Physics - Current Electricity Question with Solution | TestHub

Key Idea : Voltage across $2\Omega$ is same as voltage across arm containing $1\Omega$ and $5\Omega$ resistances. Voltage across $2\Omega$ resistance, $V=2\times 3=6\mathrm{V}$ So, voltage across lowest arm, $V_1=6\mathrm{V}$ Current across $5\Omega ,I=\frac{6}{1+5}=1\mathrm{A}$ Thus, power across $5\Omega$, $P=I^{2}R=(1{)}^2\times 5=5\mathrm{W}$