QJEEM23Phb100: JEE Main 2023 Physics - Class 12 JEEM23Phb100 Solution: Two identical heater filaments are connected first in parallel…

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:

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