Two identical heater filaments are connected first in parallel and then in series. At the same applied voltage, the ratio of heat produced in same time for parallel to series will be:

Let's consider the power $P$ dissipated by a single heater filament with resistance $R$ when connected to a voltage $V$. <br/><br/> The power dissipated by the heater filament is given by: <br/><br/> $P = \frac{V^2}{R}$ <br/><br/> When the two identical heater filaments are connected in parallel, the equivalent resistance $R_P$ is given by: <br/><br/> $$\frac{1}{R_P} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \Rightarrow R_P = \frac{R}{2}$$ <br/><br/> The total power dissipated $P_P$ by the two filaments connected in parallel is: <br/><br/> $P_P = \frac{V^2}{R_P} = \frac{V^2}{\frac{R}{2}} = 2V^2 \cdot \frac{1}{R} = 2P$ <br/><br/> When the two identical heater filaments are connected in series, the equivalent resistance $R_S$ is given by: <br/><br/> $R_S = R + R = 2R$ <br/><br/> The total power dissipated $P_S$ by the two filaments connected in series is: <br/><br/> $$P_S = \frac{V^2}{R_S} = \frac{V^2}{2R} = \frac{1}{2}V^2 \cdot \frac{1}{R} = \frac{1}{2}P$$ <br/><br/> Now, let's find the ratio of the heat produced in the same time for parallel to series connection: <br/><br/> $$\frac{P_P}{P_S} = \frac{2P}{\frac{1}{2}P} = \frac{2P \times 2}{P} = \frac{4P}{P} = 4 : 1$$