Question

Two identical heaters rated $220\text{V}-1000\text{W}$ are placed in series with each other across $220\text{V}$ supply line, then the combined power is

• A. $1000\text{W}$
• B. $2000\text{W}$
• C. $500\text{W}$
• D. $4000\text{W}$

Answer 1

The correct option is C $500\text{W}$

Let the resistance of the series combination is denoted by $R_s$ and the corresponding power by $P_s$ then

$R_s=R_1+R_2$

Here, source voltage is equal to rated voltage i.e. $V_{rated}=V$

$\therefore \frac{V^2}{P_s}=\frac{V^2}{P_1}+\frac{V^2}{P_2}$

$\Rightarrow P_s=\frac{P_1{P}_2}{P_1+P_2}=\frac{1000\times 1000}{2000}=500\text{W}$

Hence, option $\left(c\right)$ is correct.