JEE Main20241 Feb 2024Evening ShiftPhysicsCurrent ElectricityActual
In an ammeter, 5 % of the main current passes through the galvanometer. If resistance of the galvanometer is G , the resistance of ammeter will be:
Options
- AG 20
- BG 199
- C199 G
- D200 G
Correct answer
A. G 20
Step-by-step solution
As potential drop across both branches would be the same, we get I S S = I g G 95 100 I S = 5 I 100 G ⇒ S = G 19 Now, resistance of ammeter will be R A = S G S + G = G 2 19 20 G 19 ⇒ R A = G 20