Question

A bulb rated 60W, 220V is connected across 110V mains lines. The power consumed is –
A) 10W
B) 15W
C) 20W
D) 25W

Answer 1

Hint: We know that the electrical appliances are rated with a power consumption when connected to a given standard potential difference. The power consumption can vary when the voltage supplied to the appliance is deferred due to the current flow variation.

Complete answer:
The electrical appliances use the electrical energy to work out their desired functions. The power consumed by the appliances is dependent on the potential difference applied to it. We know from the Joule’s law of heating that the electrical appliances dissipate a considerable amount of energy in the form of heat. The heat energy is proportional to the voltage drop and the current through the electrical appliance with a particular resistance. The rate of this heat dissipation is the power dissipation of the given appliance. It is given as –
\[P=VI\]
In the present situation, a bulb is given the power rating of 60W at 220V mains supply. We can find the resistance of the bulb by substituting the formula for current from the Ohm’s law as –
\[\begin{align}
  & P=VI \\
 & \text{From Ohm }\!\!'\!\!\text{ s law,} \\
 & I=\dfrac{V}{R} \\
 & \Rightarrow P=\dfrac{{{V}^{2}}}{R} \\
\end{align}\]
We can substitute the given values of power rating and the voltage to find the resistance of the bulb as –
\[\begin{align}
  & P=\dfrac{{{V}^{2}}}{R} \\
 & \Rightarrow R=\dfrac{{{V}^{2}}}{P} \\
 & \Rightarrow R=\dfrac{{{220}^{2}}}{60} \\
 & \Rightarrow R=806.67\Omega \\
\end{align}\]
The resistance of the bulb will be constant even if the voltage is changed. The power consumption due to the voltage of 110V can be found as –
\[\begin{align}
  & P=\dfrac{{{V}^{2}}}{R} \\
 & \Rightarrow P=\dfrac{{{110}^{2}}}{806.67} \\
 & \therefore P=15W \\
\end{align}\]
The power consumed by the bulb when the mains supply is 110V is 15W.

The correct answer is option B.

Note:
The current through the resistor also will vary with the change in the voltage. This is the direct application of Ohm’s law in which the resistance of the material is a constant which balances the change in the current and the voltage variation for an Ohmic resistor.