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Q. No.The variation of EMF with time for four types of generators is shown in the figures. Which amongst them can be called AC voltage?
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| (a) | (b) |
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| (c) | (d) |
| 1. | (a) and (d) |
| 2. | (a), (b), (c), and (d) |
| 3. | (a) and (b) |
| 4. | only (a) |
Subtopic: AC Generator |
72%
Level 2: 60%+
NEET - 2019
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Given below are two statements.
| Assertion (A): | When a current \(I=(3+4 \sin \omega t)\) flows in a wire, then the reading of a dc ammeter connected in series is \(4\) units. |
| Reason (R): | A dc ammeter measures only the value of the current amplitude. |
| 1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
| 3. | (A) is True but (R) is False. |
| 4. | Both (A) and (R) are False. |
Subtopic: AC vs DC |
61%
Level 2: 60%+
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The instantaneous value of current in an AC circuit is \(I = 2\sin\left(100\pi t +\dfrac{\pi}{3}\right)~\text{A}\). The current will be maximum for the first time at:
| 1. | \(t = \dfrac{1}{100}s\) | 2. | \(t = \dfrac{1}{200}s\) |
| 3. | \(t = \dfrac{1}{400}s\) | 4. | \(t = \dfrac{1}{600}s\) |
Subtopic: RMS & Average Values |
65%
Level 2: 60%+
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A \(100~\Omega\) resistor is connected to a \(220~\text{V}\), \(50~\text{Hz}\) \(\text{AC}\) supply. The net power consumed over a full cycle is:
| 1. | \(484~\text{W}\) | 2. | \(848~\text{W}\) |
| 3. | \(400~\text{W}\) | 4. | \(786~\text{W}\) |
Subtopic: Power factor |
87%
Level 1: 80%+
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A \(10~\mu\text F\) capacitor is connected to a \(210~\text V,50~\text{Hz}\) source as shown in the figure. The peak current in the circuit is nearly \((\pi = 3.14)\):
1. \(0.93~\text A\)
2. \(1.20~\text A\)
3. \(0.35~\text A\)
4. \(0.58~\text A\)
1. \(0.93~\text A\)
2. \(1.20~\text A\)
3. \(0.35~\text A\)
4. \(0.58~\text A\)
Subtopic: RMS & Average Values |
56%
Level 3: 35%-60%
NEET - 2024
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If the RMS current in a \(50~\text{Hz}\) AC circuit is \(5~\text{A}\), the value of the current \(\dfrac{1}{300}~\text{s}\) after its value becomes zero is:
| 1. | \(5\sqrt2~\text{A}\) | 2. | \(5\sqrt{\dfrac32}~\text{A}\) |
| 3. | \(\sqrt{\dfrac56}~\text{A}\) | 4. | \(\dfrac{5}{\sqrt2}~\text{A}\) |
Subtopic: Different Types of AC Circuits |
67%
Level 2: 60%+
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An alternating voltage \(V=V_0\sin(\omega t)\) is applied across a circuit. As a result, a current \(I=I_0\sin\Big(\omega t+\dfrac{\pi}{2}\Big)\) flows in it. The power consumed per cycle is:
| 1. | \(\dfrac{V_0I_0}{2}\) | 2. | \(\dfrac{V_0I_0}{\sqrt2}\) |
| 3. | \(\sqrt2V_0I_0\) | 4. | zero |
Subtopic: Power factor |
77%
Level 2: 60%+
Please attempt this question first.
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The variation of an alternating current; \(I=I_0 \mathrm{sin}(\omega t)\) with time \((t)\) is shown in the figure below. The average value of the current for the half-cycle will be:
1. \(\dfrac{I_{0}}{\pi}\)
2. \(\dfrac{I_{0}}{2}\)
3. \(\dfrac{2 I_{0}}{\pi}\)
4. \(\dfrac{I_{0}}{2 \pi}\)
1. \(\dfrac{I_{0}}{\pi}\)
2. \(\dfrac{I_{0}}{2}\)
3. \(\dfrac{2 I_{0}}{\pi}\)
4. \(\dfrac{I_{0}}{2 \pi}\)
Subtopic: RMS & Average Values |
67%
Level 2: 60%+
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An alternating current given by \(I=I_0~\sin\omega t+I_0~\cos\omega t\) flows through a circuit. The RMS current is:
| 1. | \(I_0\) | 2. | \(\dfrac{I_0}{\sqrt2}\) |
| 3. | \(\sqrt2I_0\) | 4. | zero |
Subtopic: RMS & Average Values |
50%
Level 3: 35%-60%
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A light bulb is rated at \(100~\text W\) for a \(220~\text V\) AC supply. The peak voltage of the source is:
1. \(220~\text V\)
2. \(110~\text V\)
3. \(311~\text V\)
4. \(100~\text V\)
Subtopic: RMS & Average Values |
75%
Level 2: 60%+
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