The current passing through the battery in the given circuit, is:
Answer: (B) $0.5$ A
A and F are joined by a plain wire, and so are C and D. The network between B and E therefore has:
$R_{BA} = 5\ \Omega$, $R_{BD} = 2.5\ \Omega$ (through C), $R_{AE} = 3\ \Omega$ (through F), $R_{DE} = 1.5\ \Omega$, and $R_{AD} = 6\ \Omega$ as the bridge.
Check the Wheatstone condition:
$$\frac{R_{BA}}{R_{BD}} = \frac{5}{2.5} = 2, \qquad \frac{R_{AE}}{R_{DE}} = \frac{3}{1.5} = 2$$
The bridge is balanced, so no current flows through the $6\ \Omega$ resistor.
$$R_{BE} = (5 + 3) \parallel (2.5 + 1.5) = \frac{8 \times 4}{8 + 4} = \frac{8}{3}\ \Omega$$
Total resistance in the battery loop:
$$R = \frac{8}{3} + 1.5 + 5.5 + \frac{1}{3} = 3 + 7 = 10\ \Omega$$
$$I = \frac{5}{10} = 0.5\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics