A galvanometer of resistance 90 is shunted by a resistance of 10 to form an ammeter. This ammeter is then connected in series with a resistor of 41 and an ideal battery of 10 ~V . The current passing through the galvanometer coil is :

Step-by-step solution

First, find the equivalent resistance of the ammeter ( R _ A ): R _ A = R _ g S R _ g + S = 90 10 90 + 10 = 900 100 = 9 The total resistance of the circuit is: R _ eq = R + R _ A = 41 + 9 = 50 The total current drawn from the battery is: I = V R _ eq = 10 50 = 0.2 ~A = 200 ~mA Using the current divider rule, the current passing through the galvanometer coil ( I _ g ) is: I _ g = I S R _ g + S I _ g = 200 10 90 + 10 = 200 10 100 = 20 ~mA Answer: 20 ~mA