Answer to Question #92191 in Electric Circuits for Tope Joshua

Question

an electric heater is operated by applying a p.d of 50V to the resistance wire. Calculate the current carried by the wire and power rating of the electric heater. Then explain what would happen to the current and power if the applied voltage to the heater is doubled.

Answer 1

Assume that the resistance of the wire is $R$. Then the current flowing in the wire will be

$$I=\frac{V}{R}=\frac{50}{R}.$$

The power rating will be

$$P=I^2R=\Big(\frac{V}{R}\Big)^2R=\frac{V^2}{R}.$$

As we see from these two expressions, if the voltage is doubled, the current will increase twice and the power will be quadrupled. Thus, increasing the voltage is more effective than decreasing the resistance.