Alternating Current — Previous Year Questions (Class 12 Physics)
Alternating current (AC) is the standard form of electrical supply. Understanding AC circuits with resistors, inductors, and capacitors is essential for el
TL;DR: Alternating current (AC) is the standard form of electrical supply. Understanding AC circuits with resistors, inductors, and capacitors is essential f…
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Alternating current (AC) is the standard form of electrical supply. Understanding AC circuits with resistors, inductors, and capacitors is essential for el
Alternating Current — Previous Year Questions with Solutions
Q (2023, 3 marks): Define RMS (root mean square) value of AC. Derive the relationship between peak and RMS value for a sinusoidal current.
Answer: RMS value is the effective value of AC that produces the same heat in a resistor as an equivalent DC current.
For sinusoidal current i = i0 sin(ωt):
RMS value: Irms = i0/√2 ≈ 0.707 × i0
Derivation:
1. Instantaneous power: P = i^2R = i0^2 R sin^2(ωt)
2. Average power over one cycle: P_avg = ∫(i0^2 R sin^2(ωt))dt / T
3. Since ∫sin^2(ωt)dt = T/2: P_avg = i0^2 R / 2
4. For equivalent DC: P_DC = Irms^2 R
5. Equating: Irms^2 R = i0^2 R / 2
6. Therefore: Irms = i0/√2
Similarly for voltage: Vrms = V0/√2
Q (2022, 2 marks): An AC source provides a voltage V = 100 sin(100πt) V. Find the RMS voltage, peak voltage, and frequency.
Answer: Given: V = 100 sin(100πt) V
Comparing with V = V0 sin(ωt):
Peak voltage: V0 = 100 V
RMS voltage: Vrms = V0/√2 = 100/√2 = 100/1.414 ≈ 70.7 V
Angular frequency: ω = 100π rad/s
Frequency: f = ω/(2π) = 100π/(2π) = 50 Hz
Q (2023, 3 marks): Derive the impedance of an LCR series circuit and discuss the condition for resonance.
Answer: For LCR series circuit, the impedance Z is:
Z = √[R^2 + (XL - XC)^2]
Where XL = ωL (inductive reactance) and XC = 1/(ωC) (capacitive reactance)
At resonance:
- XL = XC
- ω0L = 1/(ω0C)
- ω0 = 1/√(LC)
- Resonant frequency: f0 = 1/(2π√(LC))
At resonance:
1. Impedance Z = R (minimum)
2. Current I = V/R (maximum)
3. Power factor = 1 (voltage and current in phase)
4. Energy stored in L equals energy stored in C
Resonance is used in tuning circuits for radios and communication devices.
Q (2022, 3 marks): A pure capacitor (C = 5 μF) is connected to an AC source with Vrms = 220 V and f = 50 Hz. Calculate the capacitive reactance and the RMS current.
Answer: Given:
C = 5 μF = 5 × 10^-6 F
Vrms = 220 V
f = 50 Hz
Angular frequency: ω = 2πf = 2π × 50 = 100π rad/s ≈ 314 rad/s
Capacitive reactance: XC = 1/(ωC) = 1/(314 × 5 × 10^-6)
XC = 1/(1.57 × 10^-3) ≈ 637 Ω
RMS current: Irms = Vrms/XC = 220/637 ≈ 0.345 A
Phase relationship: Current leads voltage by 90° in a pure capacitor.
Q (2023, 2 marks): Explain the concept of power factor in AC circuits. What is its significance?
Answer: Power factor (PF) = cos(φ)
Where φ is the phase difference between voltage and current.
PF = R/Z
Significance:
1. PF = 1: Voltage and current in phase (resistive circuits), all power is real.
2. PF < 1: Voltage and current out of phase (inductive or capacitive), some power is reactive.
3. Real power: P = VIrms cos(φ) (useful work done)
4. Reactive power: Q = VIrms sin(φ) (wattless power, energy oscillates)
5. Apparent power: S = VIrms
In industries:
- Low PF means higher current for same real power.
- Higher current increases transmission losses.
- Power factor correction (using capacitors) improves efficiency and reduces costs.
Q (2021, 2 marks): A 40 Ω pure resistor is connected to an AC source with Vrms = 200 V. Calculate the RMS current, average power, and peak power.
Answer: Given:
R = 40 Ω
Vrms = 200 V
RMS current: Irms = Vrms/R = 200/40 = 5 A
Average power: Pavg = Vrms × Irms × cos(0°) = 200 × 5 × 1 = 1000 W
Or: Pavg = Irms^2 × R = 5^2 × 40 = 1000 W
For sinusoidal AC:
Peak voltage: V0 = Vrms√2 = 200√2 ≈ 282.8 V
Peak current: I0 = V0/R = 282.8/40 ≈ 7.07 A
Peak power: P0 = V0 × I0 = 282.8 × 7.07 ≈ 2000 W
Frequently Asked Questions
What is the difference between apparent power, real power, and reactive power in AC circuits?
Real power (P) is the actual power consumed and converted to useful work, measured in watts (W). Reactive power (Q) oscillates between source and load without being consumed, measured in VAR. Apparent power (S) is the vector sum, measured in VA. For a circuit: S^2 = P^2 + Q^2, and power factor = P/S.
Why does a purely inductive circuit consume no real power despite having current flowing through it?
In a pure inductor, current lags voltage by 90°. Real power P = VI cos(φ) = VI cos(90°) = 0. Although current flows, the energy oscillates between the magnetic field and the source. In the first half-cycle, energy is stored in the magnetic field; in the next half-cycle, it returns to the source.
More Class 12 Physics PYQs
- Current Electricity
- Ray Optics
- Electrostatics
- Moving Charges and Magnetism
- Electromagnetic Induction
- Atoms and Nuclei
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