A 36 V battery with an internal resistance of 4 is connected to two parallel branches. Branch A contains a 21 resistor in series with an ammeter. Branch B contains a 60 resistor. The ammeter consists of a 90 galvanometer coil and a 10 shunt. The current flowing specifically through the galvanometer coil is mA .

Step-by-step solution

The equivalent resistance of the ammeter is: R_A = R_g S R_g + S = 90 10 90 + 10 = 9 Resistance of Branch A is R_ branch A = 21 + 9 = 30 Resistance of Branch B is R_ branch B = 60 The equivalent resistance of the parallel combination of Branch A and Branch B is: R_p = 30 60 30 + 60 = 20 Total resistance of the circuit including internal resistance r is: R_ total = R_p + r = 20 + 4 = 24 Main current drawn from the battery: I_ main = E R_ total = 36 24 = 1.5 A Using the current divider rule, the current through Branc