MHT CET202521 Apr 2025Evening ShiftPhysicsCurrent ElectricityActual
In the circuit shown in the figure, P R . The reading of the galvanometer remains the same with switch 'S' open or closed. Then
Options
- AI _ Q = I _ G
- BI _ Q = I _ R
- CI _ R = I _ G
- DI_P=I_G
Correct answer
C. I _ R = I _ G
Step-by-step solution
The galvanometer current I_G remains identical whether switch S is open or closed. When S is open, the galvanometer current equals the current through resistor R, so I_R = I_G . Closing S connects nodes B and D at potential V_X . Applying Kirchhoff's laws yields: I_P P = I_R R I_Q Q = I_G G I_P + I_R = I_Q + I_G Equating the open and closed expressions for I_G : V_ AC R+G = V_ AC Q(P+R) PR(Q+G) + QG(P+R) Simplifying gives P Q = R G . Substituting this ratio into the current relations confirms I_R = I_G holds in bot