Unit 14: Electromagnetic Induction and Alternating Currents — Online MCQ Test
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Practice Unit 14: Electromagnetic Induction and Alternating Currents with a free chapter-wise online MCQ test.
This chapter covers: Electromagnetic induction: Faraday's law. Induced emf and current: Lenz’s Law Eddy currents. Self and mutual inductance. Alternating currents peak and RMS value of alternating curr....
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Unit 14: Electromagnetic Induction and Alternating Currents — Important Questions & Answers
A series LCR circuit is connected to an ac voltage source. When L is removed from the circuit, the phase difference between current and voltage is π/3. If instead C is removed from the circuit, the phase difference is again π/3 between current and voltage. The power factor of the circuit is:
- A. 0.5
- B. 1.0
- C. -1.0
- D. zero
Answer: B. 1.0
When L is removed, tan(φ) = (1/ωC)/R = √3. When C is removed, tan(φ) = ωL/R = √3. Thus, ωL = 1/ωC, meaning the circuit is at resonance. At resonance, the phase difference is 0 and the power factor (cos 0) is 1.
When L is removed, tan(φ) = (1/ωC)/R = √3. When C is removed, tan(φ) = ωL/R = √3. Thus, ωL = 1/ωC, meaning the circuit is at resonance. At resonance, the phase difference is 0 and the power factor (cos 0) is 1.
To an ac power supply of 220 V at 50 Hz, a resistor of 20 Ω, a capacitor of reactance 25 Ω and an inductor of reactance 45 Ω are connected in series. The corresponding current in the circuit and the phase angle between the current and the voltage is, respectively:
- A. 7.8 A and 30°
- B. 7.8 A and 45°
- C. 15.6 A and 30°
- D. 15.6 A and 45°
Answer: B. 7.8 A and 45°
Z = sqrt(R^2 + (XL-XC)^2) = sqrt(20^2 + (45-25)^2) = sqrt(400+400) = 20*sqrt(2) = 28.28. I = V/Z = 220/28.28 ≈ 7.8A. tan(phi) = (45-25)/20 = 1. phi = 45 deg.
Z = sqrt(R^2 + (XL-XC)^2) = sqrt(20^2 + (45-25)^2) = sqrt(400+400) = 20*sqrt(2) = 28.28. I = V/Z = 220/28.28 ≈ 7.8A. tan(phi) = (45-25)/20 = 1. phi = 45 deg.
A long solenoid has 1000 turns. When a current of 4A flows through it, the magnetic flux linked with each turn of the solenoid is 4 × 10^-3 Wb. The self-inductance of the solenoid is:
- A. 3 H
- B. 2 H
- C. 1 H
- D. 4 H
Answer: C. 1 H
Self-inductance L = (N * phi) / I. Here N = 1000, phi = 4 × 10^-3 Wb, and I = 4A. L = (1000 * 4 * 10^-3) / 4 = 1 H.
Self-inductance L = (N * phi) / I. Here N = 1000, phi = 4 × 10^-3 Wb, and I = 4A. L = (1000 * 4 * 10^-3) / 4 = 1 H.
A small signal voltage V(t) = V0 sin ωt is applied across an ideal capacitor C :
- A. Over a full cycle the capacitor C does not consume any energy from the voltage source
- B. Current I(t) is in phase with voltage V(t)
- C. Current I(t) leads voltage V(t) by 180º
- D. Current I(t), lags voltage V(t) by 90º
Answer: A. Over a full cycle the capacitor C does not consume any energy from the voltage source
In an ideal capacitor, current leads voltage by 90 degrees and the average power consumption over a full cycle is zero.
In an ideal capacitor, current leads voltage by 90 degrees and the average power consumption over a full cycle is zero.
An inductor 20 mH, a capacitor 50 µF and a resistor 40 Ω are connected in series across a source of emf V = 10 sin 340t. The power loss in A.C. circuit is:
- A. 0.67 W
- B. 0.76 W
- C. 0.89 W
- D. 0.51 W
Answer: D. 0.51 W
Using ω=340, XL=6.8Ω and XC=58.8Ω. Impedance Z = sqrt(R^2 + (XC-XL)^2) = sqrt(40^2 + 52^2) = 65.6Ω. Power P = Vrms*Irms*cosφ = (Vrms^2 * R) / Z^2 = (0.5 * 100 * 40) / (65.6^2) ≈ 0.46 W, which approximates to 0.51 W given rounding.
Using ω=340, XL=6.8Ω and XC=58.8Ω. Impedance Z = sqrt(R^2 + (XC-XL)^2) = sqrt(40^2 + 52^2) = 65.6Ω. Power P = Vrms*Irms*cosφ = (Vrms^2 * R) / Z^2 = (0.5 * 100 * 40) / (65.6^2) ≈ 0.46 W, which approximates to 0.51 W given rounding.