Physics - Current Electricity Question with Solution | TestHub

$P=\frac{V^2}{R_{\mathrm{eq}}}$ $\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R}+\frac{1}{5}=\frac{5+R}{5R}{R}_{\mathrm{eq}}=\left(\frac{5R}{5+R}\right)P=30\mathrm{W}$ Substituting the values in equation (i) $$30=\frac{(10{)}^2}{\left(\frac{5R}{5+R}\right)}\Rightarrow R=10\Omega$$