MHT CET Medical202621 April 2026Morning ShiftPhysicsCurrent ElectricityActual
A galvanometer of resistance 90 is shunted by a resistance of 10 . The fraction of main current that passes through shunt is
Options
- A9 / 10
- B9 / 100
- C1 / 100
- D1 / 10
Correct answer
A. 9 / 10
Step-by-step solution
Let the main current be I , the current through the shunt be I_s , and the current through the galvanometer be I_g . Since the galvanometer and the shunt are connected in parallel, the potential difference across them is equal. I_s S = I_g G Using I = I_s + I_g , we can write I_g = I - I_s . I_s S = (I - I_s) G I_s (S + G) = I G I_s I = G S + G Given G = 90 and S = 10 : I_s I = 90 10 + 90 = 90 100 = 9 10 Answer: 9 / 10