A galvanometer of resistance 90 , is shunted by a resistance of 10 , to form an ammeter. This ammeter is connected in series with a resistor of 21 , and an ideal battery of 3 V . The current flowing through the galvanometer coil is:

Step-by-step solution

The equivalent resistance of the ammeter (galvanometer and shunt in parallel) is: R_A = G S G + S = 90 10 90 + 10 = 900 100 = 9 , The total resistance of the circuit is: R_ eq = R + R_A = 21 + 9 = 30 , The main current in the circuit is: I = V R_ eq = 3 30 = 0.1 A = 100 mA Using the current divider rule, the current through the galvanometer coil is: I_g = I S G + S = 100 10 90 + 10 = 100 10 100 = 10 mA Answer: 10 mA