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 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 each of the four arms of a wheatstone bridge is 10Omega and the resistance of the galvanomter is 500Omega . What is the equivalent resistance of the combination gt

The resistances of the four arms P, Q, R and S in a Wheatstone's bridge are 10Omega, 30Omega, 30Omega and 90Omega , respectively. The emfs and internal resistances of the cell are 7V and 5Omega respectively. If the galvanometer resistance is 50Omega , the current drawn from the cell will be

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?

In a wheatstone bridge resistance of each of the four sides is 10Omega . If the resistance of the galvanometer is also 10Omega , then effective resistance of the bridge will be

In a wheatstone bridge resistance of each of the four sides is 10Omega . If the resistance of the galvanometer is also 10Omega , then effective resistance of the bridge 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 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 of a wheatstone bridge are 100Omega,10Omega,300Omegaand 30Omega respectively . A battery of emf 1.5 V and negligible internal resistance is connected to the bridge . Calculate the current flowing through each resistance .