When a galvanometer is shunted with a resistance S , it is found that only 5 % of the main circuit current passes through the galvanometer coil. The resistance of the galvanometer ( G ) in terms of the shunt resistance ( S ) is :

Step-by-step solution

Let the main circuit current be I . The current passing through the galvanometer is I_g = 5 % of I = 5 100 I = I 20 . In a shunted galvanometer, the galvanometer and the shunt resistance are in parallel. The potential difference across them is equal: I_g G = (I - I_g) S Substituting I_g = I 20 : ( I 20 ) G = (I - I 20 ) S I 20 G = 19I 20 S Cancelling I 20 from both sides: G = 19S Answer: 19S