The current passing through the battery in the given circuit is:

1. Identify the Balanced BridgeThe central part of the network (nodes A-B-D-E) contains a $6\Omega$ resistor connected across two parallel branches. We check the resistance ratios of the arms:Top Branch Ratio: $\frac{R_{AB}}{R_{BE}}=\frac{5\Omega }{2.5\Omega }=2$ Bottom Branch Ratio: $\frac{R_{AD}}{R_{DE}}=\frac{3\Omega }{1.5\Omega }=2$ Because the ratios are equal ($\frac{5}{2.5}=\frac{3}{1.5}$), the bridge is balanced. This means the potential at node B is equal to the potential at node D. Consequently, no current flows through the $6\Omega$ resistor, and it can be removed from the calculation.2. Calculate Equivalent Resistance ($R_{eq}$)With the $6\Omega$ resistor removed, the inner network simplifies to two parallel series-arms:Upper Arm ($R_u$): $5\Omega +2.5\Omega =7.5\Omega$ Lower Arm ($R_l$): $3\Omega +1.5\Omega =4.5\Omega$ The equivalent resistance of this bridge section ($R_{bridge}$) is:
$$R_{bridge}=\frac{R_u\times R_l}{R_u+R_l}=\frac{7.5\times 4.5}{7.5+4.5}=\frac{33.75}{12}=2.8125\Omega$$
3. Find Total Circuit Resistance ($R_{total}$)Now, add the bridge resistance to the remaining series resistors in the outer loop ($1.5\Omega$, $5.5\Omega$, and $1/3\Omega$):
$$R_{total}=1.5+2.8125+5.5+0.3333\approx 10.1458\Omega$$
4. Apply Ohm’s LawUsing the battery voltage $V=5\text{ V}$:
$$I=\frac{V}{R_{total}}=\frac{5}{10.1458}\approx 0.4928\text{ A}$$
Rounding to the nearest significant figure provided in the options:$I\approx 0.5\text{ A}$ Correct Option: (2)