Two batteries, one of emf \(18\) volts and internal resistance \(2~\Omega\) and the other of emf \(12\) V and internal resistance \(1~\Omega,\) are connected as shown. The voltmeter \(\mathrm{V}\) will record a reading of:
1. \(18\) V
2. \(30\) V
3. \(14\) V
4. \(15\) V
A \(5-\)ampere fuse wire can withstand a maximum power of \(1\) watt in a circuit. The resistance of the fuse wire is:
1. | \(5\) \(\Omega\) | 2. | \(0.04~\Omega\) |
3. | \(0.2~\Omega\) | 4. | \(0.4~\Omega\) |
For the network shown in the figure below, the value of the current i is:
1.
2.
3.
4.
A car battery of emf \(12~\text{V}\) and internal resistance \(5\times 10^{-2}~\Omega\) receives a current of \(60~\text{A}\) from an external source. The terminal voltage of the battery is:
1. | \(12~\text{V}\) | 2. | \(9~\text{V}\) |
3. | \(15~\text{V}\) | 4. | \(20~\text{V}\) |
If there are two bulbs of (\(40~\text{W},200~\text{V}\)), and (\(100~\text{W},200~\text{V}\)), then the correct relation for their resistance is:
1. | \(\mathrm{R}_{40}<\mathrm{R}_{100}\) |
2. | \(\mathrm{R}_{40}>\mathrm{R}_{100}\) |
3. | \(\mathrm{R}_{40}=\mathrm{R}_{100}\) |
4. | no relation can be predicted |
When three identical bulbs are connected in series, the consumed power is 10 W. If they are now connected in parallel then the consumed power will be:
1. 30 W
2. 90 W
3. \(\frac{10}{3}\) W
4. 270 W
The current in 8 Ω resistance is (in the figure below):
1. 0.69 A
2. 0.92 A
3. 1.30 A
4. 1.6 A
The terminal potential difference of a cell is greater than its emf when:
1. | A battery of less emf is connected in its series. |
2. | A battery of higher emf is connected in its series. |
3. | A battery of higher emf is connected in its parallel. |
4. | A battery of less emf is connected in its parallel. |
A battery is charged at a potential of 15 V for 8 hours when the current flowing is 10 A. The battery on discharge supplies a current of 5 A for 15 hours. The mean terminal voltage during discharges is 14 V. The "Watt hour" efficiency of the battery is:
1. 80%
2. 90%
3. 87.5%
4. 82.5%
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. V/R
2. V/2R
3. 2V/R
4. 3V/R