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Question

Two capacitors of 1μF and 2μF are connected in series and this combination is changed upto a potential difference of 120 volt. What will be the potential difference across 1μF capacitor:

A
40volt
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B
60volt
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C
80volt
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D
120volt
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Solution

The correct option is C 80volt
Given, c1=1μf,c2=2μf,PD=120v

Ceq=c1c2c1+c2=2×12+1=23μf

We know, Q=cv Where Q is the charge, C is the capacitance of the capacitor and v is the potential difference.

Now, Qnet in circuit is equivalent capacitance of capacitors attached in the circuits is multiplied by PD

Qnet=12×23=80

Q=cv=1μfv=80v=80v

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