Question

Resistors $P,Q$ and $R$ in the circuit have equal resistances. If the battery is supplying a total power of $12\text{W}$, what is the power dissipated as heat in resistor $R$ ?

• A. $2\text{W}$
• B. $6\text{W}$
• C. $3\text{W}$
• D. $8\text{W}$

Answer 1

The correct option is A $2\text{W}$

Let resistors $P,Q$ and $R$ have resistance $r$.

The effective resistance across the source is
$R_{eff}=r+\frac{\left(r\right)\left(r\right)}{r+r}=r+\frac{r}{2}=\frac{3r}{2}$

So, net power supplied by battery is given by

$I_s^2{R}_{eff}=12$

$\Rightarrow I_s=\sqrt{\frac{12}{R_{eff}}}=\sqrt{\frac{8}{r}}\text{A}$

Since $Q$ and $R$ have equal resistance $r$, each drawn a current of $I$ which is given by

$I=\frac{1}{2}{I}_s=\sqrt{\frac{2}{r}}\text{A}$

Heat dissipation in $R$ can now be determined and is given by

$I^{2}r=\left(\frac{2}{r}\right)r=2\text{W}$

Hence, option $\left(a\right)$ is right.