[" In the Wheatstone's bridge shown,"P=2Omega,Q=3Omega,R=6Omega" and "],[S=8Omega" .In order to obtain balance,shunt resistance across "S" must be "]

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

In a Wheatstone bridge circuit, P=5 Omega, Q=6 Omega, R=10 Omega and S= 5 Omega . Find the additional resistance to be used in series with S, so that the bridge is balanced.

In Wheatstone bridge, the resistances in four arms are 10Omega,10Omega,10Omega and 20Omega . To make the bridge balance, resistance connected across 20Omega is

In a Wheatstone's network, P=2Omega, Q=2Omega, R=2Omega and S=3Omega . The resistance with which S is to be shunted in order that the bridge may be balanced is

In a Wheatstone's network , P=2Omega, Q=2Omega, R=2Omega and S=3Omega. The resistance with which S is to be shunted in order that the bridge may be balanced .

In a Wheatstone's network, P=2Omega, Q=2Omega, R=2Omega and S=3Omega . The resistance with which S is to be shunted in order that the bridge may be balanced is

In a Wheatstone's network, P=2Omega, Q=2Omega, R=2Omega and S=3Omega . The resistance with which S is to be shunted in order that the bridge may be balanced is

In figure. P=3Omega,Q=2Omega,R=6Omega,S=4Omega and X=5Omega . Calculate the current I.

In Wheatstone network, P=2Omega , Q=2 Omega , R=2Omega and S = 3 Omega . The resistance with which is to be shunted in order that the bridge may be balanced is: