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Obtain the equivalent capacitance of the netwrok in the figure given below. For a 300V supply, determine the charge and voltage across each capacitor. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/SPH_ABT_PHY_QB_XII_C02_E05_017_Q01.png" width="80%">

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Answer Text

Solution :
`C_(S)=(C_(2)C_(3))/(C_(2)+C_(3))` for a series combination of `C_(2) and C_(3)`. <br> i.e., `C_(S)=(200pF)/(2)=100pF` <Br> Capacitors `C_(1) and C_(S)` are parallel to each other. <br> `C_(P)=C_(1)+C_(S)=100pF+100pF=200pF` <br> the equivalent circuit is written as, <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/SPH_ABT_PHY_QB_XII_C02_E05_017_S01.png" width="80%"> <br> The equivalent capacitance of the circuit <br> `C_(S)=(200xx100)/(200+100)pF=(200xx1cancel(00))/(3cancel(00))pF` <br> `therefore C_(S)=(200)/(3)pF` <br> Total charge, `Q=C_(S)'V=(200)/(3)xx10^(-12)xx300` <br> i.e., `Q=2xx10^(-8)C` <br> The charges on `C_(p) and C_(4)` are the same <br> i.e., `Q_(p)=Q_(4)=2xx10^(-8)C` <br> The charge `2xx10^(-8)C` branches out across the `C_(1),C_(2) and C_(3)` combination <br> `therefore Q_(1)=Q_(2)=Q_(3)=1xx10^(-8)C and Q_(4)=2xx10^(-8)C` <br> Voltage drop in `C_(1)(=100pF)=(Q_(1))/(C_(1))=(1xx10^(-8))/(100xx10^(-12))=100V` <br> Voltage drop in 200 pF `(C_(2)&C_(3))=(1xx10^(-8))/(200xx10^(-12))V=50V` <br> Voltage drop in `C_(4)=(2xx10^(_8))/(100xx10^(-12))=200V`.

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