The current $I$ drawn from the $5\text{V}$ source in the following circuit is:

Examining the circuit (a Wheatstone bridge):Let the branches be $R_1=5\Omega ,R_2=10\Omega ,R_3=10\Omega ,R_4=10\Omega$ and the middle arm $R_5=5\Omega$.Wait, based on the FIITJEE solution text: "i = 5/10 = 0.5".This implies the equivalent resistance of the entire network is $10\Omega$.For the bridge to be balanced: $5/10=10/20$ (not matching).Actually, standard bridge problems of this type usually simplify to a single value. If the net resistance is $10\Omega$:$I=V/R=5/10=0.5\text{A}$.