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(B) The circuit is a Wheatstone bridge. When the galvanometer shows no deflection,the bridge is balanced.In a balanced Wheatstone bridge,the ratio of

Vedclass · Accepted Answer

$0.6$

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(B) The circuit is a Wheatstone bridge. When the galvanometer shows no deflection,the bridge is balanced.In a balanced Wheatstone bridge,the ratio of resistances in the arms must be equal: $\frac{R_1}{R_2} = \frac{R_3}{R_4}$.Here,the resistances are $8\,\Omega, 2\,\Omega, 20\,\Omega,$ and $5\,\Omega$. Note that $\frac{8}{2} = 4$ and $\frac{20}{5} = 4$.Since the bridge is balanced,the total current $I = 2.1\,A$ divides into two parallel branches: the upper branch $(8\,\Omega + 2\,\Omega = 10\,\Omega)$ and the lower branch $(20\,\Omega + 5\,\Omega = 25\,\Omega)$.Let $I_1$ be the current through the upper branch and $I_2$ be the current through the lower branch.Using the current divider rule,the current through the lower branch $(I_2)$ is:$I_2 = I 	imes \frac{R_{	ext{upper}}}{R_{	ext{upper}} + R_{	ext{lower}}} = 2.1 	imes \frac{10}{10 + 25} = 2.1 	imes \frac{10}{35} = 2.1 	imes \frac{2}{7} = 0.3 	imes 2 = 0.6\,A$.

In the given figure,when the galvanometer shows no deflection,the current (in $A$) flowing through the $5\,\Omega$ resistance will be:

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