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Question

Five capacitors are connected as shown in the figure. Initially, all capacitors are uncharged and S is open. When S is closed, then the potential difference between points M and N is : [assume steady state to be achieved after S closes]

A
14 V
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B
12 V
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C
10 V
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D
15 V
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Solution

The correct option is B 12 VAfter switch S is closed, all the capacitor and both batteries form a closed loop. Let that higher emf battery supplies a charge q in circuit. Now applying KVL in the circuit in anti-clockwise path, (All the capacitor will have same q charge ) ΣΔV=0 ⇒−(q4)−(q2)−(q4)−7−(q6)−(10q12)+31=0 ⇒3q+6q+3q+2q+10q12=24 ⇒24q=24×12 ∴q=24×1224=12 μC Now, considering the path M to N along direction of charge flow; VM−(q4)−7−(q6)=VN ⇒VM−VN=5q6+7 ∴VM−VN=(5×1212)+7=12 V Why this question?Concept: When a higher emf battery ispresent in a closed loop, we assume charge to besupplied in circuit from higher emf battery.Tip: In order to avoid any confusion, its better toapply KVL along direction of charge flow.

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