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Question

Two capacitors of capacitance 1 μF and 2 μF are charged to potential difference 20 V and 15 V as shown in figure. If now terminal B and C are connected together while terminal A is connected with positive terminal of battery and terminal D of capacitor connected with negative terminal of battery with an emf 30 V.


The final charges on 1 μF and 2 μF capacitors respectively will be;

A
203 μC, 803 μC
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B
503 μC, 803 μC
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C
1207 μC, 807 μC
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D
506 μC, 403 μC
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Solution

The correct option is B 503 μC, 803 μC

Before connection, charge on each capacitor:

q1=C1V1=1×20=20 μC
q2=C2V2=2×15=30 μC


Let the battery supplies charge q in circuit and the charges on capacitor will be as shown in figure. So after connection distribution of charge will be


Applying KVL in loop

(20+qC1)(30+qC2)+30=0

[taking in clockwise direction]

(20+q)1(30+q)2+30=0

2(20+q)+(30+q)=60

q=103 μC

Therefore, changes in capacitors are,

q1=20+q=20103=503 μC

Similarly, charge on 2 μF

q2=30+q

q2=30103=803 μC

Hence, option (b) is correct.
Why this question?
Tip:In questions involving two capacitors and a battery connected in parallel with them, either we can apply KVL in loop orΣΔVcapacitor=ΔVBattery
i.e., V1+V2=VBattery

Caution: While traversing in a closed loop, always follow one convention. If we consider crossing from +ve plate of capacitor to -ve plate, it can be given negative sign if potential drop is considered negative throughout.


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