In a heating arrangement, an alternating current having a peak value of \(28~\mathrm{A}\) is used. To produce the same heat energy, if direct current is used to produce the same amount of heat, then its magnitude will be:

1. about \(14~\mathrm{A}\)  2. about \(28~\mathrm{A}\)
3. about \(20~\mathrm{A}\) 4. cannot say

Subtopic:  AC vs DC |
 57%
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In the circuit shown, the AC source has a voltage
V = 20 cos(ωt) volts with ω= 2000 rad/s. The amplitude of the current will be nearest to:

1. 2 A

2. 3.3 A

3. 2/5 A

4. 5 A

Subtopic:  Different Types of AC Circuits |
 55%
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The AC source in the circuit shown in the figure produces a voltage V = 20cos(2000t) volts. Neglecting source resistance, the voltmeter and ammeter readings will be (approximately):

1. 4 V, 2.0 A

2. 0 V, 2 A

3. 0 V, 1.4 A

4. 8 V, 2.0 A

Subtopic:  Different Types of AC Circuits |
 61%
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An AC ammeter is used to measure the current in a circuit. When a given direct current passes through the circuit, the ac ammeter reads 6 A. When another alternating current passes through the circuit, the AC ammeter reads 8 A. Then the reading of this ammeter if DC and AC flow through the circuit simultaneously is:

1. 102 A

2. 14 A

3. 10 A

4. 15 A

Subtopic:  AC vs DC |
 65%
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If q is the capacitor's charge and i is the current at time t, the voltage V will be:

1. \(\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}+\mathrm{iR}-\frac{\mathrm{q}}{\mathrm{C}}=\mathrm{V}\)
2. \(\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}-\mathrm{iR}+\frac{\mathrm{q}}{\mathrm{C}}=\mathrm{V}\)
3. \(\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}+\mathrm{iR}+\frac{\mathrm{q}}{\mathrm{C}}=\mathrm{V}\)
4. \(\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}-\mathrm{iR}-\frac{\mathrm{q}}{\mathrm{C}}=\mathrm{V}\)

Subtopic:  Different Types of AC Circuits |
 55%
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A direct current of \(5~ A\) is superimposed on an alternating current \(I=10sin ~\omega t\) flowing through a wire. The effective value of the resulting current will be:

1. \(15/2~A\) 2. \(5 \sqrt{3}~A\)
3. \(5 \sqrt{5}~A\) 4. \(15~A\)
Subtopic:  AC vs DC |
 60%
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An L-C-R series circuit with 100 Ω resistance is connected to an AC source of 200 V and an angular frequency of 300 rad/s. When only the capacitance is removed, the current lags behind the voltage by 60°. When only the inductance is removed, the current leads the voltage by 60°. Calculate the power dissipated in the L-C-R circuit.

1. 200 W

2. 400 W

3. 300 W

4. Zero

Subtopic:  Power factor |
 64%
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An ideal resistance R, ideal inductance L, ideal capacitance C, and AC voltmeters V1, V2, V3 and V4 are connected to an AC source as shown. At resonance:

  572685

1. Reading in \(V_3\) = Reading in \(V_1\)
2. Reading in \(V_1\) = Reading in \(V_2\)
3. Reading in \(V_2\) = Reading in \(V_4\)
4. Reading in \(V_2\) = Reading in \(V_3\)



Subtopic:  Different Types of AC Circuits |
 85%
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A transistor-oscillator using a resonant circuit with an inductance L (of negligible resistance) and a capacitance C has a frequency f. If L is doubled and C is changed to 4C, the frequency will be:

1. f/4

2. 8f

3. f/(2√2)

4. f/2

Subtopic:  Different Types of AC Circuits |
 86%
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AIPMT - 2006
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In an L-C-R series AC circuit, the voltage across each of the components - L, C and R is 50 V. The voltage across the L-R combination will be:
1. 50 V
2. \(50 \sqrt{2} ~V\)
3. 100 V
4. 0 V 

Subtopic:  Different Types of AC Circuits |
 70%
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