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 |
The total power dissipated in watts in the circuit shown below is:
1. | 16 W | 2. | 40 W |
3. | 54 W | 4. | 4 W |
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. | \(0.4~\Omega\) |
Two cities are \(150~\text{km}\) apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is \(8\) volts and the average resistance per km is \(0.5~\text{ohm}\). The power loss in the wire is:
1. \(19.2~\text{W}\)
2. \(19.2~\text{kW}\)
3. \(19.2~\text{J}\)
4. \(12.2~\text{kW}\)
A current of \(3~\text{A}\) flows through the \(2~\Omega\) resistor shown in the circuit. The power dissipated in the \(5~\Omega\) resistor is:
1. | \(4~\text{W}\) | 2. | \(2~\text{W}\) |
3. | \(1~\text{W}\) | 4. | \(5~\text{W}\) |
If power dissipated in the 9 resistor in the circuit shown is 36 W, the potential difference across the 2 resistor will be:
1. 8 V
2. 10 V
3. 2 V
4. 4 V
If the voltage across a bulb rated \((220~\text{V}\text-100~\text{W})\) drops by \(2.5\%\) of its rated value, the percentage of the rated value by which the power would decrease is:
1. \(20\%\)
2. \(2.5\%\)
3. \(5\%\)
4. \(10\%\)
Power consumed in the given circuit is \(P_1\). On interchanging the position of \(3~\Omega\) and \(12~\Omega\) resistances, the new power consumption is \(P_2\). The ratio of \(\frac{P_2}{P_1}\) is:
1. | \(2\) | 2. | \(1 \over 2\) |
3. | \(3 \over 5\) | 4. | \(2 \over 5\) |
The power dissipated across the 8 Ω resistor in the circuit shown here is 2 W. The power dissipated in watts across the 3 Ω resistor is:
1. | 2.0 | 2. | 1.0 |
3. | 0.5 | 4. | 3.0 |
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