As shown in the figure, the resistance of a galvanometer G can be found by the half-deflection method. Here the resistance R2 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 R2=4 and thus the galvanometer resistance is found to be 6 . In this half-deflection condition the current (in mA) through the resistor R1 is: [figure]
Answer
690 to 710
Solution
Key idea: equate galvanometer current with K open and half of it with K closed to find R1.
K open: Ig= 10R1+6.
K closed: G R2= 6 410=2.4 ; I= 10R1+2.4, Ig'= 410I.
0.4 10R1+2.4= 12 10R1+6 0.8(R1+6)=R1+2.4 R1=12 .
Current through R1: I= 1012+2.4=0.694 A 694 mA.
Answer: 694 (accepted range 690 to 710)