Three capacitors of capacitance 2 pF, 3 pF and 4 pF are connected in parallel. (a) what is the total capacitance of the combination ? (b) Determine the charge on each capacitor, If the combination is connected to 100V supply.

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Answer Text

Solution :
(a) capactiances of the given capacitors are <br> `C_(1)=2pF` <br> `C_(2)=3pF` <br> `C_(3)=4pF` <br> For the parallel combination of the capacitors, equivalent capacitors C' is given by the algebric sum, <br> `C'=2+3+4=9 pF` <br> Therefore, total capacitance of the combination is 9 pf <br> (b) supply voltage , V=100 V <br> The voltage through all the three capacitor is same V=100 V <br> Charge on the capacitor of capacitance C and potential difference V is given by the relation, <br> q=Vc...(i) <br> For C=2pF <br> Charge =VC =`100xx2=200 pC=2xx10^(-10)C` <br> for C=3pF, <br> Charge =VC =100 x 3 =300 pC =`3xx10^(-10)C` <br> For C=4pF, ltBrgt charge =VC=100 x 4 =200 pC `=4xx10^(-10) c`

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