Physics - Current Electricity Question with Solution | TestHub

The power dissipated in the circuit. $$\begin{array}{c}P=\frac{V^2}{R_{\mathrm{eq}}}.....(i)\\ v=10\mathrm{volt}\\ \frac{1}{R_{\mathrm{eq}}}=\frac{1}{R}+\frac{1}{5}=\frac{5+R}{5R}\\ R_{\mathrm{eq}}=\left(\frac{5R}{5+R}\right)\\ P=30\mathrm{W}\end{array}$$ Substituting the values in equation (i) $$\begin{array}{c}30=\frac{(10{)}^2}{\left(\frac{5R}{5+R}\right)}\\ \frac{15R}{5+R}=10\Rightarrow 15R=50+10R\\ 5R=50\Rightarrow R=10\Omega \end{array}$$