Question

Two identical heaters rated $220V$, $1000W$ are placed in series with each other across $220V$ line; then the combined power is

• A. $1000W$
• B. $2000W$
• C. $500W$
• D. $4000W$

Answer 1

The correct option is C $500W$

Let the resistance of the two heaters be denoted by $R_1$ and $R_2$, respectively.

Thus,$R_1=\frac{V^2}{P_1}$ and $R_2=\frac{V^2}{P_2}$

As the heaters are connected in series so net resistance $R_n=R_1+R_2$

or $\frac{V^2}{P_n}=\frac{V^2}{P_1}+\frac{V^2}{P_2}$

or $\frac{1}{P_n}=\frac{1}{P_1}+\frac{1}{P_2}$

or $P_n=\frac{P_1{P}_2}{P_1+P_2}=\frac{1000\times 1000}{1000+1000}=500W$