c
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 connectedto \(1.5 \,\mathrm{V}\) battery whose internal resistance \(r=1\, \Omega\)

Equivalent resistance now, \(R^{\prime}=2\, \Omega+1 \,\Omega=3\, \Omega\)

Current, \(I_{\text {total }}=\frac{V}{R^{\prime}}=\frac{1.5}{3}=\frac{1}{2} \,\mathrm{A}\)

\(I_{\text {total }}=\frac{1}{2}=3 x+2 x+x=6 x\)

\(\Rightarrow x=\frac{1}{12}\)

\(\therefore \) Current through \(4\, \Omega\) resistor \(=3 x\)

\(=3 \times \frac{1}{12}=\frac{1}{4}\, A\)

Therefore, rate of Joule heating in the \(4 \,\Omega\) resistor

\(=I^{2} R=\left(\frac{1}{4}\right)^{2} \times 4=\frac{1}{4}=0.25 \,\mathrm{W}\)