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<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/SPH_ABT_PHY_QB_XII_C02_E05_016_S01.png" width="80%"> <br> Given `C_(1)=C_(2)=C_(3)=9xx10^(-12)pF` <Br> (a) For a series of combination of 'n' identical capacitors, their equivalent capacitance <br> `C_(S)=(C)/(n)=(9xx10^(-12))/(3)=3xx10^(-12)F` <br> i.e., `C_(S)=3pF` <br> (b) `V_(1):V_(2):V_(3)::(1)/(C_(1)):(1)/(C_(2)):(1)/(C_(3))` <Br> i.e., `V_(1):V_2:V_(3)::(1)/(C):(1)/(C):(1)/(C)` <br> i.e., `V_(1):V_(2):V_(3)::1:1:1` <br> `therefore V_(1)=V_(2)=V_(3)=(1)/(3)xx120V` <br> `V_(1)=V_(2)=V_(3)=40V`. <br> Thus voltage across each of the capacitance is 40 V.Related Video

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