Four resistances arranged to form a Wheatstone network are `8Omega, 12Omega,6Omega` and `27Omega` resistance, so that the bridge is balanced, is

Four resistances arranged to form a Wheatstone's network are 8Omega, 12Omega, 6 Omega, and 27Omega . The resistance that should be connected across 27Omega resistance to balance the bridge is

For resistances arranged to form a wheatstone's network are 10 Omega, 15 Omega, 6 Omega and 36 Omega . What resistance should be connected across the 36 Omega resistance to balance the bridge ?

The magnitudes of the resistances in the four arms of a wheatstone bridge are 10 Omega , 12Omega , 20Omega and 24Omega Respectively and the galvanometer resistance is 100Omega If current is sent through the bridge from a cell of emf 1.5V and 1.3Omega of Internal resistance, calculate the current through the '10 ohm' resistance.

The resistance of the four arms of a Wheatstone bridge are 1Omega , 3Omega , 2Omega and 6 Omega respectively and the resistance of the galvanometer is 100Omega . The equivalent resistance of the combination is

The resistance of the four arms of a Wheatstone bridge are P= 10Omega , Q = 100Omega , R= 40Omega and S =10Omega . What resistance in series or parallel with the last one will be required to obtain no deflection in the galvanometer?

Four resistance 4,4,4 and 12Omega form a Wheatstone’s network. Find the resistance when connected across 12Omega resistance will balance the network?

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