NEET Physics: Magnetism, Electromagnetic Induction and AC · Chapter 4

Alternating Current

What to remember

An AC voltage v = v₀ sin ωt gives rms values V_rms = V₀/√2 and I_rms = I₀/√2; household meters and ratings use rms values.
Inductive reactance X_L = ωL (current lags voltage by 90°); capacitive reactance X_C = 1/(ωC) (current leads voltage by 90°). In a series LCR circuit Z = √(R² + (X_L − X_C)²) and tan φ = (X_L − X_C)/R.
Resonance occurs at ω₀ = 1/√(LC): Z = R (minimum), current is maximum and φ = 0. The quality factor is Q = ω₀L/R.
Average power P = V_rms I_rms cos φ. Power factor cos φ = R/Z. A transformer works only on AC: V_s/V_p = N_s/N_p.

AC voltage applied to a resistor

An AC source gives v = v₀ sin ωt, where v₀ is the peak (maximum) value and ω = 2πν. Here ν is the frequency and T = 1/ν is the time period. In India, mains supply is 50 Hz.

For a resistor R: i = (v₀/R) sin ωt = i₀ sin ωt. Current and voltage are in phase.

The average of i over a full cycle is zero, but the heating effect depends on i², whose average is i₀²/2. So the average power is P = i₀²R/2 = I_rms²R. The root-mean-square (rms) value is

I_rms = I₀/√2 ≈ 0.707 I₀, V_rms = V₀/√2 ≈ 0.707 V₀

The rms value is that of a steady direct current that would produce the same heat in the same resistor. A mains supply of 220 V rms has a peak value of 220√2 ≈ 311 V. AC ammeters and voltmeters read rms values.

Phasors (in words). A phasor is a rotating arrow (vector) whose length is the peak value of the quantity and which turns anticlockwise at angular speed ω. The projection of the arrow on the vertical axis gives the instantaneous value. For a resistor, the voltage and current arrows point in the same direction. For other elements, the two arrows are separated by a fixed angle, which is the phase difference.

AC voltage applied to an inductor

For an inductor of inductance L: v = v₀ sin ωt. The self-induced emf opposes the change, so v = L di/dt. Integrating,

i = i₀ sin(ωt − π/2), with i₀ = v₀/(ωL)

The current lags the voltage by 90° (a quarter of a cycle). The opposition to the current is the inductive reactance

X_L = ωL = 2πνL, unit ohm (Ω)

X_L increases with frequency. For direct current (ν = 0), X_L = 0, so an inductor passes DC freely but blocks high-frequency AC (a choke). The average power taken by an ideal inductor over a cycle is zero. Energy is stored in the first quarter of the cycle and returned in the next.

AC voltage applied to a capacitor

For a capacitor C: charge q = Cv, so i = dq/dt = C dv/dt. Then

i = i₀ sin(ωt + π/2), with i₀ = ωC v₀

The current leads the voltage by 90°. The capacitive reactance is

X_C = 1/(ωC) = 1/(2πνC), unit ohm (Ω)

X_C decreases as frequency increases. For DC (ν = 0), X_C is infinite, so a capacitor blocks DC and passes high-frequency AC. The average power taken by an ideal capacitor over a cycle is zero.

Element	Reactance / resistance	Phase of current relative to voltage	Average power
Resistor	R	in phase (0°)	I_rms²R
Inductor	X_L = ωL	lags by 90°	zero
Capacitor	X_C = 1/(ωC)	leads by 90°	zero

A memory aid for the phase: in an inductor, voltage leads; in a capacitor, current leads.

Series LCR circuit

A resistor R, an inductor L and a capacitor C are joined in series to the source v = v₀ sin ωt. The same current flows through all three. The peak voltages are V_R = i₀R, V_L = i₀X_L and V_C = i₀X_C.

On the phasor diagram, V_L points 90° ahead of the current, V_C points 90° behind it, and V_R is along the current. So V_L and V_C are opposite, and the resultant is |V_L − V_C|. Then

v₀ = i₀ √(R² + (X_L − X_C)²)

Impedance: Z = √(R² + (X_L − X_C)²), and i₀ = v₀/Z.

Phase angle: tan φ = (X_L − X_C)/R, where φ is the angle by which voltage leads current.

Condition	Nature of the circuit	Phase
X_L > X_C	inductive	current lags voltage (φ positive)
X_L < X_C	capacitive	current leads voltage (φ negative)
X_L = X_C	purely resistive (resonance)	φ = 0

Resonance. When X_L = X_C, i.e. ωL = 1/(ωC), the angular frequency is

ω₀ = 1/√(LC), ν₀ = 1/(2π√(LC))

At resonance: Z = R is minimum; the current I = V/R is maximum; the voltage and current are in phase; the power factor is 1. The voltages across L and C are equal and opposite, so they cancel, although each can be much larger than the source voltage.

Sharpness of resonance and quality factor. A graph of current amplitude against frequency has a peak at ω₀. A narrow, tall peak is called a sharp resonance. The quality factor is

Q = ω₀L/R = 1/(ω₀CR) = (1/R)√(L/C)

Q is a pure number. The half-width of the resonance curve is about R/2L on each side, so the bandwidth is Δω = 2(R/2L) = R/L (between the points where the current falls to 1/√2 of its peak), and Q = ω₀/Δω. A larger Q means a smaller R for the same L, so the resonance is sharper and the circuit is more selective. Radio tuning uses this idea: the capacitor is varied until ω₀ matches the frequency of the wanted station.

Power in AC circuits

The instantaneous power is p = v i. Over a full cycle, the average power is

P = V_rms I_rms cos φ = I_rms² R

cos φ is the power factor, and cos φ = R/Z for a series LCR circuit.

Pure resistor: φ = 0, cos φ = 1, power is maximum.
Pure inductor or pure capacitor: φ = 90°, cos φ = 0, average power is zero.
At resonance, cos φ = 1.

Wattless current. In a circuit with φ = 90°, the current flows but no net power is used. This current is called wattless (idle) current. In general, the component of current I_rms sin φ, at 90° to the voltage, is the wattless component, and I_rms cos φ is the power component.

A low power factor is a disadvantage: for the same power and voltage, a larger current is needed, which causes more loss in the supply lines.

LC oscillations

A capacitor with charge q₀ is connected to an inductor. The charge flows, and the circuit oscillates.

At the start, all energy is in the electric field of the capacitor: U_E = q₀²/(2C). Current is zero.
As the capacitor discharges, current rises and energy moves to the inductor: U_B = ½LI².
When the capacitor is fully discharged, the current is maximum and all the energy is magnetic: ½LI₀².
The current then recharges the capacitor with opposite polarity, and the energy returns to the electric field.

The total energy q₀²/(2C) = ½LI₀² is constant if there is no resistance. The charge varies as q = q₀ cos ω₀t, with

ω₀ = 1/√(LC), ν = 1/(2π√(LC))

This is similar to a mass on a spring: L plays the role of mass and 1/C the role of the spring constant. Energy swaps between the two forms every quarter cycle, so the electric energy has twice the frequency of the charge oscillation. In real circuits, resistance damps the oscillation.

Transformers

Principle: mutual induction. A transformer has two coils, a primary and a secondary, wound on a common laminated soft-iron core. An alternating voltage across the primary produces a changing flux in the core. The same flux links the secondary and induces an emf in it. It works only with AC.

For an ideal transformer (no flux leakage, no losses), with N_p and N_s turns:

V_s/V_p = N_s/N_p = I_p/I_s

N_s > N_p: step-up transformer; voltage rises, current falls.
N_s < N_p: step-down transformer; voltage falls, current rises.
Power in = power out: V_p I_p = V_s I_s.

Efficiency: η = output power/input power = V_s I_s/(V_p I_p). Real transformers have efficiency of 90 % or more.

Loss	Cause	Reduction
Copper loss (I²R heating)	Resistance of windings	Thick copper wire of low resistance
Eddy-current loss	Induced currents in the core	Laminated core
Hysteresis loss	Repeated magnetisation and demagnetisation of the core	Core of soft iron with narrow hysteresis loop
Flux leakage	Not all flux of the primary links the secondary	Winding the coils over each other on a closed core
Humming (vibration)	Magnetostriction of the core	Firm clamping of the core

Power transmission. The power P is sent at high voltage so that the current I = P/V is small and the line loss I²R is small. At the power station, a step-up transformer raises the voltage to many kilovolts. Near the consumer, step-down transformers reduce it to the safe supply voltage (220 V). Transmitting at ten times the voltage cuts the line loss to one hundredth for the same power.

Worked examples

Example 1. A 220 V rms, 50 Hz supply is applied to an inductor of 0.1/π H. X_L = 2π × 50 × 0.1/π = 10 Ω. I_rms = 220/10 = 22 A. The current lags by 90°.

Example 2. In a series circuit, R = 30 Ω, X_L = 70 Ω, X_C = 30 Ω across 100 V rms. Z = √(30² + 40²) = 50 Ω. I = 100/50 = 2 A. cos φ = R/Z = 0.6. P = 100 × 2 × 0.6 = 120 W. Check: I²R = 4 × 30 = 120 W.

Example 3. L = 0.5 H, C = 20 μF, R = 10 Ω. ω₀ = 1/√(0.5 × 20 × 10⁻⁶) = 1/√(10⁻⁵) ≈ 316 rad s⁻¹. Q = ω₀L/R = 316 × 0.5/10 ≈ 15.8.

Example 4. A step-down transformer has 1000 turns in the primary and 50 in the secondary. Primary voltage is 2200 V. V_s = 2200 × 50/1000 = 110 V. If the secondary current is 20 A, the ideal primary current is 20 × 50/1000 = 1 A.

Example 5. A 200 V rms source is applied to a capacitor of reactance 50 Ω. I_rms = 4 A, the current leads the voltage by 90°, and the average power is zero.

Formula summary

Quantity	Formula	Units
rms value	I_rms = I₀/√2	A
Inductive reactance	X_L = ωL	Ω
Capacitive reactance	X_C = 1/(ωC)	Ω
Impedance (series LCR)	Z = √(R² + (X_L − X_C)²)	Ω
Phase angle	tan φ = (X_L − X_C)/R	none
Resonant frequency	ω₀ = 1/√(LC)	rad s⁻¹
Quality factor	Q = ω₀L/R	none
Average power	P = V_rms I_rms cos φ	W
Power factor	cos φ = R/Z	none
LC frequency	ν = 1/(2π√(LC))	Hz
Transformer	V_s/V_p = N_s/N_p = I_p/I_s	none

NEET traps

Values of 220 V and 5 A on AC meters are rms values. The peak value is √2 times larger.
The average of a sine wave over a full cycle is zero, but its rms value is not zero.
Inductor: current lags the voltage. Capacitor: current leads the voltage. Students often swap these.
An inductor opposes AC more at high frequency; a capacitor opposes AC more at low frequency.
Impedance is not R + X_L + X_C. The reactances are subtracted, because they are 180° apart.
At resonance, the impedance is minimum (equal to R), not zero and not maximum.
At resonance, the voltage across L or C can be much larger than the source voltage, though the sum of V_L and V_C is zero.
A pure inductor or capacitor draws current but no average power; the current is wattless.
A transformer cannot work on DC, because DC gives no changing flux. The secondary voltage would be zero.
In a step-up transformer, the secondary current is smaller than the primary current. Power is not stepped up.

One-liners

1. Mains frequency in India is 50 Hz.
2. I_rms = I₀/√2 ≈ 0.707 I₀.
3. In a resistor, current and voltage are in phase.
4. Inductive reactance X_L = ωL; the current lags by 90°.
5. Capacitive reactance X_C = 1/(ωC); the current leads by 90°.
6. An inductor blocks high-frequency AC; a capacitor blocks DC.
7. Series LCR impedance Z = √(R² + (X_L − X_C)²).
8. Resonant angular frequency ω₀ = 1/√(LC).
9. At resonance the current is maximum and the power factor is 1.
10. Quality factor Q = ω₀L/R; a larger Q means a sharper resonance.
11. Average power P = V_rms I_rms cos φ.
12. Power factor of a pure L or C is zero.
13. In LC oscillations energy exchanges between the electric and magnetic fields.
14. A transformer works on mutual induction and only with AC.
15. High-voltage transmission reduces the I²R loss in the lines.

Practice questions

The rms value of a household AC supply is 220 V. Taking √2 = 1.41, the peak value is about
1. 156 V
2. 220 V
3. 440 V
4. 310 V
Answer
D. 310 V
V₀ = √2 × V_rms = 1.41 × 220 ≈ 310 V.
A current i = 4 sin(100πt) ampere flows through a 10 Ω resistor. What is the average power dissipated?
1. 80 W
2. 113 W
3. 160 W
4. 40 W
Answer
A. 80 W
I_rms² = 16/2 = 8 A²; P = 8 × 10 = 80 W.
An alternating voltage is given by v = 200 sin(100πt) volt. What is its frequency?
1. 314 Hz
2. 50 Hz
3. 25 Hz
4. 100 Hz
Answer
B. 50 Hz
ω = 100π = 2πν, so ν = 50 Hz.
An inductor of inductance (1/π) H is connected to a 220 V rms, 50 Hz supply. What is the rms current?
1. 1.1 A
2. 22 A
3. 2.2 A
4. 4.4 A
Answer
C. 2.2 A
X_L = 2π × 50 × (1/π) = 100 Ω; I = 220/100 = 2.2 A.
A capacitor of capacitance (100/π) μF is connected to a 50 Hz AC source. What is its capacitive reactance?
1. 10 Ω
2. 50 Ω
3. 100 Ω
4. 200 Ω
Answer
C. 100 Ω
X_C = 1/(2πνC) = 1/(2π × 50 × 10⁻⁴/π) = 1/(10⁻²) = 100 Ω.
A series LCR circuit has L = 10 mH and C = 1 μF. What is its resonant angular frequency?
1. 5.0 × 10³ rad s⁻¹
2. 1.0 × 10⁴ rad s⁻¹
3. 1.0 × 10⁸ rad s⁻¹
4. 1.0 × 10⁵ rad s⁻¹
Answer
B. 1.0 × 10⁴ rad s⁻¹
ω₀ = 1/√(LC) = 1/√(10⁻⁸) = 10⁴ rad s⁻¹.
In a series LCR circuit, R = 6 Ω, X_L = 12 Ω and X_C = 4 Ω. What is the impedance?
1. 22 Ω
2. 17.1 Ω
3. 14 Ω
4. 10 Ω
Answer
D. 10 Ω
Z = √(36 + 64) = 10 Ω.
In a series LCR circuit, R = 30 Ω, X_L = 70 Ω and X_C = 30 Ω. What is the power factor?
1. 0.6
2. 0.75
3. 0.8
4. 0.43
Answer
A. 0.6
Z = √(900 + 1600) = 50 Ω; cos φ = R/Z = 0.6.
A series LCR circuit has L = 0.1 H, C = 10 μF and R = 5 Ω. What is its quality factor?
1. 40
2. 0.05
3. 20
4. 200
Answer
C. 20
ω₀ = 1/√(10⁻⁶) = 1000 rad s⁻¹; Q = ω₀L/R = 1000 × 0.1/5 = 20.
A transformer has 1000 turns in the primary and 50 turns in the secondary. The primary is connected to 2200 V AC. What is the secondary voltage (ideal transformer)?
1. 110 V
2. 55 V
3. 2150 V
4. 44000 V
Answer
A. 110 V
V_s = V_p × N_s/N_p = 2200 × 50/1000 = 110 V.
An ideal transformer steps 110 V up to 2200 V. The secondary current is 0.5 A. What is the primary current?
1. 20 A
2. 10 A
3. 0.5 A
4. 0.025 A
Answer
B. 10 A
Power in = power out: 110 × I_p = 2200 × 0.5, so I_p = 10 A.
In a circuit containing only a capacitor, the alternating current
1. lags the voltage by 90°
2. leads the voltage by 45°
3. is in phase with the voltage
4. leads the voltage by 90°
Answer
D. leads the voltage by 90°
For a capacitor i = i₀ sin(ωt + π/2).
The power factor of a circuit containing only an ideal inductor is
1. one
2. zero
3. infinite
4. 0.5
Answer
B. zero
φ = 90°, so cos φ = 0.
In an LC circuit, the inductance is made four times while the capacitance is unchanged. The frequency of oscillation becomes
1. half
2. double
3. four times
4. one-fourth
Answer
A. half
ν = 1/(2π√(LC)) ∝ 1/√L; with 4L the frequency is halved.
A transformer takes 2 A at 220 V in the primary and delivers 18 A at 22 V in the secondary. What is its efficiency?
1. 110 %
2. 82 %
3. 90 %
4. 99 %
Answer
C. 90 %
Output = 22 × 18 = 396 W; input = 220 × 2 = 440 W; η = 396/440 = 90 %.
At resonance, the impedance of a series LCR circuit is
1. equal to X_L + X_C
2. zero
3. equal to R and is the maximum value
4. equal to R and is the minimum value
Answer
D. equal to R and is the minimum value
With X_L = X_C the reactive terms cancel, so Z = R.
Given below are two statements: Statement I: The average value of an alternating current over a full cycle is I₀/√2. Statement II: The rms value of an alternating current is I₀/√2. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
The average over a full cycle is zero; the rms value is I₀/√2.
Given below are two statements: Statement I: The inductive reactance increases with the frequency of the source. Statement II: The capacitive reactance decreases with the frequency of the source. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
X_L = ωL rises with ω; X_C = 1/(ωC) falls with ω.
Given below are two statements: Statement I: In a capacitor, the current leads the voltage by 90°. Statement II: In an inductor, the current leads the voltage by 90°. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
In an inductor the current lags by 90°.
Given below are two statements: Statement I: The impedance of a series LCR circuit is Z = R + X_L + X_C. Statement II: At resonance the impedance of a series LCR circuit is maximum. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
Z = √(R² + (X_L − X_C)²), and at resonance Z = R, its minimum.
Given below are two statements: Statement I: At resonance the power factor of a series LCR circuit is one. Statement II: At resonance X_L = X_C. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
With X_L = X_C, Z = R and cos φ = R/Z = 1.
Given below are two statements: Statement I: A transformer works on the principle of mutual induction. Statement II: A transformer works equally well with direct current. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
DC gives no changing flux, so no emf is induced in the secondary.
Given below are two statements: Statement I: An ideal inductor consumes power on average over a full cycle. Statement II: In an ideal inductor, the energy stored in one quarter of a cycle is returned in the next quarter. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
The average power is zero because the energy is returned.
Given below are two statements: Statement I: A series LCR circuit with a high quality factor has a broad resonance curve. Statement II: The quality factor Q = ω₀L/R decreases when R decreases. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
High Q means a sharp curve, and Q rises when R falls.
Which of the following statements are correct? 1. The rms value equals the peak value divided by √2. 2. A capacitor has infinite reactance for direct current. 3. An inductor has infinite reactance for direct current.
1. 1 and 3 only
2. 1 and 2 only
3. 2 and 3 only
4. 1, 2 and 3
Answer
B. 1 and 2 only
For DC, ν = 0, so X_L = 0 and X_C is infinite.
Which of the following statements are correct? 1. A step-up transformer raises the voltage and lowers the current. 2. A transformer works with direct current. 3. A soft iron core reduces hysteresis loss.
1. 1 and 3 only
2. 1 and 2 only
3. 2 and 3 only
4. 1, 2 and 3
Answer
A. 1 and 3 only
A transformer needs AC, so statement 2 is wrong.
Assertion (A): A capacitor blocks direct current. Reason (R): The capacitive reactance X_C = 1/(ωC) becomes infinite for ω = 0.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
For DC the reactance is infinite, so no steady current flows.
Assertion (A): At resonance in a series LCR circuit, the voltage across the inductor can be larger than the source voltage. Reason (R): At resonance the impedance of the circuit is maximum.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
R is false: Z = R is the minimum at resonance.
Assertion (A): Electrical power is transmitted at high voltage. Reason (R): A transformer works on the principle of mutual induction.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both are true, but mutual induction does not explain why high voltage is used.
Assertion (A): A transformer can be used to step up a direct voltage. Reason (R): A transformer needs a changing magnetic flux in its core to induce a secondary emf.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
A is false: DC produces no changing flux.
Assertion (A): The core of a transformer is made of laminated sheets. Reason (R): Soft iron has a narrow hysteresis loop.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both true; lamination cuts eddy currents, which the hysteresis loop does not explain.
Assertion (A): Transmitting power at high voltage reduces the loss in the lines. Reason (R): For a given power, a higher voltage means a smaller current, so I²R is smaller.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Loss = I²R = (P/V)²R.
Match List I with List II. List I: a. Pure resistor b. Pure inductor c. Pure capacitor d. Series LCR with X_L > X_C List II: i. Current leads voltage by 90° ii. Current in phase with voltage iii. Current lags voltage by 90° iv. Current lags voltage by less than 90°
1. a-iii, b-ii, c-i, d-iv
2. a-ii, b-iii, c-iv, d-i
3. a-ii, b-iii, c-i, d-iv
4. a-ii, b-i, c-iii, d-iv
Answer
C. a-ii, b-iii, c-i, d-iv
Resistor in phase; inductor lags 90°; capacitor leads 90°; an inductive circuit lags by φ < 90°.
Match List I with List II. List I: a. Inductive reactance b. Capacitive reactance c. Impedance of series LCR d. Resonant angular frequency List II: i. 1/ωC ii. 1/√(LC) iii. √(R² + (X_L − X_C)²) iv. ωL
1. a-iv, b-i, c-ii, d-iii
2. a-iii, b-i, c-iv, d-ii
3. a-i, b-iv, c-iii, d-ii
4. a-iv, b-i, c-iii, d-ii
Answer
D. a-iv, b-i, c-iii, d-ii
Standard formulae.
Match List I with List II. List I: a. Quality factor b. Power factor c. Average power d. rms current List II: i. I₀/√2 ii. ω₀L/R iii. R/Z iv. V_rms I_rms cos φ
1. a-ii, b-iii, c-i, d-iv
2. a-iii, b-ii, c-iv, d-i
3. a-ii, b-iv, c-iii, d-i
4. a-ii, b-iii, c-iv, d-i
Answer
D. a-ii, b-iii, c-iv, d-i
Q = ω₀L/R; cos φ = R/Z; P = V_rms I_rms cos φ; I_rms = I₀/√2.
Match List I with List II. List I: a. X_L for L = 1/π H at 50 Hz b. X_C for C = 200/π μF at 50 Hz c. Z for R = 12 Ω, X_L − X_C = 5 Ω d. Z of a series LCR at resonance with R = 20 Ω List II: i. 13 Ω ii. 20 Ω iii. 50 Ω iv. 100 Ω
1. a-iii, b-iv, c-i, d-ii
2. a-iv, b-iii, c-ii, d-i
3. a-iv, b-iii, c-i, d-ii
4. a-iv, b-i, c-iii, d-ii
Answer
C. a-iv, b-iii, c-i, d-ii
a: 2π × 50/π = 100 Ω. b: 1/(100 × 2 × 10⁻⁴) = 50 Ω. c: √(144 + 25) = 13 Ω. d: Z = R = 20 Ω.
Match List I with List II. List I: a. Step-up transformer b. Step-down transformer c. Copper loss d. Eddy-current loss List II: i. Laminated core ii. N_s > N_p iii. N_s < N_p iv. Thick wire of low resistance
1. a-ii, b-iii, c-i, d-iv
2. a-ii, b-iii, c-iv, d-i
3. a-iii, b-ii, c-iv, d-i
4. a-ii, b-i, c-iv, d-iii
Answer
B. a-ii, b-iii, c-iv, d-i
N_s > N_p raises voltage. Copper loss falls with thick wire; eddy loss falls with lamination.
Match List I with List II. List I: a. Capacitor fully charged, current zero b. Capacitor discharged, current maximum c. Frequency of LC oscillations d. Mechanical analogue of L List II: i. Entire energy in the inductor ii. Entire energy in the capacitor iii. Mass iv. 1/(2π√(LC))
1. a-ii, b-iii, c-iv, d-i
2. a-i, b-ii, c-iv, d-iii
3. a-ii, b-i, c-iv, d-iii
4. a-ii, b-i, c-iii, d-iv
Answer
C. a-ii, b-i, c-iv, d-iii
Energy is electric at maximum charge and magnetic at maximum current; L corresponds to mass.
A series LCR circuit with R = 30 Ω, X_L = 70 Ω and X_C = 30 Ω is connected to a 100 V rms source. What average power is consumed?
1. 150 W
2. 200 W
3. 160 W
4. 120 W
Answer
D. 120 W
Z = 50 Ω; I = 2 A; P = I²R = 4 × 30 = 120 W.
A series LCR circuit has R = 40 Ω and is at resonance at 50 Hz, with X_L = X_C = 20 Ω. If the source frequency is doubled to 100 Hz, the impedance becomes
1. 50 Ω
2. 70 Ω
3. 40 Ω
4. 30 Ω
Answer
A. 50 Ω
At 100 Hz: X_L = 40 Ω and X_C = 10 Ω. Z = √(40² + 30²) = 50 Ω.
A power of 100 kW is sent through a line of total resistance 5 Ω at 10 kV. What is the power lost in the line?
1. 500 W
2. 20 MW
3. 50 kW
4. 5 kW
Answer
A. 500 W
I = P/V = 10 A; loss = I²R = 100 × 5 = 500 W.
A bulb is joined in series with a capacitor. The bulb will glow when the combination is connected to
1. a DC source of any voltage
2. neither AC nor DC
3. a DC source only if the capacitor is large
4. an AC source of suitable voltage
Answer
D. an AC source of suitable voltage
A capacitor blocks steady DC but passes AC.
A capacitor of 100 μF charged to 100 V is connected across an ideal coil of inductance 1 H. What is the maximum current in the coil?
1. 10 A
2. 1 A
3. 0.1 A
4. 0.5 A
Answer
B. 1 A
Energy: ½CV² = ½LI²; I = V√(C/L) = 100 × √(10⁻⁴) = 1 A.
A series LCR circuit has L = 0.4 H, C = 10 μF and R = 20 Ω. What is its quality factor?
1. 100
2. 25
3. 10
4. 0.1
Answer
C. 10
ω₀ = 1/√(4 × 10⁻⁶) = 500 rad s⁻¹; Q = 500 × 0.4/20 = 10.
A series LCR circuit has R = 10 Ω, X_L = 15 Ω and X_C = 5 Ω. The phase difference between the voltage and the current is
1. 45°, and the voltage leads the current
2. 45°, and the current leads the voltage
3. 30°, and the voltage leads the current
4. 60°, and the voltage leads the current
Answer
A. 45°, and the voltage leads the current
tan φ = (15 − 5)/10 = 1, so φ = 45°. X_L > X_C, so the current lags the voltage.

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1.Which of the following devices is used to convert alternating current into direct current?
2.Alternating current was invented by _______.
3.The charge flowing through a circuit in 10 minutes is 300 C. Find the current drawn by the circuit.
4.The correct relation between charge(Q) following through a conductor, time(t) for which the charge follows and the current (I) following through the conductor is:
5.The force experienced by a current carrying conductor in a magnetic field is maximum when the angle between the direction of current and the direction of magnetic field is
6.The formula of current I = _______ /time (t).
7.The generation of heat in a conductor due to electric current is known as:
8.The electric current flowing through a metallic wire is directly proportional to the potential difference V across its ends, provided its _____ remains the same.
9.The electric ______ between two points in an electric circuit carrying some current is the work done to move a unit charge from one point to the other.
10.The electric power consumed by a resistor of resistance R, carrying current I, is _______

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