As shown in the figure, the resistance of a galvanometer G can be found by the half-deflection method. Here the resistance R_2 is adjusted such that when the key K is closed the deflection in the galvanometer becomes half of the value as compared to when K is open. Half-deflection is obtained at R_2 = 4\ and thus the galvanometer resistance is found to be 6\ . In this half-deflection condition the current (in mA) through the resistor R_1 is:
A. A
Let the resistance of the galvanometer be G and the series resistance be R_1. When the key K is open, the current through the galvanometer is: I_1 = V R_1 + G When the key K is closed, the total resistance of the circuit becomes: R_ eq = R_1 + G R_2 G + R_2 The total current from the battery is I = V R_ eq . The current through the galvanometer is: I_2 = I R_2 G + R_2 = V R_1 + G R_2 G + R_2 R_2 G + R_2 = V R_2 R_1(G + R_2) + G R_2 In the half-deflection condition, I_2 = I_1 2 : V R_2 R_1(G + R_2) + G R_2 = V 2(R_1 + G) Cross-multiplying and simplifying: 2 R_2 (R_1 + G) = R_1(G + R_2) + G R_2 2 R_1 R_2 + 2 G R_2 = R_1 G + R_1 R_2 + G R_2 R_1 R_2 + G R_2 = R_1 G R_1(G - R_2) = G R_2 R_1 = G R_2 G - R_2 Given G = 6\ and R_2 = 4\ , we can find R_1: R_1 = 6 4 6 - 4 = 24 2 = 12\ In the half-deflection condition (key K closed), the total equivalent resistance of the circuit is: R_ eq = 12 + 6 4 6 + 4 = 12 + 2.4 = 14.4\ The current through the resistor R_1 is the total current from the battery: I = V R_ eq = 10 14.4 = 100 144 = 25 36 \ A Converting the current to mA: I = 25 36 1000\ mA = 25000 36 \ mA 694.44\ mA
Related: Physics — Current Electricity · All PYQ Banks