Q. No.Consider the circuit shown in the figure below. The current \(I_3\) is equal to:
1. \(5\) A
2. \(3\) A
3. \(-3\) A
4. \(\frac{-5}{6}\) A
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For measurement of potential difference, the potentiometer is preferred in comparison to the voltmeter because:
1. the potentiometer is more sensitive than the voltmeter.
2. the resistance of the potentiometer is less than
the voltmeter.
3. the potentiometer is cheaper than the voltmeter.
4. the potentiometer does not take current from the circuit.
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In the Wheatstone's bridge (shown in the figure below) \(X=Y\) and \(A>B\). The direction of the current between \(a\) and \(b\) will be:
| 1. | from \(a\) to \(b\). |
| 2. | from \(b\) to \(a\). |
| 3. | from \(b\) to \(a\) through \(c\). |
| 4. | from \(a\) to \(b\) through \(c\). |
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A resistance of 4 Ω and a wire of length 5 metres and resistance 5 Ω are joined in series and connected to a cell of e.m.f. 10 V and internal resistance 1 Ω. A parallel combination of two identical cells is balanced across 300 cm of the wire. The e.m.f. E of each cell is:
1. 1.5 V
2. 3.0 V
3. 0.67 V
4. 1.33 V
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In the circuit given below, the emf of the cell is \(2\) volt and the internal resistance is negligible. The resistance of the voltmeter is \(80\) ohm. The reading of the voltmeter will be:
1. \(0.80\) volt
2. \(1.60\) volt
3. \(1.33\) volt
4. \(2.00\) volt
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In the circuit shown below, \(E_1 = 4.0~\text{V}\), \(R_1 = 2~\Omega\), \(E_2 = 6.0~\text{V}\), \(R_2 = 4~\Omega\) and \(R_3 = 2~\Omega\). The current \(I_1\) is:
1. \(1.6\) A
2. \(1.8\) A
3. \(1.25\) A
4. \(1.0\) A
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The potential difference across \(8~\Omega\) resistance is \(48~\text V\) as shown in the figure below. The value of potential difference across \(X\) and \(Y\) points will be:
1. \(160~\text V\)
2. \(128~\text V\)
3. \(80~\text V\)
4. \(62~\text V\)
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What is the equivalent resistance between terminals \(A\) and \(B\) of the network?
| 1. | \(\dfrac{57}{7}~\Omega\) | 2. | \(8~\Omega\) |
| 3. | \(6~\Omega\) | 4. | \(\dfrac{57}{5}~\Omega\) |
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The effective resistance between points \(P\) and \(Q\) of the electrical circuit shown in the figure is:
| 1. | \(\frac{2 R r}{\left(R + r \right)}\) | 2. | \(\frac{8R\left(R + r\right)}{\left( 3 R + r\right)}\) |
| 3. | \(2r+4R\) | 4. | \(\frac{5R}{2}+2r\) |
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\(12\) cells each having the same emf are connected in series with some cells wrongly connected. The arrangement is connected in series with an ammeter and two similar cells which are in series. Current is \(3~\text{A}\) when cells and battery aid each other and is \(2~\text{A}\) when cells and battery oppose each other. The number of cells wrongly connected is/are:
1. \(4\)
2. \(1\)
3. \(3\)
4. \(2\)
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