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Capacitance of capacitor `C_(1)` is 100 pF. <br> Capacitance of capacitor `C_(2)` is 200 pF <br> Capacitance of capacitor `C_(3)` is 200 pF <br> Capacitance of capacitor `C_(4)` is 100 pF <br> Supply potential, V=300 V <br> Capacitor `C_(2)` and `C_(3)` are connected in series. let their equivalent capacitance be C''. <br> `:. 1/(C')=1/200+1/200=2/200` <br> C'=100 pF <br> Capacitors `C_(1)` and C' are in parallel. let their equivalent capacitances be C'' <br> `:. C''=C'+C_(1)` <br> `=100+100=200 pF` <br> `C''` and `C_(4)` are connected in series . let their equivalent capacitance be C. <br> `:. 1/C=1/(C'')+1/(C_(4))` <br> `=1/200+1/100=(2+1)/200` <br> `C=200/3 pF` <br> Hence, the equivalent capacitance of the circuit is `200/3 pf` <br> potential difference across C''=V'' <br> potential difference across `C_(4)=V_(4)` <br> `:. V''+V_(4)=V=300V` <br> Charge on `C_(4)` is given by <br> `Q_(4)=CV` <br> `=200/3xx10^(-12) xx300` <br> `=2xx10^(-8)C` <br> `:. V_(4)=(Q_(4))/(C_(4))` <br> `=(2xx10^(-8))/(100xx10^(-12)) =200 V` <br> `:.` Voltage across `C_(1)` is given by <br> `V_(1)=V-V_(4)` <br> =300-200=100 V <br> Hence , potential difference `V_(1)` across `C_(1)` is 100 V <br> charge on `C_(1)` is given by <br> `Q_(1)=C_(1)V_(1)` <br> `=100xx10^(-12)xx100` <br> `=10^(-8)C` <br> `C_(2)` and `C_(3)` having same capacitances have a potential difference of 100 V together. since `C_(2)` adn `C_(3)` are in series, the potential difference across `C_(2)` adn `C_(3)` given by <br> `V_(2)=V_(3)=50 V` <br> Therefore, charge on `C_(4)` is given by <br> `Q_(2)=C_(2)V_(2)` <br> `=200xx10^(-12) xx50 ` <br> `=10^(-8)C` <br> And charge on `C_(3)` is given by <br> `Q_(3)=C_(3)V_(3)` <br> `=200xx10^(-12)xx50` <br> `=10^(-3)C` <br> Hence, the equivalent capacitance of the given circuit is `200/3 pF` with .Related Video

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