Question

A heater A gives out $300\text{W}$ of heat when connected to a $200\text{V}$ d.c. supply. A second heater B gives out $600\text{W}$ when connected to a $200\text{V}$ d.c. supply. If a series combination of the two heaters is connected to a $200\text{V}$ d.c. supply the heat output will be (in watt).

Answer 1

Given,

For $\text{A}\to P_1=300\text{W};V_1=200\text{volt}$

For $\text{B}\to P_2=600\text{W};V_2=200\text{volt}$

We know that, $P=\frac{V^2}{R}$

Then, $R_{\text{A}}=\frac{(200{)}^2}{300}$

$R_{\text{B}}=\frac{(200{)}^2}{600}$

In series combination,

$R_{\text{eq}}=R_{\text{A}}+R_{\text{B}}$

Now net power in series is,

$P_{\text{eq}}=\frac{V^2}{R_{\text{eq}}}$

$P_{\text{eq}}=\frac{(200{)}^2}{\frac{(200{)}^2}{300}+\frac{(200{)}^2}{600}}$

$P_{\text{eq}}=200\text{watt}$