Physics - Current Electricity Question with Solution | TestHub

Wire has a resistance $1\Omega {\mathrm{cm}}^{-1}$. $\therefore$ Resistance of $40\mathrm{cm}$ of wire $=40\Omega$ and resistance of $60\mathrm{cm}$ of wire $=60\Omega$ No current flows through the galvanometer since the bridge in balanced. Hence the branch containing the galvanometer can be removed. For balanced bridge $\frac{4}{40}=\frac{Y}{60}$ $$\therefore \mathrm{Y}=\frac{4\times 60}{40}=6\Omega$$ $4\Omega$ and $6\Omega$ are in series. Their equivalent resistance is $10\Omega$. $40\Omega$ and $60\Omega$ are in series. Their equivalent resistance is $100\Omega$. $10\Omega$ and $100\Omega$ are in parallel Their equivalent resistance is $\frac{1000}{110}\Omega$. $\therefore$ The current $I=\frac{\mathrm{V}}{\mathrm{R}}=\frac{6\times 110}{1000}=0.66\mathrm{A}$