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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

  1. A1
  2. B4 5
  3. C5 4
  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   &

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