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InterviewSolution
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Three capacitors of capacities `8muF,8muFand4muF` are connected in a series and a potential difference of 120 volt is maintained across the combination. Calculate the charge on capacitor of capacity `4muF`. |
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Answer» Let the capacitors be `C_(1)=8muF,C_(2)=8muFandC_(3)=4muF`. All the capacitors are connected in series, so, the equivalent capacitance `C_(s)` can be calculated as `(1)/(C_(s))=(1)/(C_(1))+(1)/(C_(2))+(1)/(C_(3))` `(1)/(C_(s))=(1)/(8)+(1)/(8)+(1)/(4)` `(1)/(C_(s))=(1+1+2)/(8)` `C_(s)=2muF` In series combination, `Q_(1)=Q_(2)=Q_(3)` But `Q=C_(s)V` `:.` Charge on capacitor of capacity `4muF` can be calculated as `Q=2xx10^(-6)xx120` `=240xx10^(-6)C=240muC` |
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