In the figure, the four arms of a Wheatstone bridge have the following resistances AB = 100Omega, BC = 10Omega, C...

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.

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.

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 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

Three resistance P, Q, R each of 2 Omega and an unknown resistance S from the four amrs of a Wheatstone's bridge circuit. When a resistance of 6 Omega is connected in parallel to S the bridge gets balanced. What is the value of S ?