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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 V
After 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

VMVN=5q6+7

VMVN=(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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