MHT CET202619 April 2026Evening ShiftPhysicsCurrent ElectricityActual
When a galvanometer is shunted by a resistance 'S', its current capacity increases 'n' times. If the same galvanometer is shunted by another resistance ' S^1 ', its current capacity will increase to ' n^1 '. The value of n in terms of n^1 , S and S^1 is
Options
- An^1 + S S^1
- BS(n^1 - 1) - S^1 S
- C(n^1 + 1)S^1 S
- DS + S^1(n^1 - 1) S
Correct answer
D. S + S^1(n^1 - 1) S
Step-by-step solution
Let the resistance of the galvanometer be G and its full scale deflection current be I_g . When shunted by S , the new current capacity is I = n I_g . Since the galvanometer and the shunt are in parallel, the potential difference across them is equal: I_g G = (I - I_g) S I_g G = (n I_g - I_g) S G = (n - 1) S When shunted by S^1 , the new current capacity is I^1 = n^1 I_g . Similarly, the potential difference across them is equal: I_g G = (I^1 - I_g) S^1 I_g G = (n^1 I_g - I_g) S^1 G = (n^1 - 1) S^1 Equating the two