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Two capacitors of capacitances 10muF and...

Two capacitors of capacitances `10muF and 20muF` are connected in series across a potential difference of 100V. The potential difference across each capacitor is respectively

A

66.67 V, 33.33 V

B

60 V, 40 V

C

50 V, 50 V

D

90 V, 10 V

Text Solution

Verified by Experts

The correct Answer is:
A

`V_(1)=Q/C_(1)=(VC_(2))/(C_(1)+C_(2))=(100xx20xx10^(-6))/(30xx10^(-6))`
`=66.67V`
`therefore" "V_(2) = V-V_(1)=100-66.67=33.33V`
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NIKITA PUBLICATION-ELECTROSTATICS-Multiple Choice Questions
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  16. The equivalent capacity between the points A and B in the following fi...

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  17. The p.d. across the capacitance of 2muF in the figure along with is

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  18. A circuit is shown in the figure below. Find out the charge of the con...

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  19. Four condensers each of capacity 4muF are connected as shown in the ad...

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