The charge flowing across the cell on closing the switch is equal to:
Correct answer: B. CV/2
- A. CV
- B. CV/2
- C. 2CV
- D. zero
Explanation
When the switch is kept open: Total capacitance (C) = ( 1/C + 1/C)-1 = (2/C)-1 = C/2 Total charge (Q) = total capacitance x total voltage = C/2 x V = CV/2 When the switch is closed: {Capacitor 2 gets short circuited and all the current flows through the switch path} Total capacitance (C') = C Total charge (Q') = C' x V = CV Charge flowing through cell (ΔQ) = Q' - Q = CV - CV/2 ΔQ = CV/2 .
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