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 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

In a Wheatstone bridge resistance connected the bridge is balanced , when the resistance are in the ratio

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

A currrent of 0.1 A enters a Wheatstone bridge from a source of the electricity . The resistance of each of the first three arms of the bridge is 10Omega and the resistance of its fourth arm is 20Omega . If the resistance of the galvanometer be 100Omega what will be the galvanometer current ?

The resistance of each arm of Wheatstone Bridge is 10 Omega . A resistance of 10 Omega is connected in series with the galvanometer then equivalent resistance across battery will be

The resistance of each arm of the wheat stone bridge is 10Omega . A resistance of 10Omega is connected in series with galvanometer then the equivalent resistance across the battery will be:-

The resistance of each arm of the wheat stone bridge is 10Omega . A resistance of 10Omega is connected in series with galvanometer then the equivalent resistance across the battery will be:-

The resistance of each arm of the wheat stone bridge is 10Omega . A resistance of 10Omega is connected in series with galvanometer then the equivalent resistance across the battery will be

The resistance of each arm of the wheat stone bridge is 10Omega . A resistance of 10Omega is connected in series with galvanometer then the equivalent resistance across the battery 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