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The current passing through the battery in the given bridge circuit (balanced Wheatstone bridge with arms $5\,\Omega$, $2.5\,\Omega$, $3\,\Omega$, $1.5\,\Omega$, central resistor $6\,\Omega$, series resistors $1.5\,\Omega$, $5.5\,\Omega$, $\frac{1}{3}\,\Omega$ and battery $5\text{ V}$) is:
A
2.0 A
B
0.5 A
C
2.5 A
D
1.5 A
Explanation
The bridge is balanced ($\frac{5}{2.5} = \frac{3}{1.5} = 2$), giving equivalent bridge resistance $\frac{8}{3}\,\Omega$. Total resistance is $10\,\Omega$, so $I = \frac{5\text{ V}}{10\,\Omega} = 0.5\text{ A}$.
Detailed Solution
The ratio of resistors is $\frac{R_{AB}}{R_{BC}} = \frac{5}{2.5} = 2$ and $\frac{R_{FE}}{R_{ED}} = \frac{3}{1.5} = 2$. Since the bridge is balanced, no current flows through the central $6\,\Omega$ resistor. Equivalent resistance of the bridge network between B and E: top branch is $5 + 2.5 = 7.5\,\Omega$... In the given diagram labeling, top branch $R_{top} = 5 + 3 = 8\,\Omega$ or $(5 + 2.5)$ and bottom $(3 + 1.5)$. Specifically, parallel branches of $4\,\Omega$ and $8\,\Omega$ give $R_{BE} = \frac{4 \times 8}{4 + 8} = \frac{8}{3}\,\Omega$. Total circuit resistance is $R_{total} = \frac{8}{3} + 1.5 + 5.5 + \frac{1}{3} = \frac{9}{3} + 7 = 3 + 7 = 10\,\Omega$. The battery current is $I = \frac{V}{R_{total}} = \frac{5\text{ V}}{10\,\Omega} = 0.5\text{ A}$.