What should be the bill for the month of March for a heater of resistance 60.5 Ω connected to 220 V mains? The cost of energy is Rs.2.5 per kW h and the heater is used for 3 hours daily.

Given,

Resistance (R) of the heater = 60.5 Ω

Potential difference (V) = 220 V.

The power (P) of the heater is

$$\begin{gathered} \mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}} \\ \mathrm{P}=\frac{220\times 220}{60.5}=\frac{48400}{60.5} \\ =\frac{48400}{60.5} \\ \mathrm{P}=800 \mathrm{W} \end{gathered}$$

Energy consumed by the heater in a day = P × t

= 800 W × 3 h

= 2400 W h

There are 31 days in the month of March

∴ Energy consumed by the heater in 31 days

= 2400 × 31

= 74400 Wh = 74.4 kWh

⇒ For 1 kWh, the charge is 2.5 rupees.

So, for 74.4 kWh, the charge will be = 74.4 × 2.5

= 186 rupees.