Two capacitors of capacitances `10muF and 20muF` are connected in series across a potential difference of 100V. The potential difference across each c
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
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Correct Answer - 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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