In a Wheatstone bridge P = Q = 10 ohm and R = S = 15 `Omega` and G = 20 `Omega` and a cell of e.m.f. 1.5 V is connected in circuit. The current drawn from the cell is:-

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.

When a resistance of 12 Omega is connected with a cell of emf 1.5 V, 0.1A current flows through the resistance. Internal resistance of the cell is—

The resistance of the four arms A, B, C and D of a wheatstone bridge as shown in the figure are 10 Omega , 20 Omega , 30 Omega and 60 Omega respectively . The emf and internal resistance of the cell are 15 V and 2.5 Omega respectively . If the galvanometer resistance is 80 Omega then the net current drawn from the cell will be

The resistance of the four arms A, B, C and D of a wheatstone bridge as shown in the figure are 10 Omega , 20 Omega , 30 Omega and 60 Omega respectively . The emf and internal resistance of the cell are 15 V and 2.5 Omega respectively . If the galvanometer resistance is 80 Omega then the net current drawn from the cell will be

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 four resistances of a wheatstone brige are 1Omega each. When the bridge is balaced,the cell has an emf of 1 volt.The current drawn from the cell is

The emf of a cell E is 15 V as shown in the figure with an internal resistance of 0.5Omega . Then the value of the current drawn from the cell is