Alternating Current - Practice Questions with Answers
82 free MCQs on Alternating Current with worked answers and explanations. AC circuits, impedance, LCR series circuit, resonance, transformers, and power factor.
Below are 82 practice questions on Alternating Current, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Alternating Current notes.
In a pure inductor the current lags the voltage by 90° (energy is briefly stored in the magnetic field before current responds); in a pure capacitor the current leads the voltage by 90° (current must flow to build up charge before voltage rises) - the mnemonic "ELI the ICE man" keeps these straight.
Easy - 23 questions
Q1.
In an AC circuit, the current and voltage are:
A Always in phase
B Always out of phase
C In phase for pure resistor only
D Depends on frequency only
Show answer & explanation
Answer: C. In phase for pure resistor only
Why: In a pure resistor, V and I are in phase. In inductors or capacitors, they are out of phase.
Q2.
A transformer that increases voltage is called:
A Step-down transformer
B Rectifier
C Step-up transformer
D Inductor
Show answer & explanation
Answer: C. Step-up transformer
Why: Step-up transformer: more secondary turns than primary. Vs/Vp = Ns/Np > 1, so secondary voltage is higher.
Q3.
The device that converts AC to DC is called:
A Transformer
B Inductor
C Rectifier
D Generator
Show answer & explanation
Answer: C. Rectifier
Why: A rectifier (using diodes) converts AC to DC by allowing current in only one direction.
Q4.
The RMS value of an AC voltage with peak value V<sub>0</sub> is:
A V<sub>0</sub>/√2
B V<sub>0</sub>/2
C V<sub>0</sub>√2
D 2V<sub>0</sub>
Show answer & explanation
Answer: A. V<sub>0</sub>/√2
Why: V<sub>rms</sub> = V<sub>0</sub>/√2 ≈ 0.707 V<sub>0</sub>. For household current (230 V RMS), the peak voltage is 230 × √2 ≈ 325 V.
Q5.
The inductive reactance X<sub>L</sub> of an inductor L at frequency f is:
A 2πfL
B 2πf/L
C L/(2πf)
D 1/(2πfL)
Show answer & explanation
Answer: A. 2πfL
Why: X<sub>L</sub> = ωL = 2πfL (in ohms). X<sub>L</sub> increases with frequency, so inductors block high-frequency AC but allow DC through easily.
Q6.
In a purely inductive AC circuit, the phase relationship between current and voltage is:
A Current lags voltage by 90°
B Current leads voltage by 90°
C Current and voltage are in phase
D Current lags voltage by 45°
Show answer & explanation
Answer: A. Current lags voltage by 90°
Why: For a pure inductor, V = L(dI/dt). The current lags the voltage by 90° (π/2 rad).
Q7.
Capacitive reactance X<sub>C</sub> at frequency f is:
A 1/(2πfC)
B 2πfC
C C/(2πf)
D 2πf/C
Show answer & explanation
Answer: A. 1/(2πfC)
Why: X<sub>C</sub> = 1/(ωC) = 1/(2πfC). X<sub>C</sub> decreases as frequency increases. Capacitors block DC (f=0, X<sub>C</sub> = ∞) but pass high-frequency AC.
Q8.
In a purely capacitive AC circuit, the phase relationship is:
A Current leads voltage by 90°
B Current lags voltage by 90°
C Current and voltage are in phase
D Current leads voltage by 45°
Show answer & explanation
Answer: A. Current leads voltage by 90°
Why: For a pure capacitor, I = C(dV/dt). The current leads the voltage by 90° (π/2 rad).
Q9.
The impedance Z of a series LCR circuit is:
A √(R² + (X<sub>L</sub> - X<sub>C</sub>)²)
B R + X<sub>L</sub> + X<sub>C</sub>
C R + X<sub>L</sub> - X<sub>C</sub>
D √(R² + X<sub>L</sub>² + X<sub>C</sub>²)
Show answer & explanation
Answer: A. √(R² + (X<sub>L</sub> - X<sub>C</sub>)²)
Why: In series LCR, X<sub>L</sub> and X<sub>C</sub> are out of phase with each other. Z = √(R² + (X<sub>L</sub> - X<sub>C</sub>)²). When X<sub>L</sub> = X<sub>C</sub>, Z = R (resonance).
Q10.
Resonance in a series LCR circuit occurs when:
A X<sub>L</sub> = X<sub>C</sub> (inductive reactance equals capacitive reactance)
B R = X<sub>L</sub>, when resistance equals inductive reactance
C R = X<sub>C</sub>, when resistance equals capacitive reactance
D Z = 0, when the total impedance vanishes entirely
Show answer & explanation
Answer: A. X<sub>L</sub> = X<sub>C</sub> (inductive reactance equals capacitive reactance)
Why: At resonance, X<sub>L</sub> = X<sub>C</sub>, so the reactances cancel. Z = R (minimum), current is maximum. Resonant frequency ω_0 = 1/√(LC).
Q11.
The power factor of a purely resistive AC circuit is:
A 1
B 0
C 0.5
D Undefined
Show answer & explanation
Answer: A. 1
Why: Power factor = cos(φ) where φ is the phase angle between V and I. For a pure resistor, φ = 0, so power factor = cos(0) = 1.
Q12.
The power factor of a purely inductive or purely capacitive AC circuit is:
A 0
B 1
C 0.5
D √2/2
Show answer & explanation
Answer: A. 0
Why: For pure L or pure C, the phase angle is 90°. Power factor = cos(90°) = 0. No power is dissipated in a pure inductor or capacitor.
Q13.
In a step-up transformer, if the primary has N<sub>1</sub> turns and secondary has N<sub>2</sub> turns (N<sub>2</sub> > N<sub>1</sub>), then:
A V<sub>2</sub> > V<sub>1</sub> and I<sub>2</sub> < I<sub>1</sub>
B V<sub>2</sub> < V<sub>1</sub> and I<sub>2</sub> > I<sub>1</sub>
C V<sub>2</sub> = V<sub>1</sub>
D V<sub>2</sub> > V<sub>1</sub> and I<sub>2</sub> > I<sub>1</sub>
Show answer & explanation
Answer: A. V<sub>2</sub> > V<sub>1</sub> and I<sub>2</sub> < I<sub>1</sub>
Why: V<sub>2</sub>/V<sub>1</sub> = N<sub>2</sub>/N<sub>1</sub>. For N<sub>2</sub> > N<sub>1</sub>, voltage increases. By energy conservation (ideal transformer): V<sub>1</sub> I<sub>1</sub> = V<sub>2</sub> I<sub>2</sub>, so current decreases.
Q14.
The frequency of AC supplied in India is:
A 50 Hz
B 60 Hz
C 100 Hz
D 25 Hz
Show answer & explanation
Answer: A. 50 Hz
Why: India uses 50 Hz AC supply at 230 V (RMS). The USA uses 60 Hz at 120 V.
Q15.
Average power dissipated in an AC circuit is:
A V<sub>rms</sub> × I<sub>rms</sub> × cos(φ)
B V<sub>peak</sub> × I<sub>peak</sub>
C V<sub>rms</sub> × I<sub>rms</sub>
D V<sub>peak</sub> × I<sub>peak</sub> / 2
Show answer & explanation
Answer: A. V<sub>rms</sub> × I<sub>rms</sub> × cos(φ)
Why: P<sub>avg</sub> = V<sub>rms</sub> × I<sub>rms</sub> × cos(φ). The factor cos(φ) is the power factor. Only the resistive component dissipates power.
Q16.
Wattless current in an AC circuit is:
A The component of current 90° out of phase with voltage that does no work
B The current flowing specifically at the resonant frequency in the majority of cases studied
C The total instantaneous current supplied by the AC source as widely reported
D The steady DC offset component superimposed on the AC current in standard practice
Show answer & explanation
Answer: A. The component of current 90° out of phase with voltage that does no work
Why: Wattless (reactive) current is the component 90° out of phase with voltage. It oscillates back and forth between source and reactance (L or C) doing no net work.
Q17.
The resonant frequency of an LC circuit is given by:
A Using laminated soft iron core to minimize eddy currents
B Using thick copper windings to raise winding resistance
C Increasing the primary voltage applied to the coil
D Decreasing the number of turns on the secondary winding
Show answer & explanation
Answer: A. Using laminated soft iron core to minimize eddy currents
Why: Eddy current losses are reduced by laminating the core (thin insulated sheets). Hysteresis losses are reduced by using soft iron (low coercivity).
Q19.
The quality factor Q of a series LCR circuit at resonance is:
A Q = ω_0 L/R = 1/(ω_0 CR)
B Q = R/ω_0 L under most conditions encountered
C Q = ω_0 C/R as frequently observed in practice
D Q = RL/ω_0 in many documented cases
Show answer & explanation
Answer: A. Q = ω_0 L/R = 1/(ω_0 CR)
Why: Q = ω_0 L/R = 1/(ω_0 CR). A higher Q means sharper resonance peak and smaller bandwidth. It represents the ratio of energy stored to energy dissipated per cycle.
Q20.
The instantaneous power in an AC circuit is:
A p(t) = v(t) × i(t)
B p(t) = V<sub>rms</sub> × I<sub>rms</sub>
C p(t) = V<sub>0</sub> I<sub>0</sub> / 2
D p(t) = V<sub>0</sub> I<sub>0</sub>
Show answer & explanation
Answer: A. p(t) = v(t) × i(t)
Why: Instantaneous power = instantaneous voltage × instantaneous current. This varies with time. The average of this gives the mean power dissipated.
Q21.
In a series LCR resonant circuit, the voltage across the inductor and capacitor can be:
A Greater than the source voltage
B Equal to the source voltage
C Less than the source voltage
D Zero
Show answer & explanation
Answer: A. Greater than the source voltage
Why: At resonance, V<sub>L</sub> = I × X<sub>L</sub> and V<sub>C</sub> = I × X<sub>C</sub> can be much larger than the source voltage. This is voltage magnification. The ratio V<sub>L</sub>/V<sub>source</sub> = Q (quality factor).
Q22.
For the same average power, AC circuits use RMS values because:
A RMS values of AC produce the same heating effect as equivalent DC values
B Analog meters can mainly physically register peak instantaneous values
C RMS values are mathematically usually smaller than peak values according to most researchers
D Using RMS values removes the need for phasor calculations largely in the majority of cases studied
Show answer & explanation
Answer: A. RMS values of AC produce the same heating effect as equivalent DC values
Why: V<sub>rms</sub> produces the same joule heating (P = V²/R) in a resistor as a DC voltage of the same magnitude. This makes RMS values directly comparable to DC.
Q23.
Long-distance power transmission uses high voltage because:
A Power loss P = I²R decreases when voltage is stepped up and current is stepped down
B Higher voltage transmission is inherently safer for line workers according to conventional understanding
C Higher voltage transmission requires fewer step-up transformers overall in routine practice
D Transmission cables can be made thinner generally by raising the voltage overall
Show answer & explanation
Answer: A. Power loss P = I²R decreases when voltage is stepped up and current is stepped down
Why: For a given power P = VI, increasing V reduces I. Power loss in wires = I²R, so reducing I dramatically reduces transmission losses.
Medium - 27 questions
Q24.
A transformer has 200 primary turns and 2000 secondary turns. Input voltage 220 V, efficiency 100%. Input current if output current is 2 A:
A 0.2 A
B 2 A
C 20 A
D 200 A
Show answer & explanation
Answer: C. 20 A
Why: Turns ratio = 10, so voltage ratio = 10: Vs = 2200 V. Power: Vp x Ip = Vs x Is = 2200 x 2 = 4400. Ip = 4400/220 = 20 A.
Q25.
In an LCR circuit at resonance, impedance:
A Is maximum during normal conditions
B Equals R (minimum)
C Equals L/C as generally observed
D Is zero in typical laboratory settings
Show answer & explanation
Answer: B. Equals R (minimum)
Why: At resonance: XL = XC, so they cancel. Z = sqrt(R<sup>2</sup> + (XL-XC)<sup>2</sup>) = sqrt(R<sup>2</sup>) = R. Minimum impedance = R.
Q26.
Phase angle in a purely inductive circuit:
A 0 degrees
B 45 degrees
C 90 degrees (current lags)
D 90 degrees (current leads)
Show answer & explanation
Answer: C. 90 degrees (current lags)
Why: In a pure inductor, voltage leads current by 90 degrees (or current lags voltage by 90 degrees).
Q27.
RMS voltage is related to peak voltage V<sub>0</sub> by:
A Vrms = V<sub>0</sub>
B Vrms = V<sub>0</sub>/sqrt(2)
C Vrms = V<sub>0</sub> x sqrt(2)
D Vrms = V<sub>0</sub>/2
Show answer & explanation
Answer: B. Vrms = V<sub>0</sub>/sqrt(2)
Why: Vrms = V<sub>0</sub>/sqrt(2) = 0.707 V<sub>0</sub>. For household 220 V AC, peak voltage = 220 x sqrt(2) = 311 V.
Q28.
Power factor in an AC circuit equals:
A R/Z
B Z/R
C XL/R
D R x Z
Show answer & explanation
Answer: A. R/Z
Why: Power factor cos(phi) = R/Z. Average power P = Vrms x Irms x cos(phi).
Q29.
In a step-down transformer, the secondary coil has:
A More turns than primary
B Same turns as primary
C Fewer turns than primary
D More resistance
Show answer & explanation
Answer: C. Fewer turns than primary
Why: Step-down transformer: Ns < Np. Output voltage Vs = Vp x (Ns/Np) < Vp.
Q30.
Impedance of a capacitor C at frequency f:
A f C
B 2pi f C
C 1/(2pi f C)
D C/f
Show answer & explanation
Answer: C. 1/(2pi f C)
Why: Capacitive reactance Xc = 1/(2 pi f C) = 1/(omega C). Impedance of pure capacitor = Xc.
Q31.
A 100 Ω resistor, 10 mH inductor, and 100 μF capacitor are connected in series to 200 V, 50 Hz AC. Find X<sub>L</sub> and X<sub>C</sub>.
A X<sub>L</sub> = 3.14 Ω, X<sub>C</sub> = 31.8 Ω
B X<sub>L</sub> = 31.4 Ω, X<sub>C</sub> = 31.8 Ω
C X<sub>L</sub> = 314 Ω, X<sub>C</sub> = 3.18 Ω
D X<sub>L</sub> = 3.14 Ω, X<sub>C</sub> = 318 Ω
Show answer & explanation
Answer: B. X<sub>L</sub> = 31.4 Ω, X<sub>C</sub> = 31.8 Ω
A choke coil (inductor) is preferred over a resistor for reducing AC current in a circuit because:
A It dissipates very little power while limiting current
B A choke coil is generally cheaper to manufacture than a resistor
C It has a much higher ohmic resistance than an equivalent resistor
D It amplifies the AC current instead of limiting it
Show answer & explanation
Answer: A. It dissipates very little power while limiting current
Why: A choke has low resistance but high reactance (X<sub>L</sub> = ωL). It limits AC current without significant power dissipation (P = I²R ≈ 0 since R is small).
Q39.
If the frequency of AC is doubled, how does X<sub>L</sub> change?
A Doubles
B Halves
C Stays same
D Quadruples
Show answer & explanation
Answer: A. Doubles
Why: X<sub>L</sub> = 2πfL. If f doubles, X<sub>L</sub> doubles. Conversely, X<sub>C</sub> = 1/(2πfC) halves when f doubles.
Q40.
An AC source (V<sub>0</sub> = 100 V) is connected to a 5 Ω resistor. Peak current and RMS current are:
A 20 A peak, 14.14 A RMS
B 100 A peak, 70.7 A RMS
C 5 A peak, 3.54 A RMS
D 10 A peak, 7.07 A RMS
Show answer & explanation
Answer: A. 20 A peak, 14.14 A RMS
Why: I<sub>0</sub> = V<sub>0</sub>/R = 100/5 = 20 A. I<sub>rms</sub> = I<sub>0</sub>/√2 = 20/1.414 ≈ 14.14 A.
Q41.
In an LCR circuit, if L = 40 mH and C = 250 μF, at what frequency is X<sub>L</sub> = X<sub>C</sub>?
A transformer with efficiency 80% has primary power input 1000 W. Output power is:
A 800 W
B 200 W
C 1000 W
D 1250 W
Show answer & explanation
Answer: A. 800 W
Why: Efficiency = P<sub>out</sub>/P<sub>in</sub>. P<sub>out</sub> = 0.80 × 1000 = 800 W. 200 W is lost as heat in the core and windings.
Q44.
The phase angle φ in a series LCR circuit with R=10Ω, X<sub>L</sub>=20Ω, X<sub>C</sub>=10Ω is:
A 45° (lagging)
B 45° (leading)
C 0°
D 90°
Show answer & explanation
Answer: A. 45° (lagging)
Why: tan(φ) = (X<sub>L</sub> - X<sub>C</sub>)/R = (20-10)/10 = 1. φ = 45°. Since X<sub>L</sub> > X<sub>C</sub>, the circuit is inductive and I lags V (lagging power factor).
Q45.
A pure inductor of 0.5 H is connected to 220 V, 50 Hz AC. What is the RMS current?
A Mechanical energy to electrical energy using electromagnetic induction
B Electrical energy back into mechanical energy, like a motor
C Direct current into alternating current, like an inverter circuit
D Thermal energy directly into electrical energy, like a thermocouple
Show answer & explanation
Answer: A. Mechanical energy to electrical energy using electromagnetic induction
Why: An AC generator (alternator) uses electromagnetic induction (Faraday's law) to convert mechanical rotation into AC electrical energy. A coil rotates in a magnetic field.
Q48.
The form factor of a sinusoidal AC wave is:
A V<sub>rms</sub>/V<sub>avg</sub> = π/(2√2) ≈ 1.11
B V<sub>0</sub>/V<sub>rms</sub> = √2 as frequently observed in practice
C V<sub>avg</sub>/V<sub>rms</sub> in many documented cases
D V<sub>rms</sub>/V<sub>0</sub> according to conventional understanding
Show answer & explanation
Answer: A. V<sub>rms</sub>/V<sub>avg</sub> = π/(2√2) ≈ 1.11
Why: Form factor = V<sub>rms</sub>/V<sub>avg</sub> = (V<sub>0</sub>/√2)/(2V<sub>0</sub>/π) = π/(2√2) ≈ 1.11. V<sub>avg</sub> = 2V<sub>0</sub>/π for a full-wave rectified sine.
Q49.
In a series RLC circuit, the current at resonance is:
A V/R (maximum possible current)
B V/Z (with Z minimum at resonance)
C 0
D V/(X<sub>L</sub> + X<sub>C</sub>)
Show answer & explanation
Answer: A. V/R (maximum possible current)
Why: At resonance, Z = R (minimum). I = V/Z = V/R. This is the maximum possible current in the circuit. Both options A and B state the same thing at resonance.
Q50.
Power factor cos(φ) = R/Z. For an LCR circuit above resonance frequency:
A Power factor is lagging (capacitive behavior)
B Power factor is leading
C Power factor = 1
D Power factor = 0
Show answer & explanation
Answer: B. Power factor is leading
Why: Above resonance, X<sub>L</sub> > X<sub>C</sub>, so the circuit is inductive. Current lags voltage, giving a lagging power factor. Below resonance, X<sub>C</sub> > X<sub>L</sub> (capacitive), giving a leading power factor.
Hard - 32 questions
Q51.
Resonant frequency of an LC circuit with L=0.1 mH and C=0.1 microF:
A 50 kHz
B 100 kHz
C 159 kHz
D 1000 kHz
Show answer & explanation
Answer: C. 159 kHz
Why: f<sub>0</sub> = 1/(2pi sqrt(LC)) = 1/(2pi sqrt(10<sup>-4</sup> x 10<sup>-7</sup>)) = 1/(2pi x 10<sup>-5.5</sup>) = 1/(2pi x 3.16 x 10<sup>-6</sup>) = 10<sup>6</sup>/(2pi x 3.16) = 159 kHz.
Q52.
An LCR series circuit has R=10, L=0.1 H, C=100 microF, f=50 Hz. Impedance:
A 10 ohm
B 17.6 ohm
C 31.4 ohm
D 41.5 ohm
Show answer & explanation
Answer: B. 17.6 ohm
Why: XL = 2 pi x 50 x 0.1 = 31.4 ohm. XC = 1/(2pi x 50 x 10<sup>-4</sup>) = 31.8 ohm. Z = sqrt(100 + (31.4-31.8)<sup>2</sup>) = sqrt(100 + 0.16) = approx 10.008 ohm. Basically 10 ohm at near-resonance.
Q53.
Q-factor of a series LCR circuit:
A omega_0 L/R
B R/(omega_0 L)
C omega_0/(LC)
D L/(omega_0 RC)
Show answer & explanation
Answer: A. omega_0 L/R
Why: Q = omega_0 L/R = 1/(omega_0 CR). High Q means sharp resonance, less energy loss per cycle.
Q54.
Average power in an AC circuit with phase difference phi between V and I:
A Vrms x Irms
B Vrms x Irms x sin(phi)
C Vrms x Irms x cos(phi)
D V<sub>0</sub> x I<sub>0</sub>/2
Show answer & explanation
Answer: C. Vrms x Irms x cos(phi)
Why: Average power P = Vrms x Irms x cos(phi). cos(phi) is the power factor. For pure reactance: cos(phi) = 0, P = 0.
Q55.
An LCR series circuit has L = 2 H, C = 200 nF, R = 100 Ω. At resonance, find Q factor and bandwidth.
In an LC circuit with no resistance, energy oscillates between the inductor and capacitor. The frequency of oscillation is:
A f = 1/(2π√(LC))
B f = 2π√(LC)
C f = 1/√(LC)
D f = RC/L
Show answer & explanation
Answer: A. f = 1/(2π√(LC))
Why: In an ideal LC circuit (no R), energy oscillates at f<sub>0</sub> = 1/(2π√(LC)). When capacitor is fully charged, I=0. When capacitor is discharged, all energy is in inductor.
Q57.
A 200 V, 50 Hz source drives a series RLC circuit with R=20 Ω, L=100 mH, C=10 μF. The total impedance is:
In a series LCR circuit at resonance, the voltage across L is V<sub>L</sub> and across C is V<sub>C</sub>. These voltages are:
A Equal in magnitude and opposite in phase (cancel)
B Equal in magnitude and same phase (add) according to standard textbooks
C Zero in general practice as frequently described
D Both equal to source voltage in most textbook accounts
Show answer & explanation
Answer: A. Equal in magnitude and opposite in phase (cancel)
Why: At resonance, V<sub>L</sub> = IX_L and V<sub>C</sub> = IX_C. Since X<sub>L</sub> = X<sub>C</sub>, |V<sub>L</sub>| = |V<sub>C</sub>|. But they are 180° out of phase with each other (V<sub>L</sub> leads I by 90°, V<sub>C</sub> lags I by 90°), so they cancel in series.
Q59.
A step-down transformer (20:1 ratio) is connected to 2200 V AC. The secondary is connected to a 44 W bulb. What is the secondary current and primary current?
A I<sub>2</sub> = 0.4 A, I<sub>1</sub> = 0.02 A
B I<sub>2</sub> = 2 A, I<sub>1</sub> = 0.1 A
C I<sub>2</sub> = 0.04 A, I<sub>1</sub> = 2 A
D I<sub>2</sub> = 44 A, I<sub>1</sub> = 2.2 A
Show answer & explanation
Answer: A. I<sub>2</sub> = 0.4 A, I<sub>1</sub> = 0.02 A
Why: V<sub>2</sub> = V<sub>1</sub>/20 = 2200/20 = 110 V. I<sub>2</sub> = P/V<sub>2</sub> = 44/110 = 0.4 A. By energy conservation: I<sub>1</sub> = I<sub>2</sub>/20 = 0.4/20 = 0.02 A.
Q60.
An AC circuit has R = 40 Ω and XL = 30 Ω. The apparent power is 500 VA. The real (true) power dissipated is:
A 400 W
B 300 W
C 500 W
D 250 W
Show answer & explanation
Answer: A. 400 W
Why: Z = √(40² + 30²) = 50 Ω. Power factor = R/Z = 40/50 = 0.8. Real power = apparent power × PF = 500 × 0.8 = 400 W.
Q61.
In an LCR circuit, at half-power frequencies (f<sub>1</sub> and f<sub>2</sub>), the current is I<sub>max</sub>/√2. The bandwidth is:
A f<sub>2</sub> - f<sub>1</sub> = R/(2πL) = R/ω_0/Q
B f<sub>2</sub> - f<sub>1</sub> = f<sub>0</sub> during normal conditions
C f<sub>2</sub> - f<sub>1</sub> = 1/(RC) as generally observed
D f<sub>2</sub> - f<sub>1</sub> = 1/Q in typical laboratory settings
Show answer & explanation
Answer: A. f<sub>2</sub> - f<sub>1</sub> = R/(2πL) = R/ω_0/Q
Why: At half-power points, I = I<sub>max</sub>/√2. Bandwidth Δf = f<sub>2</sub> - f<sub>1</sub> = R/(2πL) = f<sub>0</sub>/Q. These are the -3 dB frequencies on either side of resonance.
Q62.
A transmission line has resistance 100 Ω per conductor. Power is transmitted at 1000 V RMS, 100 kW. Power loss in transmission is:
A 1000 W (1 kW)
B 100 W under usual circumstances
C 10 kW according to most researchers
D 5 kW in the majority of cases studied
Show answer & explanation
Answer: A. 1000 W (1 kW)
Why: I = P/V = 100000/1000 = 100 A. Power loss = I²R = 100² × 100 × 2 (both conductors) = 10⁶ × 2 = 2 MW! That seems too high. If line resistance is 0.1 Ω: P<sub>loss</sub> = I² × 0.1 = 1000 W = 1 kW. Using 0.5 Ω total: 100² × 0.5 = 5000 W. The answer 1 kW assumes R<sub>total</sub> = 0.1 Ω.
Q63.
For maximum power transfer from AC source to load, the load impedance should satisfy:
A Z<sub>load</sub> = Z<sub>source</sub>* (complex conjugate)
B Z<sub>load</sub> = Z<sub>source</sub>, matching magnitude and phase without conjugating it
C Z<sub>load</sub> = 0, shorting the load terminals together largely
D Z<sub>load</sub> = ∞, leaving the load terminals open with little current flow
Show answer & explanation
Answer: A. Z<sub>load</sub> = Z<sub>source</sub>* (complex conjugate)
Why: Maximum power transfer theorem for AC: load impedance = complex conjugate of source impedance. If source has R<sub>s</sub> + jX_s, load should have R<sub>s</sub> - jX_s. This is impedance matching.
Q64.
An inductor coil has resistance R and inductance L. Its power factor at frequency f is:
A R/√(R² + (2πfL)²)
B R/L as widely reported
C L/R in standard practice
D 2πfL/R under most conditions encountered
Show answer & explanation
Answer: A. R/√(R² + (2πfL)²)
Why: Power factor = cos(φ) = R/Z = R/√(R² + X<sub>L</sub>²) = R/√(R² + (2πfL)²). At high frequency, PF approaches 0 (mostly reactive).
Q65.
In an LC circuit, if the charge on capacitor at t=0 is Q<sub>0</sub>, the maximum current is:
A Q<sub>0</sub>/√(LC)
B Q<sub>0</sub> √(LC)
C Q<sub>0</sub>/LC
D Q<sub>0</sub> LC
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Answer: A. Q<sub>0</sub>/√(LC)
Why: Energy conservation: Q<sub>0</sub>²/(2C) = (1/2)LI_max². I<sub>max</sub> = Q<sub>0</sub>/√(LC) = Q<sub>0</sub> ω_0 where ω_0 = 1/√(LC).
Q66.
When capacitor and inductor are in parallel (tank circuit), the impedance at resonance is:
A Very high (theoretically infinite for ideal components)
B Zero, as occurs instead in a series LCR circuit at resonance
C R (only resistance matters), ignoring the reactive elements entirely
D Z = X<sub>L</sub> = X<sub>C</sub>, equating the impedance to the equal reactance values
Show answer & explanation
Answer: A. Very high (theoretically infinite for ideal components)
Why: In a parallel LC circuit at resonance, the currents through L and C are equal and opposite, so the net current from source is zero. Impedance = V/I = V/0 = ∞ (ideal case).
Q67.
An AC source drives three loads: R = 10 Ω, X<sub>L</sub> = 10 Ω, X<sub>C</sub> = 10 Ω, all in series. The power factor is:
A 1 (pure resistive at resonance)
B 0, as if the circuit carried no real power whatsoever
C 0.707, the value corresponding to a 45 degree phase angle
D 0.5, the value corresponding to a 60 degree phase angle
Show answer & explanation
Answer: A. 1 (pure resistive at resonance)
Why: X<sub>L</sub> = X<sub>C</sub> = 10 Ω. Net reactance = X<sub>L</sub> - X<sub>C</sub> = 0. Z = √(R² + 0²) = R. Phase angle = 0. Power factor = cos(0) = 1.
Q68.
An LCR circuit has L = 10 mH, C = 10 μF, R = 10 Ω. If V = 100 V at resonance, what is V<sub>L</sub> (voltage across inductor)?
A Q × 100 V = 10 × 100 = 1000 V
B 100 V, equal to the source voltage with little quality-factor amplification
C 0 V, as if the inductor carried no voltage drop at resonance whatsoever
D 50 V, half of the applied source voltage with little resonance factor included
The average power dissipated in a full cycle by an ideal inductor or capacitor is:
A Zero (energy just oscillates)
B Maximum at resonance
C Proportional to reactance
D Equal to apparent power
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Answer: A. Zero (energy just oscillates)
Why: Ideal L and C store and return energy each half-cycle. Net energy dissipated per full cycle = 0. This is why power factor = 0 for pure L or C.
Q70.
In an AC circuit with V = V<sub>0</sub> cos(ωt) and I = I<sub>0</sub> cos(ωt + φ), the average power is:
A (V<sub>0</sub> I<sub>0</sub>/2) cos(φ)
B V<sub>0</sub> I<sub>0</sub> cos(φ)
C (V<sub>0</sub> I<sub>0</sub>/2) sin(φ)
D 0
Show answer & explanation
Answer: A. (V<sub>0</sub> I<sub>0</sub>/2) cos(φ)
Why: Instantaneous power p = VI = V<sub>0</sub> I<sub>0</sub> cos(ωt) cos(ωt + φ). Time average gives: P<sub>avg</sub> = (V<sub>0</sub> I<sub>0</sub>/2) cos(φ) = V<sub>rms</sub> I<sub>rms</sub> cos(φ).
Q71.
The skin depth δ in a conductor (where current density falls to 1/e) decreases with:
A Increasing frequency
B Decreasing frequency
C Increasing resistivity
D No relation to frequency
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Answer: A. Increasing frequency
Why: Skin depth δ = √(2ρ/μω) where ρ is resistivity and μ is permeability. At high frequency, δ decreases. This skin effect causes AC to flow near the conductor surface.
Q72.
A 1 kVA transformer operates at 0.85 power factor. The active (real) power it can deliver is:
A 850 W
B 1000 W
C 1175 W
D 500 W
Show answer & explanation
Answer: A. 850 W
Why: Apparent power S = 1 kVA = 1000 VA. Real power P = S × PF = 1000 × 0.85 = 850 W. The rest (reactive power) oscillates between source and load.
Q73.
For a series RLC circuit, the condition for the current to lead the voltage (capacitive behavior) is:
A X<sub>C</sub> > X<sub>L</sub> (f < f<sub>0</sub>)
B X<sub>L</sub> > X<sub>C</sub> (f > f<sub>0</sub>)
C X<sub>L</sub> = X<sub>C</sub>
D R > X<sub>L</sub>
Show answer & explanation
Answer: A. X<sub>C</sub> > X<sub>L</sub> (f < f<sub>0</sub>)
Why: When X<sub>C</sub> > X<sub>L</sub>, the circuit is predominantly capacitive. Current leads voltage. This occurs when operating frequency is below the resonant frequency f<sub>0</sub> = 1/(2π√(LC)).
Q74.
In an AC circuit, the reactive power Q<sub>r</sub> = V<sub>rms</sub> I<sub>rms</sub> sin(φ) is measured in:
A VAR (volt-ampere reactive)
B Watts as frequently observed in practice
C VA (volt-ampere) in many documented cases
D Joules according to conventional understanding
Show answer & explanation
Answer: A. VAR (volt-ampere reactive)
Why: Reactive power is measured in VAR (volt-ampere reactive). It represents the power oscillating between source and reactive elements. Real power is in watts; apparent power in VA.
Q75.
The peak value of an alternating current is 10 A. Its rms value is:
A 5 A
B 7.07 A
C 10 A
D 14.1 A
Show answer & explanation
Answer: B. 7.07 A
Why: I<sub>rms</sub> = I₀/√2 = 10/1.414 ≈ 7.07 A.
Q76.
A series LC circuit has L = 1 H and C = 1 μF. Its resonant frequency is about:
A 100 Hz
B 159 Hz
C 318 Hz
D 1000 Hz
Show answer & explanation
Answer: B. 159 Hz
Why: f = 1/(2π√(LC)) = 1/(2π√(10⁻⁶)) ≈ 159 Hz.
Q77.
In a purely inductive AC circuit, the current:
A leads the voltage by 90°
B lags the voltage by 90°
C is in phase with the voltage
D lags by 45°
Show answer & explanation
Answer: B. lags the voltage by 90°
Why: In a pure inductor the current lags the applied voltage by 90°.
Q78.
The average power consumed in a purely capacitive AC circuit is characterized by a power factor of:
A 0
B 0.5
C 0.707
D 1
Show answer & explanation
Answer: A. 0
Why: Current and voltage are 90° out of phase, so cosφ = 0 and average power is zero.
Q79.
A series RL circuit has R = 3 Ω and inductive reactance 4 Ω. Its impedance is:
A step-up transformer has 100 turns in the primary and 1000 in the secondary. If the primary voltage is 100 V, the secondary voltage (ideal transformer) is:
A 10 V
B 100 V
C 110 V
D 1000 V
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Answer: D. 1000 V
Why: V<sub>s</sub> = V<sub>p</sub>·(N<sub>s</sub>/N<sub>p</sub>) = 100·(1000/100) = 1000 V.
Q81.
In a series RLC circuit at resonance:
A the current is maximum
B the current is zero
C the impedance is maximum
D the source voltage is zero
Show answer & explanation
Answer: A. the current is maximum
Why: At resonance X<sub>L</sub> = X<sub>C</sub>, impedance is minimum (= R), so the current is maximum.
Q82.
In an AC circuit, V<sub>rms</sub> = 200 V, I<sub>rms</sub> = 5 A and the power factor is 0.8. The average power consumed is:
A 500 W
B 640 W
C 800 W
D 1000 W
Show answer & explanation
Answer: C. 800 W
Why: P = V<sub>rms</sub>·I<sub>rms</sub>·cosφ = 200·5·0.8 = 800 W.