Calculate I for the given circuit diagram.

It is a balanced Wheatstone bridge. Hence no current flows through resistance of arm BDand the resistance of arm BD is ineffective
$\therefore \quad \frac{R_{A B}}{R_{B C}}=\frac{R_{A D}}{R_{D C}}$ $\Rightarrow \frac{® \Omega}{® \Omega}=\frac{® \Omega}{๑ \Omega}$ Resistance of arm $A B C=5 \Omega+5 \Omega=10 \Omega$ Resistance of arm $A D C=5 \Omega+5 \Omega=10 \Omega$ The effective resistance between $A$ and $C$ is $\frac{1}{R_{A C}}=\frac{1}{10}+\frac{1}{10}$ or $R_{A C}=5 \Omega$ Now the equivalent circuit becomes $\therefore$ The equivalent resistance of the circuit
$\frac{1}{R}=\frac{1}{5}+\frac{1}{5} \quad$ or $\quad R=2.5 \Omega$ Current in the circuit, $I=\frac{V}{R}=\frac{25}{2.5}=10 \mathrm{~A}$