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:

A \(5-\)ampere fuse wire can withstand a maximum power of \(1\) watt in a circuit. The resistance of the fuse wire is:

The total power dissipated in watts in the circuit shown below is:

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:
    

If power dissipated in the 9 $\mathrm{\Omega }$ resistor in the circuit shown is 36 W, the potential difference across the 2 $\mathrm{\Omega }$ resistor will be:

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. \(10\%\)

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

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:
     

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: