Chapter 3 · Effective value of AC & the transformer: practice questions
Unit 1 — Dynamic Electricity and Electromagnetism · Physics. These are 10 questions from the Magmo3 bank on this lesson, from easy to hard. Solve each one, then open "Show answer and explanation". Magmo3 has 72 questions on this lesson, each with an explanation.
The effective (root-mean-square, rms) value of an alternating current is best defined as:
Athe maximum instantaneous value reached by the current during a cycle
Bthe value of a steady direct current that produces the same heating effect in the same resistance during the same time
Cthe average value of the current taken over one complete cycle
Dthe value indicated by a moving-coil ammeter
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Correct answer: (B) the value of a steady direct current that produces the same heating effect in the same resistance during the same time
Explanation: The effective (rms) value is the steady direct current that gives the same heating effect in the same resistance in the same time.
Question 2Easy
The periodic time of an alternating current is 0.02 s. Its frequency is equal to:
A50 Hz
B20 Hz
C100 Hz
D0.02 Hz
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Correct answer: (A) 50 Hz
Explanation: f = 1/T = 1/0.02 = 50 Hz.
Question 3Easy
In a step-up transformer, the number of turns of the secondary coil compared with that of the primary coil is...
Aequal
Bgreater
Calways exactly double
Dsmaller
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Correct answer: (B) greater
Explanation: A step-up transformer raises the voltage, which requires Nₛ greater than Nₚ since Vₛ/Vₚ = Nₛ/Nₚ.
Question 4Medium
The average value of a sinusoidal alternating current taken over one complete cycle is equal to:
AI_max
BI_max / √2
C2 I_max / π
Dzero
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Correct answer: (D) zero
Explanation: Over one complete cycle the current is positive and negative for equal times, so its average value is zero.
Question 5Medium
The instantaneous induced e.m.f. of an AC generator equals zero at the instant when:
Athe magnetic flux is changing at its greatest rate
Bthe magnetic flux linking the coil is zero
Cthe coil's plane is parallel to the magnetic field lines
Dthe magnetic flux linking the coil is maximum
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Correct answer: (D) the magnetic flux linking the coil is maximum
Explanation: The e.m.f. vanishes when the flux linking the coil is maximum (plane perpendicular to the field), since the rate of change of flux is then zero.
Question 6Medium
The primary coil of an ideal transformer has 1200 turns and is connected to a 240 V alternating source. To obtain an effective output voltage of 20 V, the number of turns of the secondary coil must be...
A144 turns
B14400 turns
C100 turns
D60 turns
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Correct answer: (C) 100 turns
Explanation: Nₛ = Nₚ x Vₛ/Vₚ = 1200 x (20/240) = 100 turns.
Question 7Medium
A transformer of efficiency 90 % is supplied with an electric power of 2000 W at its primary coil. The power lost (dissipated) inside the transformer equals...
A2000 W
B20 W
C200 W
D1800 W
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Correct answer: (C) 200 W
Explanation: The lost power = (1 - efficiency) x input = 0.10 x 2000 = 200 W.
Question 8Hard
The primary coil of an ideal transformer has 1100 turns and is connected to a 220 V alternating supply. The secondary coil feeds a lamp rated '12 V, 36 W', which operates normally. The number of secondary turns and the current in the primary coil are…
A6 turns and 0.30 A
B60 turns and 0.16 A
C600 turns and 0.16 A
D60 turns and 3.0 A
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Correct answer: (B) 60 turns and 0.16 A
Explanation: Nₛ = 1100 × 12/220 = 60 turns; for an ideal transformer the input power equals 36 W, so Iₚ = 36/220 ≈ 0.16 A (3 A is the secondary current).
Question 9Hard
An AC generator produces a peak e.m.f. of 200 V. The instantaneous value of the e.m.f. after one sixth of the periodic time has elapsed, measured from the zero position, is equal to:
A200 V
B141 V
C100 V
D173 V
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Correct answer: (D) 173 V
Explanation: e.m.f. = emf_max sin(ωt); after T/6, ωt = π/3, so e.m.f. = 200 × sin 60° = 173 V.
Question 10Hard
A power station delivers 20 kW of electric power to a transmission line of total resistance 10 ohm at an effective voltage of 1000 V. The power lost as heat in the line equals...
A4 kW
B0.4 kW
C10 kW
D2 kW
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Correct answer: (A) 4 kW
Explanation: The line current is I = P/V = 20000/1000 = 20 A, so the loss = I squared R = 20 squared x 10 = 4000 W = 4 kW.