- 1(current)
Q. No.The AC voltage across a resistance can be measured using a:
| 1. | hot wire voltmeter |
| 2. | moving coil galvanometer |
| 3. | potential coil galvanometer |
| 4. | moving magnet galvanometer |
Subtopic: RMS & Average Values |
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Level 2: 60%+
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An alternating current is given by:
\(i=i_1\sin\omega t+i_2\cos \omega t. \)
What is the RMS value of the current?
| 1. | \( \dfrac{1}{\sqrt{2}}\left(i_1^2+i_2^2\right)^{1/2} \) | 2. | \(\dfrac{1}{\sqrt{2}}\left(i_1+i_2\right)^2 \) |
| 3. | \( \dfrac{1}{2}\left(i_1^2+i_2^2\right)^{1/2} \) | 4. | \( \dfrac{1}{\sqrt{2}}\left(i_1+i_2\right) \) |
Subtopic: RMS & Average Values |
75%
Level 2: 60%+
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Find the peak current and resonant frequency of the following circuit (as shown in the figure).
1. \(0.2\) A and \(50\) Hz
2. \(0.2\) A and \(100\) Hz
3. \(2\) A and \(100\) Hz
4. \(2\) A and \(50\) Hz
Subtopic: RMS & Average Values |
66%
Level 2: 60%+
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The current flowing through an AC circuit is given by
\(I=5 \sin (120 \pi t)~\text{A}\)
How long will the current take to reach the peak value starting from zero?
1. \(\frac{1}{60}~\text{s} \)
2. \(60~\text{s} \)
3. \(\frac{1}{120}~\text{s} \)
4. \(\frac{1}{240}~\text{s} \)
\(I=5 \sin (120 \pi t)~\text{A}\)
How long will the current take to reach the peak value starting from zero?
1. \(\frac{1}{60}~\text{s} \)
2. \(60~\text{s} \)
3. \(\frac{1}{120}~\text{s} \)
4. \(\frac{1}{240}~\text{s} \)
Subtopic: RMS & Average Values |
74%
Level 2: 60%+
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A direct current of \(4\) A and an alternating current of peak value \(4\) A flow through resistance of \(3~\Omega\) and \(2~\Omega\) respectively. The ratio of heat produced in the two resistances in same interval of time will be:
1. \(3:2\)
2. \(3:1\)
3. \(3:4\)
4. \(4:3\)
1. \(3:2\)
2. \(3:1\)
3. \(3:4\)
4. \(4:3\)
Subtopic: RMS & Average Values |
68%
Level 2: 60%+
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In the AC circuit shown in the figure, the value of \(I_{rms}\) is equal to:
1. \(2\) A
2. \(2\sqrt{2}\) A
3. \(4\) A
4. \(\sqrt{2}\) A
1. \(2\) A
2. \(2\sqrt{2}\) A
3. \(4\) A
4. \(\sqrt{2}\) A
Subtopic: RMS & Average Values |
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Level 1: 80%+
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In a series \(RLC\) circuit, \(R=80~\Omega\), \(X_L=100~\Omega\), \(X_C=40~\Omega\). If the source voltage is \(2500 \cos (628 t)\) V, the peak current is equal to:
1. \(5\) A
2. \(25 \) A
3. \(10\) A
4. \(12\) A
1. \(5\) A
2. \(25 \) A
3. \(10\) A
4. \(12\) A
Subtopic: RMS & Average Values |
93%
Level 1: 80%+
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- 1(current)
Q. No.
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