Two 500 W heaters when connected in series across a source of constant voltage V supply Q joules of heat in a certain time t. When the heaters are connected in parallel across the same source, the amount of heat energy supplied in the same time t will be

Let R be the resistance of each heater. When they are connected in series, the resistance of the combination is 2 R. Therefore, heat supplied in time t is

                     $Q=\frac{V^{2}t}{2R}$

When they are connected in parallel, the resistance of the combination is R' = R/2. Therefore, heat supplied in time t is 

                   $Q\text{'}=\frac{V^{2}t}{R\text{'}}=\frac{2V^{2}t}{R}$

Thus Q' = 4Q.