a magnetic field.
Reason : An emf can be induced by changing the magnetic field.
Answer. B

Question. Assertion : Eddy currents are produced in any metallic conductor when magnetic flux is changed around it.
Reason : Electric potential determines the flow of charge
Answer. B

CASE STUDY BASED QUESTION

1. Electromagnetic induction is defined as the production of an electromotive force across an electric conductor in the changing magnetic field. The discovery of induction was done by Michael Faraday in the year 1831. Electromagnetic induction finds many applications such as in electrical components which includes transformers, inductors, and other devices such as electric motors and generators. Alternating current is defined as an electric current which reverse in direction periodically. In most of the electric power circuits, the waveform of alternating current is the sine wave.

Question. How to increase the energy stored in an inductor by four times?
(a) By doubling the current
(b) This is not possible
(c) By doubling the inductance
(d) By making current 2 times
Answer. A

Question. Consider an inductor whose linear dimensions are tripled and the total number of turns per unit length is kept constant, what happens to the self-inductance?
(a) 9 times
(b) 3 times
(c) 27 times
(d) 13 times
Answer. B

Question. Lenz law is based on which of the following conservation
(a) Charge
(b) Mass
(c) Momentum
(d) Energy
Answer. D

Question. What will be the acceleration of the falling bar magnet which passes through the ring such that the ring is held horizontally and the bar magnet is dropped along the axis of the ring?
(a) It depends on the diameter of the ring and the length of the magnet
(b) It is equal due to gravity
(c) It is less than due to gravity
(d) It is more than due to gravity
Answer. C

2. Self Induction. When a current I flows through a coil, flux linked with it is φ = LI, where L is a constant known as self inductance of the coil. Any charge in current sets up an induced emf in the coil. Thus, self inductance of a coil is the induced emf set up in it when the current passing through it changes at the unit rate. It is a measure of the opposition to the growth or the decay of current flowing through the coil. Also, value of self inductance depends on the number of turns in the solenoid, its area of cross-section and the permeability of its core material.

Question. The inductance in a coil plays the same role as 
(a) inertia in mechanics
(b) energy in mechanics
(c) momentum in mechanics
(d) force in mechanics
Answer. A

Question. A current of 2.5 A flows through a coil of inductance 5 H. The magnetic flux linked with the coil is 
(a) 0.5 Wb
(b) 12.5 Wb
(c) zero
(d) 2 Wb
Answer. B

Question. The inductance L of a solenoid depends upon its radius R as
(a) L ∝ R
(b) L ∝ 1/R
(c) L ∝ R2
(d) L ∝ R3
Answer. C

Question. The unit of self-inductance is
(a) Weber ampere
(b) Weber-1 ampere
(c) Ohm second
(d) Farad
Answer. C

Question. The induced emf in a coil of 10 henry inductance in which current varies from 9 A to 4 A in 0.2 second is
(a) 200 V
(b) 250 V
(c) 300 V
(d) 350 V
Answer. B

NUMERICALS

Question. Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V is induced, calculate the self-induction of the circuit.
Answer. 4H

Question. A long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside normal to the axis of the solenoid. The current carried by the solenoid changes steadily from 2 A to 4 A in 0.1 s, what is the induced emf in the loop while the current is changing? 
Answer. M= 120X3.14x10^-9 H and e₂ = 7.5x 10^-6V

Question. An air cored solenoid with length 30 cm, area of cross-section 25 cm² and number of turns 500 carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10^-3 s. How much is the average back emf induced across the ends of the open switch in the circuit ? Ignore the variation in magnetic field near the ends of the solenoid.
Answer. 6.5V

Question. A rectangular loop of sides 8 cm × 2 cm with a small cut is stationary in a uniform magnetic field produced by an electromagnet. If the current feeding the electromagnet is gradually reduced so that the magnetic field decreases from its initial value of 0·3 T at the rate of 0·02 Ts^-1. If the cut is joined and the loop has a resistance of 1·6 Ω, how much power is dissipated by the loop as heat ? What is the source of this power?
Answer. 6.4 x 10^-10 W

Question. A 0.5 m long solenoid of 10 turns/cm has area of cross-section 1 cm² . Calculate the voltage induced across its ends if the current in the solenoid is changed from 1A to 2A in 0.1s.
Answer. 0.628mV

IELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS

TOPICS TO BE COVERED

Electromagnetic induction; Faraday's laws induced emf and current; Lenz's law, Eddy currents self and mutual inductance.

Need for displacement current Alternating currents, peak and rms value of altering current/voltage

Reactance and Impedance. Lc oscillations (qualitative treatment only).

LCR series circuit; Resaonance; Power in AC circuits, wattless current.

AC generator and transformer.

KEY POINTS

QUESTIONS

VERY SHORT ANSWER QUESTIONS (1 Mark)

Question 1. Why core of a transformer is laminated?
Answer: The core of a transformer is laminated to minimize energy losses due to eddy currents by increasing the electrical resistance of the core path.
In simple words: Lamination breaks up large eddy current loops, reducing heat wastage.

Exam Tip: Mention the term "eddy currents" and "heat loss" to secure full marks.

Question 2. What is the direction of induced currents in metal rings 1 and 2 seen from the top when current I in the wire is increasing steadily?
Answer: Ring 1: Clockwise
Ring 2: Anticlockwise
In simple words: According to Lenz's law, the rings oppose the increase in magnetic flux by creating opposing fields.

Exam Tip: Apply right-hand thumb rule and Lenz's law carefully.

Question 3. In which of the following cases will the mutual inductance be (i) minimum (ii) maximum?
Answer: (i) Minimum in case (a) where axes are perpendicular or coils are far apart.
(ii) Maximum in case (c) where coils are coaxially wound close together.
In simple words: Mutual inductance is highest when magnetic flux linkage is maximum and lowest when coupling is weak.

Exam Tip: Mutual inductance depends on orientation, distance, and core material.

Question 4. In a series L-C-R circuit, voltages across inductor, capacitor, and resistor are \(V_L\), \(V_C\) and \(V_R\) respectively. What is the phase difference between (i) \(V_L\) and \(V_R\) (ii) \(V_L\) and \(V_C\)?
Answer: (i) The phase difference between \(V_L\) and \(V_R\) is \(\frac{\pi}{2}\) radians.
(ii) The phase difference between \(V_L\) and \(V_C\) is \(\pi\) radians.
In simple words: Inductor voltage leads resistor voltage by 90 degrees, and opposes capacitor voltage by 180 degrees.

Exam Tip: Remember phasor relationships: \(V_L\) leads current by \(90^\circ\), \(V_C\) lags current by \(90^\circ\), and \(V_R\) is in phase with current.

Question 5. Why can't transformer be used to step up or step down dc voltage?
Answer: A transformer works on the principle of electromagnetic induction which requires a changing magnetic flux. Direct current (DC) produces a constant magnetic field, so no emf is induced in the secondary coil.
In simple words: DC does not change with time, hence no induced voltage in the secondary coil.

Exam Tip: Highlight that transformers require alternating magnetic flux.

Question 6. In an a.c. circuit, instantaneous voltage and current are \(V = 200 \sin 300 t\) volt and \(i = 8 \cos 300 t\) ampere respectively. What is the average power dissipated in the circuit?
Answer: Given \(V = 200 \sin 300 t\)
\(i = 8 \cos 300 t = 8 \sin\left(300 t + \frac{\pi}{2}\right)\)
Phase difference \(\phi = \frac{\pi}{2}\)
Average power \(P_{av} = V_{rms} I_{rms} \cos\left(\frac{\pi}{2}\right) = 0\)
In simple words: When voltage and current have a phase difference of 90 degrees, the circuit is purely reactive and average power is zero.

Exam Tip: Pure inductor or capacitor circuits consume zero average power (wattless current).

Question 7. Sketch a graph that shows change in reactance with frequency of a series LCR circuit.
Answer: The net reactance \(X = |X_L - X_C|\) varies parabolically with frequency, dropping to zero at the resonant frequency \(\nu_r\) and then increasing.
In simple words: Reactance decreases up to resonance and increases thereafter.

Exam Tip: Label axes clearly with reactance \(X\) on y-axis and frequency \(\nu\) on x-axis.

Question 8. A coil A is connected to an A.C. ammeter and another coil B to a source of alternating e.m.f. What will be the reading in ammeter if a copper plate is introduced between the coils as shown.
Answer: The reading of the ammeter will become zero.
In simple words: The copper plate shields the magnetic field due to eddy currents set up in it, cutting off flux linkage.

Exam Tip: Eddy currents in a metal sheet cause electromagnetic shielding.

Question 9. In a circuit instantaneously voltage and current are \(V = 150 \sin 314 t\) volt and \(i = 12 \cos 314 t\) ampere respectively. Is the nature of circuit is capacitive or inductive?
Answer: Current \(i = 12 \cos 314 t = 12 \sin\left(314 t + \frac{\pi}{2}\right)\)
Since current leads voltage by \(\frac{\pi}{2}\), the nature of the circuit is purely capacitive.
In simple words: Current leading voltage implies a capacitive circuit.

Exam Tip: CILE: Capacitor - Current leads EMF.

Question 10. In a series L-C-R circuit \(V_L = V_C \neq V_R\). What is the value of power factor?
Answer: Since \(V_L = V_C\), the circuit is in resonance.
At resonance, impedance \(Z = R\), so phase angle \(\phi = 0^\circ\).
Power factor \(\cos\phi = \cos 0^\circ = 1\).
In simple words: At resonance, voltage and current are in phase, making power factor unity.

Exam Tip: Resonance condition implies maximum current and unity power factor.

Question 11. In an inductor L, current passed \(I_0\) and energy stored in it is U. If the current is now reduced to \(I_0/2\), what will be the new energy stored in the inductor?
Answer: Energy stored in an inductor is given by \(U = \frac{1}{2} L I^2\)
When current becomes \(I' = \frac{I_0}{2}\),
\(U' = \frac{1}{2} L \left(\frac{I_0}{2}\right)^2 = \frac{1}{4} \left(\frac{1}{2} L I_0^2\right) = \frac{U}{4}\)
In simple words: Energy is proportional to the square of current, so halving current reduces energy to one-fourth.

Exam Tip: Always check whether energy depends linearly or quadratically on the variable.

Question 12. A rectangle loop \(a\,b\,c\,d\) of a conducting wire has been changed into a square loop \(a'\,b'\,c'\,d'\) as shown in figure. What is the direction of induced current in the loop?
Answer: Clockwise.
In simple words: As the loop area decreases, magnetic flux decreases, and current flows to oppose this decrease.

Exam Tip: Use Lenz's law correlating area changes with flux variations.

Question 13. Twelve wires of equal lengths are connected in the form of a skeleton of a cube, which is moving with a velocity \(\vec{V}\) in the direction of magnetic field \(\vec{B}\). Find the emf in each arm of the cube.
Answer: The emf in each arm of the cube is zero.
In simple words: Motional emf requires motion perpendicular to the magnetic field. Here motion is parallel to the field.

Exam Tip: Formula for motional emf is \(e = (\vec{v} \times \vec{B}) \cdot \vec{l}\). If \(\vec{v}\parallel\vec{B}\), emf is zero.

Question 14. Current versus frequency (\(I - \nu\)) graphs for two different series L-C-R circuits have been shown in adjoining diagram. \(R_1\) and \(R_2\) are resistances of the two circuits. Which one is greater - \(R_1\) or \(R_2\)?
Answer: \(R_1 > R_2\)
In simple words: Higher resistance results in a lower peak current at resonance due to greater damping.

Exam Tip: Peak current at resonance \(I_0 = \frac{V}{R}\). Sharper resonance means smaller resistance.

Question 15. Why do we prefer carbon brushes than copper in an a.c. generator?
Answer: Carbon brushes are self-lubricating, cause less wear and tear of the slip rings, and have high resistance to corrosion compared to copper.
In simple words: Carbon is softer, reduces friction, and prevents sparking.

Exam Tip: Mention low friction and corrosion-free nature.

Question 16. What are the values of capacitive and inductive reactance in a dc circuit?
Answer: For DC, frequency \(\nu = 0\).
Capacitive reactance \(X_C = \frac{1}{2\pi\nu C} = \infty\) (infinity)
Inductive reactance \(X_L = 2\pi\nu L = 0\)
In simple words: Capacitors block DC completely, while inductors offer zero resistance to steady DC.

Exam Tip: Remember \(X_C\) is inversely proportional to frequency, and \(X_L\) is directly proportional.

Question 17. Give the direction of the induced current in a coil mounted on an insulating stand when a bar magnet is quickly moved along the axis of the coil from one side to the other as shown in figure.
Answer: As the magnet approaches, the current is clockwise (looking from the magnet side). As it passes through and moves away, the current reverses to anticlockwise.
In simple words: Approaching north pole induces north polarity (anticlockwise facing magnet), and receding south pole induces opposite current.

Exam Tip: Lenz's law dictates that the induced current opposes the motion of the magnet.

Question 18. In figure, the arm PQ is moved from \(x = 0\) to \(x = 2b\) with constant speed V. Consider the magnet field as shown in figure. Write (i) direction of induced current in rod (ii) polarity induced across rod.
Answer: (i) Direction of induced current is from Q to P.
(ii) Polarity: End P becomes positive and end Q becomes negative.
In simple words: Using Fleming's Right Hand Rule, positive charges accumulate at end P.

Exam Tip: Apply Fleming's Right-Hand Rule for generator action.

Question 19. A wire moves with some speed perpendicular to a magnetic field. Why is emf induced across the rod?
Answer: Lorentz force acting on the free charge carriers of the conducting wire causes them to drift to one end, hence developing a potential difference across it.
In simple words: Magnetic force pushes electrons to one end, creating an emf.

Exam Tip: Mention magnetic Lorentz force \(F = q(v \times B)\).

Question 20. Predict the polarity of the capacitor in the situation described in the figure below.
Answer: Plate 'a' will be negative with respect to plate 'b'.
In simple words: Induced current flows in a direction that opposes the approaching north pole of the magnet.

Exam Tip: Use Lenz's law to determine induced current direction.

Question 21. A circular coil rotates about its vertical diameter in a uniform horizontal magnetic field. What is the average emf induced in the coil?
Answer: Zero.
In simple words: Over a complete rotation, positive and negative halves of the induced emf cancel out.

Exam Tip: Average value over one complete cycle of alternating quantity is always zero.

Question 22. Define RMS Value of Current.
Answer: The RMS (Root Mean Square) value of alternating current is defined as that value of steady direct current which would generate the same amount of heat in a given resistor in a given time as is generated by the alternating current over a full cycle.
In simple words: Effective heating value of AC equivalent to DC.

Exam Tip: Mathematically, \(I_{rms} = \frac{I_0}{\sqrt{2}}\).

SHORT ANSWER QUESTIONS (2 Marks)

Question 1. An ac source of rms voltage V is put across a series combination of an inductor L, capacitor C and a resistor R. If \(V_L\), \(V_C\) and \(V_R\) are the r ms voltage across L, C and R respectively then why is \(V \neq V_L + V_C + V_R\)? Write correct relation among \(V_L\), \(V_C\) and \(V_R\).
Answer: Voltages across L, C, and R are not in the same phase. \(V_L\) leads current by \(90^\circ\), \(V_C\) lags current by \(90^\circ\), and \(V_R\) is in phase with current.
Correct relation: \(V = \sqrt{V_R^2 + (V_L - V_C)^2}\).
In simple words: Voltages must be added vectorially, not algebraically, due to phase differences.

Exam Tip: Draw the phasor diagram to justify vector addition.

Question 2. A bar magnet is falling with some acceleration 'a' along the vertical axis of a coil as shown in fig. What will be the acceleration of the magnet (whether \(a > g\) or \(a < g\)) if (a) coil ends are not connected to each other? (b) coil ends are connected to each other?
Answer: (a) When ends are not connected, no current is induced, so acceleration \(a = g\).
(b) When ends are connected, induced current creates a magnetic field that opposes the fall (Lenz's law), so acceleration \(a < g\).
In simple words: Closed circuit induces a repulsive force slowing down the falling magnet.

Exam Tip: Connect current induction with opposing magnetic force.

Question 3. The series L-C-R circuit shown in fig. is in resonance state. What is the voltage across the inductor?
Answer: At resonance, \(X_L = X_C\).
Current \(I = \frac{V}{R}\).
Voltage across inductor \(V_L = I \cdot X_L = \frac{V}{R} \cdot \omega L\).
In simple words: Voltage across inductor equals current multiplied by inductive reactance.

Exam Tip: Note that \(V_L\) can be greater than the supply voltage \(V\) at resonance.

Question 4. The divisions marked on the scale of an a.c. ammeter are not equally spaced. Why?
Answer: An AC ammeter works on the thermal effect of electric current, where heat produced \(H \propto I^2\). Because of the square law dependence, scale divisions are crowded at the lower end and wider apart at the higher end.
In simple words: Deflection is proportional to the square of current, making the scale non-linear.

Exam Tip: Mention heating effect \(H \propto I^2\).

Question 5. Circuit shown here uses an airfield parallel plate capacitor. A mica sheet is now introduced between the plates of capacitor. Explain with reason the effect on brightness of the bulb B.
Answer: Introducing a mica sheet increases capacitance (\(C = K C_0\)). Capacitive reactance \(X_C = \frac{1}{\omega C}\) decreases. Consequently, total impedance \(Z\) decreases, current in the circuit increases, and the brightness of bulb B increases.
In simple words: Mica increases capacitance, lowering impedance and making the bulb glow brighter.

Exam Tip: Trace the chain: Dielectric increases \(C \rightarrow\) decreases \(X_C \rightarrow\) decreases \(Z \rightarrow\) increases \(I\).

Question 6. In the figure shown, coils P and Q are identical and moving apart with same velocity V. Induced currents in the coils are \(I_1\) and \(I_2\). Find \(I_1/I_2\).
Answer: Induced emf \(E = Blv\).
For coil P, length in field is \(b\), so \(E_1 = \frac{B v b}{R}\).
For coil Q, length cutting field is \(l\), so \(E_2 = \frac{B v l}{R}\).
Ratio \(\frac{I_1}{I_2} = \frac{b}{l}\).
In simple words: Induced current is proportional to the cutting length of the conductor.

Exam Tip: Use motional emf formula with the correct effective length.

Question 7. A \(1.5\,\mu\text{F}\) capacitor is charged to \(5\,\text{V}\). The charging battery is then disconnected, and a \(12\,\text{mH}\) coil is connected in series with the capacitor so that LC Oscillations occur. What is the maximum current in the coil? Assume that the circuit has no resistance.
Answer: By conservation of electromagnetic energy:
Maximum electrostatic energy = Maximum magnetic energy
\(\frac{1}{2} \frac{q_0^2}{C} = \frac{1}{2} L I_0^2\)
\(I_0 = q_0 \sqrt{\frac{1}{L C}} = (C V_0) \frac{1}{\sqrt{L C}} = V_0 \sqrt{\frac{C}{L}}\)
Given \(C = 1.5\,\mu\text{F} = 1.5 \times 10^{-6}\,\text{F}\), \(L = 12\,\text{mH} = 12 \times 10^{-3}\,\text{H}\), \(V_0 = 5\,\text{V}\).
\(I_0 = 5 \sqrt{\frac{1.5 \times 10^{-6}}{12 \times 10^{-3}}} = 5 \sqrt{1.25 \times 10^{-4}} = 5 \times 0.01118 \approx 63.7\,\text{mA}\).
In simple words: Equate maximum electrostatic energy of capacitor to maximum magnetic energy of inductor to find peak current.

Exam Tip: Substitute values carefully with proper powers of 10.

Question 8. The self inductance of the motor of an electric fan is \(10\,\text{H}\). What should be the capacitance of the capacitor to which it should be connected in order to impart maximum power at \(50\,\text{Hz}\)?
Answer: For maximum power, the circuit must be in resonance: \(X_L = X_C\)
\(\omega L = \frac{1}{\omega C} \implies C = \frac{1}{\omega^2 L} = \frac{1}{(2\pi\nu)^2 L}\)
\(C = \frac{1}{(2 \times 3.14 \times 50)^2 \times 10} = \frac{1}{(314)^2 \times 10} \approx 1.01 \times 10^{-6}\,\text{F} = 1\,\mu\text{F}\).
In simple words: Maximum power occurs at resonance when inductive and capacitive reactances cancel out.

Exam Tip: Use resonance condition formula \(\omega = \frac{1}{\sqrt{LC}}\).

Question 9. How does an inductor behave in a DC circuit after the current reaches to steady state? Justify.
Answer: An inductor behaves like a simple conducting wire (zero resistance / short circuit) in steady state DC.
Reason: In steady state, current is constant (\(\frac{dI}{dt} = 0\)), so induced emf \(e = -L\frac{dI}{dt} = 0\).
In simple words: Constant DC current produces no back emf, so the inductor acts as a plain wire.

Exam Tip: Highlight that steady current means zero rate of change of current.

Question 10. How does an inductor behave in an AC circuit at very high frequency?
Answer: At very high frequency, inductive reactance \(X_L = 2\pi\nu L\) becomes extremely large (approaching infinity). Thus, the inductor acts like an open circuit.
In simple words: High frequency AC is blocked by an inductor.

Exam Tip: Relate frequency directly to reactance \(X_L\).

Question 11. An electric bulb is connected in series with an inductor and an AC source. When switch is closed and after sometime an iron rod is inserted into the interior of inductor. How will the brightness of bulb be affected? Justify your answer.
Answer: The brightness of the bulb decreases.
Justification: Inserting an iron rod increases the magnetic permeability and self-inductance \(L\). This increases inductive reactance \(X_L\), total impedance \(Z\), and reduces the circuit current.
In simple words: Iron core increases inductance, reducing current and dimming the bulb.

Exam Tip: Core insertion increases inductance \(L\) and reactance \(X_L\).

Question 12. Show that in the free oscillation of an LC circuit, the sum of energies stored in the capacitor and the inductor is constant with time.
Answer: Total energy \(U = \frac{1}{2} L I^2 + \frac{q^2}{2 C}\)
Differentiating with respect to time \(t\):
\(\frac{dU}{dt} = \frac{1}{2} L (2I \frac{dI}{dt}) + \frac{1}{2C} (2q \frac{dq}{dt}) = I \left( L\frac{dI}{dt} + \frac{q}{C} \right)\)
From Kirchhoff's loop rule for LC circuit, \(L\frac{dI}{dt} + \frac{q}{C} = 0\), hence \(\frac{dU}{dt} = 0\), proving total energy is constant.
In simple words: Energy oscillates between electrical and magnetic forms without loss.

Exam Tip: Use time derivative and energy conservation principles.

Question 13. Show that the potential difference across the LC combination is zero at the resonating frequency in series LCR circuit
Answer: Potential difference across L is \(V_L = I X_L\).
Potential difference across C is \(V_C = I X_C\).
Net voltage across LC combination \(V_{LC} = V_L - V_C = I(X_L - X_C)\).
At resonance, \(X_L = X_C\), therefore \(V_{LC} = 0\).
In simple words: At resonance, inductive and capacitive voltages cancel out completely.

Exam Tip: Emphasize opposite phase relationship between \(V_L\) and \(V_C\).

Question 14. How does a capacitor behave in a DC circuit after the steady state? Explain your answer.
Answer: A capacitor acts as an open key in a DC circuit after reaching steady state.
Explanation: Once fully charged, no further current flows through the capacitor because the dielectric prevents charge transport across plates, and capacitive reactance is infinite for DC.
In simple words: Fully charged capacitor blocks direct current entirely.

Exam Tip: Steady state DC means \(X_C = \infty\).

Question 15. For circuits used for transmitting electric power, a low power factor implies large power loss in transmission. Explain.
Answer: Power transmitted is \(P = V I \cos\phi\), so current \(I = \frac{P}{V \cos\phi}\).
For a given power and voltage, if power factor \(\cos\phi\) is low, current \(I\) must be large. Power loss in transmission lines is given by \(I^2 R\), leading to massive heat losses.
In simple words: Low power factor demands higher current, increasing \(I^2 R\) line losses.

Exam Tip: Connect low power factor to high current and \(I^2 R\) heat loss.

Question 16. An applied Voltage signal consists of a superposition of DC Voltage and an AC Voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the DC signal will appear across C where as AC signal will appear across L.
Answer: For DC, frequency is zero, so capacitive reactance \(X_C = \infty\) (blocks DC) and inductive reactance \(X_L = 0\) (passes DC easily). Thus, DC appears across the capacitor C.
For high frequency AC, \(X_L\) is very high (blocks AC) and \(X_C\) is very low (passes AC). Thus, high frequency AC appears across the inductor L.
In simple words: Inductors block high frequencies and pass DC; capacitors block DC and pass high frequencies.

Exam Tip: Use frequency behavior of \(X_L\) and \(X_C\).

Question 17. A bar magnet M is dropped so that is falls vertically through the coil C. The graph obtained for voltage produced across the coil Vs time is shoin figure. (i) Explain the shape of the graph (ii) Why is the negative peak longer than the positive peak?
Answer: (i) As the magnet enters, flux increases inducing one polarity; as it exits, flux changes in opposite direction inducing reverse polarity.
(ii) The negative peak is longer and higher because the magnet accelerates as it falls under gravity, increasing the speed of flux change (\(\frac{d\phi}{dt}\)) upon exit.
In simple words: Faster exit velocity increases induced emf magnitude and shortens time duration.

Exam Tip: Acceleration due to gravity causes exit speed to exceed entry speed.

Question 18. What is the Significance of Q-factor in a series LCR resonant circuit?
Answer: Quality factor (Q-factor) measures the sharpness of resonance and selectivity of the circuit. Higher Q-factor indicates a sharper resonance curve and better tuning capability.
In simple words: High Q means better frequency selectivity and sharper tuning.

Exam Tip: Formula is \(Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 C R}\).

Question 19. How does mutual inductance of a pair of coils kept coaxially at a distance in air change when (i) the distance between the coils is increased? (ii) an iron rod is kept between them?
Answer: (i) When distance is increased, magnetic flux linkage decreases, so mutual inductance decreases.
(ii) When an iron rod is placed between them, magnetic permeability increases, enhancing flux linkage and thereby increasing mutual inductance.
In simple words: Spreading coils apart reduces mutual inductance; adding an iron core boosts it.

Exam Tip: Mutual inductance depends on geometry, separation, and core permeability.

SHORT ANSWER QUESTIONS (3 Marks)

Question 1. Obtain an expression for the self inductance of a straight solenoid of length l and radius r (\(l \gg r\)).
Answer: Consider a solenoid of length \(l\(, radius \(r\), total turns \(N\), and number of turns per unit length \(n = \frac{N}{l}\).
Magnetic field inside the solenoid is \(B = \mu_0 n I\).
Magnetic flux linked per turn = \(B \times \text{Area} = (\mu_0 n I) (\pi r^2)\).
Total flux linkage \(\Phi = N (\mu_0 n I \pi r^2) = (n l) \mu_0 n I \pi r^2 = \mu_0 n^2 l \pi r^2 I\).
Since \(\Phi = L I\), the self inductance is \(L = \mu_0 n^2 \pi r^2 l\).
In simple words: Derive by calculating magnetic field, total flux linkage, and equating to \(LI\).

Exam Tip: State Ampere's law assumption for uniform magnetic field inside long solenoids.

Question 2. Distinguish between : (i) resistance and reactance (ii) reactance and impedance.
Answer: (i) Resistance is the opposition offered by a conductor to both DC and AC flow, independent of frequency, and causes power dissipation. Reactance is opposition offered by inductance or capacitance to AC, dependent on frequency, with no net power dissipation.
(ii) Reactance is opposition offered solely by L or C. Impedance is the total effective opposition of an LCR circuit combining resistance and net reactance (\(Z = \sqrt{R^2 + (X_L - X_C)^2}\)).
In simple words: Resistance consumes power; reactance does not; impedance is the vector sum of both.

Exam Tip: Use clear comparative points: frequency dependence and power loss.

Question 3. In a series L-C-R circuit \(X_L\), \(X_C\) and R are the inductive reactance, capacitive reactance and resistance respectively at a certain frequency f. If the frequency of a.c. is doubled, what will be the values of reactances and resistance of the circuit?
Answer: Resistance \(R\) remains unchanged with frequency.
Inductive reactance \(X_L = 2\pi f L\). When frequency is doubled (\(2f\)), \(X_L' = 2X_L\) (doubled).
Capacitive reactance \(X_C = \frac{1}{2\pi f C}\). When frequency is doubled, \(X_C' = \frac{X_C}{2}\) (halved).
In simple words: Resistance stays same, inductive reactance doubles, and capacitive reactance halves.

Exam Tip: Recall \(X_L \propto f\) and \(X_C \propto \frac{1}{f}\).

Question 4. What are eddy currents? Write their any four applications.
Answer: Eddy currents are induced circulating electric currents set up in bulk conductors when subjected to a changing magnetic field.
Applications:
1. Electromagnetic damping in dead-beat galvanometers.
2. Induction furnaces for melting metals.
3. Magnetic braking in induction trains.
4. Speedometers of automobiles.
In simple words: Circulating currents induced in metal blocks by changing magnetic flux.

Exam Tip: Mention Foucault currents as another name and list practical utility.

Question 5. In a series L-R 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 \(P_2\). Find \(P_1/P_2\).
Answer: Initially, for L-R circuit: \(R = X_L\).
Impedance \(Z_1 = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + R^2} = \sqrt{2}R\).
Power factor \(P_1 = \cos\phi_1 = \frac{R}{Z_1} = \frac{R}{\sqrt{2}R} = \frac{1}{\sqrt{2}}\).
When capacitor is added such that \(X_L = X_C\), the circuit is at resonance: \(Z_2 = R\).
Power factor \(P_2 = \cos\phi_2 = \frac{R}{R} = 1\).
Ratio \(\frac{P_1}{P_2} = \frac{1/\sqrt{2}}{1} = \frac{1}{\sqrt{2}}\).
In simple words: Initial power factor is \(1/\sqrt{2}\) and becomes 1 at resonance.

Exam Tip: Power factor formula is \(\cos\phi = \frac{R}{Z}\).

Question 6. Instantaneous value of a.c. through an inductor L is \(e = e_0 \cos\omega t\). Obtain an expression for instantaneous current through the inductor. Also draw the phasor diagram.
Answer: Induced emf \(e = -L \frac{di}{dt} = e_0 \cos\omega t\)
\(di = \frac{e_0}{L} \cos\omega t \, dt\)
Integrating both sides: \(i = \frac{e_0}{L} \int \cos\omega t \, dt = \frac{e_0}{\omega L} \sin\omega t = i_0 \sin\omega t\)
where \(i_0 = \frac{e_0}{\omega L}\).
In simple words: Integrating the voltage equation gives a sine wave current lagging by 90 degrees.

Exam Tip: Inductor voltage leads current by \(90^\circ\).

Question 7. In an inductor of inductance L, current passing is \(I_0\). Derive an expression for energy stored in it. In what forms is this energy stored?
Answer: Small work done in establishing current against back emf: \(dw = -e \, i \, dt = (L \frac{di}{dt}) i \, dt = L i \, di\)
Total work done \(W = \int_0^{I_0} L i \, di = \frac{1}{2} L I_0^2\)
This energy is stored in the form of a magnetic field within the inductor.
In simple words: Energy is stored as a magnetic field inside the inductor coil.

Exam Tip: Integrate power over time to obtain energy formula.

Question 8. Which of the following curves may represent the reactance of a series LC combination.
Answer: [Ans. : (b)]
In simple words: Net reactance curve passes through zero at resonance frequency.

Exam Tip: Total reactance is \(X_L - X_C\), which changes sign at resonance.

Question 9. A sinusoidal e.m.f. device operates at amplitude \(E_0\) and frequency \(\nu\) across a purely (1) resistive (2) capacitive (3) inductive circuit. If the frequency of driving source is increased. How would (a) amplitude \(E_0\) and (b) amplitude \(I_0\) increase, decrease or remain same in each case?
Answer: (a) Amplitude \(E_0\) remains same in all cases as it depends on the source.
(b) Amplitude \(I_0\):
- Resistive circuit: \(I_0 = \frac{E_0}{R}\) remains same.
- Capacitive circuit: \(I_0 = \frac{E_0}{X_C} = E_0(2\pi\nu C)\) increases.
- Inductive circuit: \(I_0 = \frac{E_0}{X_L} = \frac{E_0}{2\pi\nu L}\) decreases.
In simple words: Source voltage amplitude stays constant; current amplitude changes based on frequency-dependent reactance.

Exam Tip: Check formulas for \(X_L\) and \(X_C\) against frequency changes.

Question 10. The figure shows, in (a) a sine curved \(\delta(t) = \sin\omega t\) and three other sinusoidal curves \(A(t)\), \(B(t)\) and \(C(t)\) each of the form \(\sin(\omega t - \phi)\). (a) Rank the three curves according to the value of \(\phi\), most positive first and most negative last (b) Which curve corresponds to which phase as in (b) of the figure? (c) which curve leads the others? [Ans. : (a) C, B, A; (b) 1, A; 2, B; (c) A]
Answer: (a) Ranking by phase \(\phi\): C, B, A.
(b) Curve correspondence matches 1, A; 2, B.
(c) Curve A leads the others.
In simple words: Phase comparison determines which sinusoidal curve leads or lags on the time axis.

Exam Tip: Observe zero-crossings and peaks to determine phase lag/lead.

Question 11. In an LC circuit, resistance of the circuit is negligible. If time period of oscillation is T them : (i) at what time is the energy stored completely electrical (ii) at what time is the energy stored completely magnetic (iii) at what time is the total energy shared equally between the inductor and capacitor.
Answer: (i) Electrical energy is completely stored at \(t = 0, \frac{T}{2}, \frac{3T}{2}, \dots\)
(ii) Magnetic energy is completely stored at \(t = \frac{T}{4}, \frac{3T}{4}, \frac{5T}{4}, \dots\)
(iii) Energy is shared equally at \(t = \frac{T}{8}, \frac{3T}{8}, \frac{5T}{8}, \dots\)
In simple words: Electrical and magnetic energies oscillate out of phase by a quarter period.

Exam Tip: Relate energy states to sinusoidal charge and current variations.

Question 12. An alternating voltage of frequency f is applied across a series LCR circuit. Let \(f_r\) be the resonance frequency for the circuit. Will the current in the circuit lag, lead or remain in phase with the applied voltage when (i) \(f > f_r\) (ii) \(f < f_r\)? Explain your answer in each case.
Answer: (i) When \(f > f_r\), \(X_L > X_C\), so \(V_L > V_C\) (\(V_L - V_C > 0\)). Current will lag the voltage.
(ii) When \(f < f_r\), \(X_L < X_C\), so \(V_L < V_C\) (\(V_L - V_C < 0\)). Current will lead the voltage.
In simple words: Higher frequency makes the circuit inductive (current lags); lower frequency makes it capacitive (current leads).

Exam Tip: Inductive dominance causes lagging current; capacitive dominance causes leading current.

Question 13. Figure (a), (b), (c) Show three alternating circuits withe equal currents. If frequency of alternating emf be increased, what will be the effect on current in the three cases? Explain.
Answer: (i) Circuit (a - Resistive): No effect, resistance is independent of frequency.
(ii) Circuit (b - Inductive): Current will decrease because inductive reactance \(X_L = 2\pi f L\) increases with frequency.
(iii) Circuit (c - Capacitive): Current will increase because capacitive reactance \(X_C = \frac{1}{2\pi f C}\) decreases with frequency.
In simple words: Resistor current is unaffected; inductor current drops; capacitor current rises with frequency.

Exam Tip: Recall frequency dependence of \(R\), \(X_L\), and \(X_C\).

LONG ANSWER QUESTIONS (5 Marks)

Question 1. Draw a labelled diagram to explain the principle and working of an a.c. generator. Deduce the expression for emf generated. Why cannot the current produced by an a.c. generator be measured with a moving coil ammeter?
Answer: Principle: Based on electromagnetic induction; when a coil rotates in a magnetic field, magnetic flux linked with it changes, inducing an emf.
Derivation: Let a coil of \(N\) turns and area \(A\) rotate with angular velocity \(\omega\) in magnetic field \(B\). Flux \(\phi = N B A \cos(\omega t)\). Induced emf \(e = -\frac{d\phi}{dt} = N B A \omega \sin(\omega t) = E_0 \sin(\omega t)\).
Reason: Moving coil ammeters measure average current over a full cycle, which is zero for AC.
In simple words: AC generator rotates a coil in a magnetic field to generate sinusoidal emf via Faraday's law.

Exam Tip: Clearly state Faraday's law and flux equation \(\Phi = NBA\cos\omega t\).

Question 2. Explain, with the help of a neat and labelled diagram, the principle, construction and working of a transformer.
Answer: Principle: Mutual induction between two coils.
Construction: Laminated soft iron core with primary and secondary coils wound on it.
Working: Alternating current in primary produces changing magnetic flux, which induces alternating emf in secondary.
Relation: \(\frac{E_s}{E_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s} = K\).
In simple words: Transformers step up or step down AC voltages using mutual induction.

Exam Tip: Mention ideal transformer assumption \(P_{in} = P_{out}\).

Question 3. An L-C circuit contains inductor of inductance L and capacitor of capacitance C with an initial charge \(q_0\). The resistance of the circuit is negligible. Let the instant the circuit is closed be \(t = 0\). (i) What is the total energy stored initially? (ii) What is the maximum current through inductor? (iii) What is frequency at which charge on the capacitor will oscillate? (iv) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?
Answer: (i) Initial total energy \(U = \frac{q_0^2}{2C}\).
(ii) Maximum current \(I_0 = \frac{q_0}{\sqrt{LC}}\).
(iii) Oscillation frequency \(\nu = \frac{1}{2\pi\sqrt{LC}}\).
(iv) Total initial energy \(\frac{q_0^2}{2C}\) is eventually dissipated as heat.
In simple words: Initial electrostatic energy converts to magnetic energy and eventually dissipates entirely as heat through the resistor.

Exam Tip: Use energy conservation principles for LC oscillations.

Question 4. An a.c. \(i = i_0 \sin\omega t\) is passed through a series combination of an inductor (L), a capacitor (C) and a resistor (R). Use the phasor diagram to obtain expressions for the (a) impedance of the circuit and phase angle between voltage across the combination and current passed in it. Hence show that (i) leads the voltage when \(\omega < \frac{1}{\sqrt{LC}}\) (ii) is in phase with voltage when \(\omega = \frac{1}{\sqrt{LC}}\).
Answer: Using phasor addition of voltages: \(V = \sqrt{V_R^2 + (V_L - V_C)^2} = I \sqrt{R^2 + (X_L - X_C)^2}\).
Impedance \(Z = \frac{V}{I} = \sqrt{R^2 + (X_L - X_C)^2}\).
Phase angle \(\tan\phi = \frac{X_L - X_C}{R}\).
(i) When \(\omega < \frac{1}{\sqrt{LC}}\), \(X_C > X_L\), so current leads voltage.
(ii) When \(\omega = \frac{1}{\sqrt{LC}}\), \(X_L = X_C\), so current is in phase with voltage.
In simple words: Phasor diagrams combine resistive and reactive voltages vectorially to find impedance and phase.

Exam Tip: Draw clear right-angled voltage triangles in phasor diagrams.

Question 5. Write two differences in each of resistance, reactance and impedance for an ac circuit. Derive an expression for power dissipated in series LCR circuit.
Answer: Differences:
- Resistance: Independent of frequency, causes active power loss.
- Reactance: Dependent on frequency, zero average power loss.
- Impedance: Total opposition combining R and X, determines peak AC flow.
Power Derivation: Instantaneous power \(P = V I = [V_0 \sin(\omega t)] [I_0 \sin(\omega t - \phi)]\).
Average power \(P_{av} = V_{rms} I_{rms} \cos\phi\).
In simple words: Resistance dissipates power; reactance stores and returns energy; impedance is the net AC opposition.

Exam Tip: Average power formula is \(P = VI\cos\phi\) where \(\cos\phi\) is the power factor.

NUMERICALS

Question 1. In a series C-R circuit, applied voltage is \(V = 110 \sin 314 t\) volt. What is the (i) The peak voltage (ii) Average voltage over half cycle?
Answer: (i) Peak voltage \(V_0 = 110\,\text{V}\)
(ii) Average voltage over half cycle \(V_{av} = \frac{2V_0}{\pi} = \frac{2 \times 110 \times 7}{22} = 70\,\text{V}\).
In simple words: Peak voltage is read directly from the sine equation; average voltage over half cycle is \(\frac{2V_0}{\pi}\).

Exam Tip: Remember average over full cycle is zero, but over half cycle it is \(\frac{2V_0}{\pi}\).

Question 2. Magnetic flux linked with each turn of a 25 turns coil is 6 milliweber. The flux is reduced to 1 mWb in 5 s. Find induced emf in the coil.
Answer: Given \(N = 25\), \(\phi_1 = 6\,\text{mWb} = 6 \times 10^{-3}\,\text{Wb}\), \(\phi_2 = 1\,\text{mWb} = 1 \times 10^{-3}\,\text{Wb}\), \(\Delta t = 5\,\text{s}\).
Induced emf \(E = -N \frac{\Delta\phi}{\Delta t} = -25 \frac{(1 - 6) \times 10^{-3}}{5} = -25 \frac{-5 \times 10^{-3}}{5} = 25 \times 10^{-3}\,\text{V} = 0.025\,\text{V}\).
In simple words: Apply Faraday's law \(E = -N \frac{\Delta\phi}{\Delta t}\).

Exam Tip: Watch out for units (milliwebers to weavers conversion).

Question 3. The current through an inductive circuit of inductance \(4\,\text{mH}\) is \(i = 12 \cos 300 t\) ampere. Calculate : (i) Reactance of the circuit. (ii) Peak voltage across the inductor.
Answer: Given \(L = 4\,\text{mH} = 4 \times 10^{-3}\,\text{H}\), \(\omega = 300\,\text{rad/s}\), \(I_0 = 12\,\text{A}\).
(i) Inductive reactance \(X_L = \omega L = 300 \times 4 \times 10^{-3} = 1.2\,\Omega\).
(ii) Peak voltage \(V_0 = I_0 X_L = 12 \times 1.2 = 14.4\,\text{V}\).
In simple words: Multiply peak current by inductive reactance to find peak voltage.

Exam Tip: Formula for inductive reactance is \(X_L = \omega L\).

Question 4. A power transmission line feeds input power at \(2400\,\text{V}\) to a step down ideal transformer having 4000 turns in its primary. What should be number of turns in its secondary to get power output at \(240\,\text{V}\)?
Answer: Given \(V_p = 2400\,\text{V}\), \(V_s = 240\,\text{V}\), \(N_p = 4000\).
Using transformation ratio: \(\frac{V_s}{V_p} = \frac{N_s}{N_p}\)
\(N_s = N_p \left(\frac{V_s}{V_p}\right) = 4000 \times \frac{240}{2400} = 400\,\text{turns}\).
In simple words: Voltage ratio equals turns ratio for an ideal transformer.

Exam Tip: Step-down transformer has \(V_s < V_p\) and \(N_s < N_p\).

Question 5. The magnetic flux linked with a closed circuit of resistance \(8\,\Omega\) varies with time according to the expression \(\phi = (5t^2 - 4t + 2)\) where \(\phi\) is in milliweber and t in second. Calculate the value of induced current at \(t = 15\,\text{s}\).
Answer: \(\phi = 5t^2 - 4t + 2\,\text{mWb} = (5t^2 - 4t + 2) \times 10^{-3}\,\text{Wb}\)
Induced emf \(E = -\frac{d\phi}{dt} = -(10t - 4) \times 10^{-3}\,\text{V}\)
At \(t = 15\,\text{s}\): \(E = -(10(15) - 4) \times 10^{-3} = -146 \times 10^{-3}\,\text{V} = -0.146\,\text{V}\).
Induced current \(I = \frac{|E|}{R} = \frac{0.146}{8} = 0.01825\,\text{A}\).
In simple words: Differentiate flux equation with respect to time to find emf, then divide by resistance.

Exam Tip: Be careful with milli prefix (\(10^{-3}\)) during differentiation.

Question 6. A capacitor, a resistor and \(\frac{4}{\pi}\) henry inductor are connected in series to an a.c. source of \(50\,\text{Hz}\). Calculate capacitance of capacitor if the current is in phase with voltage.
Answer: Current in phase with voltage means resonance condition: \(X_L = X_C\)
\(2\pi f L = \frac{1}{2\pi f C} \implies C = \frac{1}{4\pi^2 f^2 L}\)
Given \(L = \frac{4}{\pi}\,\text{H}\), \(f = 50\,\text{Hz}\).
\(C = \frac{1}{4 \pi^2 (50)^2 (\frac{4}{\pi})} = \frac{1}{40000 \pi \pi^2 \dots} = \frac{1}{40000 \times \pi \times \frac{4}{\pi}} = \frac{1}{160000} = 6.25 \times 10^{-6}\,\text{F} = 6.25\,\mu\text{F}\).
In simple words: Equate inductive and capacitive reactances at resonance to solve for capacitance.

Exam Tip: Resonance formula is \(\omega^2 = \frac{1}{LC}\).

Question 7. A series C-R circuit consists of a capacitance \(16\,\text{mF}\) and resistance \(8\,\Omega\). If the input a.c. voltage is \((200\,\text{V}, 50\,\text{Hz})\), calculate (i) voltage across capacitor and resistor. (ii) Phase by which voltage lags/leads current.
Answer: Given \(C = 16\,\text{mF} = 16 \times 10^{-3}\,\text{F}\), \(R = 8\,\Omega\), \(V = 200\,\text{V}\), \(f = 50\,\text{Hz}\).
Capacitive reactance \(X_C = \frac{1}{2\pi f C} = \frac{1}{2 \times 3.14 \times 50 \times 16 \times 10^{-3}} = \frac{1}{5.024} \approx 0.199\,\Omega\).
Impedance \(Z = \sqrt{R^2 + X_C^2} = \sqrt{8^2 + (0.199)^2} \approx 8\,\Omega\).
Current \(I = \frac{V}{Z} = \frac{200}{8} = 25\,\text{A}\).
(i) \(V_C = I X_C = 25 \times 0.199 = 4.975\,\text{V}\); \(V_R = I R = 25 \times 8 = 200\,\text{V}\).
(ii) Phase angle \(\tan\phi = \frac{X_C}{R} = \frac{0.199}{8} = 0.0249 \implies \phi \approx 1.43^\circ\) (voltage lags current).
In simple words: Calculate impedance, current, individual component voltages, and phase angle.

Exam Tip: In a RC circuit, voltage lags current by phase angle \(\phi\).

Question 8. A rectangular conducting loop of length l and breadth b enters a uniform magnetic field B as shown below. The loop is moving at constant speed v and at \(t = 0\) it just enters the field B. Sketch the following graphs for the time interval \(t = 0\) to \(t = \frac{3l}{v}\): (i) Magnetic flux - time (ii) Induced emf - time (iii) Power - time. Resistance of the loop is R.
Answer: (i) Magnetic flux increases linearly from \(t = 0\) to \(t = \frac{l}{v}\), stays constant till \(t = \frac{2l}{v}\), and decreases linearly to zero at \(t = \frac{3l}{v}\).
(ii) Induced emf is constant positive during entry, zero inside, and constant negative during exit.
(iii) Power \(P = \frac{E^2}{R}\) is constant non-zero during entry and exit, and zero inside.
In simple words: Flux changes only when entering or leaving the magnetic field.

Exam Tip: Graph shapes correspond to flux slopes during boundary crossings.

Question 9. A charged \(8\,\text{mF}\) capacitor having charge \(5\,\text{mC}\) is connected to a \(5\,\text{mH}\) inductor. What is : (i) the frequency of current oscillations? (ii) the frequency of electrical energy oscillations in the capacitor? (iii) the maximum current in the inductor? (iv) the magnetic energy in the inductor at the instant when charge on capacitor is \(4\,\text{mC}\)?
Answer: Given \(C = 8\,\mu\text{F}\) (assuming \(\mu\text{F}\)), \(q_0 = 5\,\text{mC}\), \(L = 5\,\text{mH}\).
(i) Frequency \(\nu = \frac{1}{2\pi\sqrt{LC}}\).
(ii) Frequency of electrical energy oscillation \(\nu_E = 2\nu\).
(iii) Maximum current \(I_0 = \frac{q_0}{\sqrt{LC}}\).
(iv) Magnetic energy \(U_B = U_{total} - U_E = \frac{q_0^2}{2C} - \frac{q^2}{2C} = \frac{5^2 - 4^2}{2C} \times 10^{-6} = \frac{9}{2 \times 8 \times 10^{-6}} \dots\)
In simple words: Energy oscillates at twice the frequency of current.

Exam Tip: Energy oscillates twice per cycle of current.

Question 10. A \(31.4\,\Omega\) resistor and \(0.1\,\text{H}\) inductor are connected in series to a \(200\,\text{V}\), \(50\,\text{Hz}\) ac source. Calculate (i) the current in the circuit (ii) the voltage (rms) across the inductor and the resistor. (iii) Is the algebraic sum of voltages across inductor and resistor more than the source voltage? If yes, resolve the paradox.
Answer: Given \(R = 31.4\,\Omega\), \(L = 0.1\,\text{H}\), \(V = 200\,\text{V}\), \(f = 50\,\text{Hz}\).
\(X_L = 2\pi f L = 2 \times 3.14 \times 50 \times 0.1 = 31.4\,\Omega\).
Impedance \(Z = \sqrt{R^2 + X_L^2} = \sqrt{(31.4)^2 + (31.4)^2} = 31.4\sqrt{2} \approx 44.4\,\Omega\).
(i) Current \(I = \frac{V}{Z} = \frac{200}{44.4} = 4.5\,\text{A}\).
(ii) \(V_L = I X_L = 4.5 \times 31.4 = 141.3\,\text{V}\); \(V_R = I R = 4.5 \times 31.4 = 141.3\,\text{V}\).
(iii) Yes, algebraic sum (\(141.3 + 141.3 = 282.6\,\text{V}\)) exceeds source voltage because \(V_L\) and \(V_R\) are out of phase by \(90^\circ\) and must be added vectorially: \(V = \sqrt{V_R^2 + V_L^2} = 200\,\text{V}\).
In simple words: Vector addition resolves the apparent voltage sum paradox.

Exam Tip: Phasor addition accounts for phase differences in AC circuits.

Question 11. A square loop of side \(12\,\text{cm}\) with its sides parallel to X and Y-axis is moved with a velocity of \(8\,\text{cm/s}\) in positive x-direction. Magnetic field exists in z-directions. (i) Determine the direction and magnitude of induced current if field changes with \(10^{-3}\,\text{Tesla/cm}\) along negative x-direction. (ii) Determine the direction and magnitude of induced current if field changes with \(10^{-3}\,\text{Tesla/s}\).
Answer: (i) Rate of change of flux = induced emf \(= (0.12)^2 \times 10^{-3} \times 8 = 11.52 \times 10^{-5}\,\text{Wb/s}\).
(ii) Rate of change of flux = induced emf \(= (0.12)^2 \times 10^{-3} = 1.44 \times 10^{-5}\,\text{Wb/s}\).
In simple words: Calculate rate of change of magnetic flux to find induced emf.

Exam Tip: Use \(\frac{d\Phi}{dt} = A \frac{dB}{dt}\) or spatial field gradient multiplied by velocity.

HOTS

VERY SHORT ANSWER QUESTIONS (1 Mark)

Question 1. A conducting rod PQ is in motion at speed v in uniform magnetic field as shown in Fig. What are the polarities at P and Q?
Answer: P is negative (-), Q is positive (+).
In simple words: Use Fleming's Right-Hand Rule to determine charge accumulation.

Exam Tip: Apply Lorentz force on free electrons inside the rod.

Question 2. A long straight wire with current i passes (without touching) three square wire loops with edge lengths 2L, 1.5L land L. The loops are widely spaced (so as to not affect one another). Loops 1 and 3 are symmetric about the long wire. Rank the loops according to the size of the current induced in them if current i is (a) constant and (b) increasing greatest first.
Answer: (a) Constant current: Zero induced current in all loops.
(b) Increasing current: Loop 3 > Loop 2 > Loop 1 (or based on proximity and area flux changes).
In simple words: Steady current induces no emf; increasing current induces emf proportional to flux change rate.

Exam Tip: Faraday's law requires changing flux for induced current.

Question 3. In an L-C circuit, current is oscillating with frequency \(4 \times 10^6\,\text{Hz}\). What is the frequency with which magnetic energy is oscillating?
Answer: Frequency of magnetic energy oscillation is twice the current oscillation frequency: \(\nu_m = 2\nu = 2 \times 4 \times 10^6 = 8 \times 10^6\,\text{Hz}\).
In simple words: Energy peaks twice per complete cycle of current oscillation.

Exam Tip: Energy oscillates at double the electrical frequency.

SHORT ANSWER QUESTIONS (2 Marks)

Question 1. Two circular conductors are perpendicular to each other as shown in figure. If the current is changed in conductor B, will a current be induced in the conductor A?
Answer: No current is induced in coil A since angle \(\theta = 90^\circ\), meaning magnetic flux linked with coil A due to coil B remains zero.
In simple words: Perpendicular magnetic fields produce zero flux linkage across mutually orthogonal coils.

Exam Tip: Flux linkage is proportional to \(\cos\theta\); at \(90^\circ\), flux is zero.

NUMERICALS

Question 1. Figure shows a wire ab of length l which can slide on a U-shaped rail of negligible resistance. The resistance of the wire is R. The wire is pulled to the right with a constant speed v. Draw an equivalent circuit diagram representing the induced emf of a battery. Find the current in the wire using this diagram.
Answer: Equivalent circuit consists of an emf source \(\varepsilon = B l v\) connected in series with resistance \(R\).
Current in the wire \(I = \frac{\varepsilon}{R} = \frac{B l v}{R}\).
In simple words: Moving conductor acts like a battery of emf \(Blv\) connected across resistance \(R\).

Exam Tip: Treat motional emf as a source voltage in the equivalent circuit.