JEE Main20238 Apr 2023Morning ShiftPhysicsCurrent ElectricityActual
In this figure the resistance of the coil of galvanometer G is 2 Ω . The emf of the cell is 4 V . The ratio of potential difference across C 1 and C 2 is
Options
- A1
- B4 5
- C5 4
- D3 4
Correct answer
B. 4 5
Step-by-step solution
Current flowing through both capacitors will be zero at steady state. The potential difference across capacitor C 1 will be the sum of the potential difference across the resistor 6   Ω and the galvanometer resistance. Mathematically, V C 1 = i 6 Ω + R G       . . . 1 Similarly, the potential difference across capacitor C 2 will be the sum of the potential difference across the resistor 8   Ω and the galvanometer resistance. Mathematically, V C 2 = i R G + 8 Ω   &