JEE Main 2023 — Current Electricity Question with Solution

Let's consider the power $$P$$ dissipated by a single heater filament with resistance $$R$$ when connected to a voltage $$V$$.

The power dissipated by the heater filament is given by:

$$P=\frac{V^2}{R}$$

When the two identical heater filaments are connected in parallel, the equivalent resistance $$R_P$$ is given by:

$$\frac{1}{R_P}=\frac{1}{R}+\frac{1}{R}=\frac{2}{R}\Rightarrow R_P=\frac{R}{2}$$

The total power dissipated $$P_P$$ by the two filaments connected in parallel is:

$$P_P=\frac{V^2}{R_P}=\frac{V^2}{\frac{R}{2}}=2V^2\cdot \frac{1}{R}=2P$$

When the two identical heater filaments are connected in series, the equivalent resistance $$R_S$$ is given by:

$$R_S=R+R=2R$$

The total power dissipated $$P_S$$ by the two filaments connected in series is:

$$P_S=\frac{V^2}{R_S}=\frac{V^2}{2R}=\frac{1}{2}{V}^2\cdot \frac{1}{R}=\frac{1}{2}P$$

Now, let's find the ratio of the heat produced in the same time for parallel to series connection:

$$\frac{P_P}{P_S}=\frac{2P}{\frac{1}{2}P}=\frac{2P\times 2}{P}=\frac{4P}{P}=4:1$$