- 1(current)
Q. No.The net resistance of the circuit between \(A\) and \(B\) is:
| 1. | \(\dfrac{8}{3}~\Omega\) | 2. | \(\dfrac{14}{3}~\Omega\) |
| 3. | \(\dfrac{16}{3}~\Omega\) | 4. | \(\dfrac{22}{3}~\Omega\) |
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| 1. | \(7R\) | 2. | \(5R\) |
| 3. | \(4R\) | 4. | \(3R\) |
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In the circuit shown in the figure below, the current supplied by the battery is:
1. \(2~\text A\)
2. \(1~\text A\)
3. \(0.5~\text A\)
4. \(0.4~\text A\)
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In a Wheatstone bridge, all four arms have equal resistance \(R.\) If the resistance of the galvanometer arm is also \(R,\) the equivalent resistance of the combination is:
| 1. | \(R/4\) | 2. | \(R/2\) |
| 3. | \(R\) | 4. | \(2R\) |
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In the circuit shown in the figure below, if the potential difference between \(B\) and \(D\) is zero, then value of the unknown resistance \(X\) is:
| 1. | \(4~\Omega\) | 2. | \(2~\Omega\) |
| 3. | \(3~\Omega\) | 4. | EMF of a cell is required to find the value of \(X\) |
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Three resistances \(\mathrm P\), \(\mathrm Q\), and \(\mathrm R\), each of \(2~\Omega\) and an unknown resistance \(\mathrm{S}\) form the four arms of a Wheatstone bridge circuit. When the resistance of \(6~\Omega\) is connected in parallel to \(\mathrm{S}\), the bridge gets balanced. What is the value of \(\mathrm{S}\)?
| 1. | \(2~\Omega\) | 2. | \(3~\Omega\) |
| 3. | \(6~\Omega\) | 4. | \(1~\Omega\) |
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| 1. | \(0.2~\text{A}\) | 2. | \(0.1~\text{A}\) |
| 3. | \(2.0~\text{A}\) | 4. | \(1.0~\text{A}\) |
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For the network shown in the figure below, the value of the current \(i\) is:
1. \(\frac{18V}{5}\)
2. \(\frac{5V}{9}\)
3. \(\frac{9V}{35}\)
4. \(\frac{5V}{18}\)
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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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Five equal resistances each of resistance \(R\) are connected as shown in the figure below. A battery of \(V\) volts is connected between \(A\) and \(B\). The current flowing in \(AFCEB \) will be:
1. \(\dfrac{V}{R}\)
2. \(\dfrac{V}{2R}\)
3. \(\dfrac{2V}{R}\)
4. \(\dfrac{3V}{R}\)
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- 1(current)
Q. No.
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