Class 12 Physics Chapter 7: Alternating Current worksheet with answer key
A 20-mark worksheet on Alternating Current for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.
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Answer: (d) leads the source voltage by 45°
tan φ = (XC − XL)/R = (50 − 10)/40 = 1, so φ = +45°. φ is positive because XC > XL: the circuit is predominantly capacitive and the current leads.
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Answer: (b) 100 Hz
XL = V/I = 200/2.0 = 100 Ω. XL = 2πνL gives ν = 100/(2π × 0.50/π) = 100/1.0 = 100 Hz. 628 is ω = XL/L in rad s⁻¹, not ν; 141 Hz takes 2.0 A as the peak current; 71 Hz takes 200 V as the peak voltage.
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Answer: (a) lags the voltage by π/2
For a pure inductor i = im sin(ωt − π/2): the current reaches its maximum a quarter period after the voltage, so it lags by π/2. A lead of π/2 is the capacitor's case; in phase is the resistor's.
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Answer: (c) 90 Ω
XL = 2πνL is directly proportional to ν: XL = 30 × (150/50) = 90 Ω. 10 Ω treats XL as inversely proportional to ν; 270 Ω squares the ratio.
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Answer: (a) 325 V and −325 V
The meter reads the rms value V = 230 V. The peak is vm = √2 V = 1.414 × 230 ≈ 325 V, and the sinusoidal voltage swings from +325 V to −325 V. Taking the reading as the peak gives ±230 V; dividing by √2 instead of multiplying gives ±163 V; an ac voltage is not one-signed.
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XL = 2πνL, so L = 40/(2π × 50) = 0.127 H ≈ 0.13 H. XL is proportional to ν, so at 200 Hz it is 40 × 4 = 160 Ω.
- L = XL/2πν = 0.13 H (1)
- XL ∝ ν, so 160 Ω at 200 Hz (1)
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In a capacitor the current leads the voltage by π/2, a quarter of a period. T = 1/50 s = 20 ms, so the current reaches its maximum T/4 = 5.0 ms before the voltage.
- current first (leads by π/2) (1)
- T/4 = 5.0 ms (1)
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No. VL is π/2 ahead of VR, so the source voltage is the hypotenuse of a right triangle: V² = VR² + VL². VL = √(100² − 60²) = 80 V. Subtracting 60 from 100 treats the two voltages as in phase.
- voltages add as phasors at right angles, not as numbers (1)
- VL = √(100² − 60²) = 80 V (1)
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Kirchhoff's loop rule: v − L di/dt = 0, so di/dt = (vm/L) sin ωt. Integrating, i = −(vm/ωL) cos ωt + constant; the constant is zero because the current oscillates symmetrically about zero. Since −cos ωt = sin(ωt − π/2), i = im sin(ωt − π/2) with im = vm/ωL: the current lags the voltage by π/2. The inductive reactance XL = ωL, so im = vm/XL.
- v = L di/dt from the loop rule (1)
- integration to i = −(vm/ωL) cos ωt, constant zero (1)
- i = im sin(ωt − π/2): lag of π/2; XL = ωL (1)
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The antenna picks up signals of many frequencies and acts as the source; the capacitance is varied until the resonant frequency 1/2π√(LC) equals the wanted station's frequency, whose current is then maximum. C = 1/(4π²ν²L) = 1/(4π² × (1.0 × 10⁶)² × 0.25 × 10⁻³) ≈ 1.0 × 10⁻¹⁰ F = 100 pF. Since ν ∝ 1/√C, doubling ν needs C reduced to a quarter, about 25 pF.
- resonant frequency matched to the station; its current is maximum (1)
- C ≈ 100 pF (1)
- C reduced to one quarter, ≈ 25 pF (1)
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(a) ω₀ = 1/√(LC) = 1/√(0.40 × 2.5 × 10⁻⁶) = 1/(1.0 × 10⁻³) = 1000 rad s⁻¹. (b) At resonance I = V/R and VC = IXC = V(ω₀L)/R = QV, so Q = ω₀L/R = 10: R = 1000 × 0.40/10 = 40 Ω. (c) At resonance cos φ = 1 and Z = R: P = V²/R = 50²/40 = 62.5 W, all of it in R.
- ω₀ = 1000 rad s⁻¹ (1)
- VC/V = ω₀L/R = 10 gives R = 40 Ω (1)
- P = V²/R = 62.5 W (1)
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