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, $I_{\text{total }}=\frac{V}{R^{\prime }}=\frac{1.5}{3}=\frac{1}{2}\mathrm{A}$ $$\begin{array}{rl}& I_{\text{total }}=\frac{1}{2}=3x+2x+x=6x\\ & \Rightarrow x=\frac{1}{12}\end{array}$$ $\therefore$ 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 $$=I^{2}R={\left(\frac{1}{4}\right)}^2\times 4=\frac{1}{4}=0.25\mathrm{W}$$