Physics - Current Electricity Question with Solution | TestHub

Resistors $4\Omega ,6\Omega$ and $12\Omega$ are connected in parallel, its equivalent resistance $(R)$ is given by

$$\frac{1}{R}=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}\Rightarrow R=\frac{12}{6}=2\Omega$$
Again $R$ is connected to $1.5\mathrm{V}$ battery whose internal resistance $r=1\Omega$.
Equivalent resistance now,
$$R^{\prime }=2\Omega +1\Omega =3\Omega$$
Current, $I_{\text{total }}=\frac{V}{R^{\prime }}=\frac{1.5}{3}=\frac{1}{2}\mathrm{A}$$$\begin{array}{r}\begin{array}{rl}& I_{\text{total }}=\frac{1}{2}=3x+2x+x=6x\\ & \Rightarrow x=\frac{1}{12}\end{array}\end{array}$$$\therefore$ Current through $4\Omega$ resistor $=3x$$$=3\times \frac{1}{12}=\frac{1}{4}\mathrm{A}$$
Therefore, rate of Joule heating in the $4\Omega$ resistor
$$=I^{2}R={\left(\frac{1}{4}\right)}^2\times 4=\frac{1}{4}=0.25\mathrm{W}$$