The resistances of the four arms (picture) of Wheatstone bridge are as follows - AB = 100 Omega, BC = 10 Omega, CD ...

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?

The four arms of a Wheatstone bridge (Fig. 3.26) have the following resistances: AB = 100 Omega, BC = 10 Omega, CD = 5Omega, and DA = 60Omega . A galvanometer of 15W resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.

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 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

Compare the heat developed in the four arms of a balanced. Wheatstone bridge. The resistance of the arms are 10Omega ,100Omega,40Omega and 400Omega . Also indicate in which arm the heat produced will be the minimum and in which the maximum.