In the Wheatstone's bridge shown, $P = 2\,\Omega ,$ $Q = 3\,\Omega ,$ $R = 6\,\Omega $ and $S = 8\,\Omega $. In order to obtain balance, shunt resistance across '$S$' must be .............. $\Omega$
Medium
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(d) Let the value of shunt be $r$. Hence the equivalent resistance of branch containing $S$ will be $\frac{{Sr}}{{S + r}}$
In balance condition, $\frac{P}{Q} = \frac{{Sr/(S + r)}}{R}$. This gives $r = 8\,\Omega $
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