InterviewSolution
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InterviewSolution
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For the given capacitor configuration (a) Find the charges on each capacitor (b) potential difference across them ( c) energy stored in each capacitor . |
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Answer» <P> Solution :Capacitor b and c in parallel combination`C_(p) =C_(b) + C_(c) = (6+2) mu F = 8 muF `  Capacitor a, `c_(p)` and d are in series combination , so the resulatant copacitance `(1)/(C_(s))=(1)/(C_(a))+(1)/(C_(cp))+(1)/(C_(d))=(1)/(8) +(1)/(8) +(1)/(8) = (3)/(8)` `C_(s)= (8)/(3) muF `  (a) CHARGE on each capacitor , Charge on capacitor a `Q_(a) =C_(s) V= (8)/(3) xx9` `Q_(a)= 24muC` Capacitor b and c in parallel Charge on capacitor `b , Q_(b) =(6)/(3)xx9=18` `Q_(b) = 18muC` Charge on capacitor` c , Q_(c) = (2)/(3)xx9=6 ` `Q_(c) = 6 muC` (b) Potential difference across each capacitor `V= (q)/(C)` Capacitor `C_(a), V_(a) =(q_(a))/(C_(a))= (24xx10^(-6))/(8xx10^(-6))=3V` Capacitor `C_(b), V_(b)= (qb)/(C_(b))= (18xx10^(-6))/(6xx10^(-6))=3V` Capacitor `C_(c), V_(c) = (q_(c))/(C_(c))= (6xx10^(-6))/(2xx10^(-6))= 3 V ` Capacitor `C_(d), V_(d)= (q_(d))_/(C_(d))= (24xx10^(-6))/(8xx10^(-6))=3V ` (c) Energy stores in a capacitor `U=(1)/(2) CV^(2)` Energy in capacitor `C_(a), U_(a)=(1)/(2) C_(a)V_(a)^(2)=(1)/(2)xx8xx10^(-6) xx(3)^(2)` `U_(a)= 36 muj` Capacitor `C_(b), U_(b)= (1)/(2)C_(b)V_(b)^(2)=(1)/(2)xx6xx10^(-6)xx(3)^(2)` `U_(a)=27 muj` Capacitor `C_(c) , U_(c)=(1)/(2)C_(c)V_(c)^(2)=(1)/(2)xx2xx10^(-6)xx(3)^(2)` `U_(a)=9muJ ` Capacitor `C_(d), U_(d)=(1)/(2)C_(d)V_(d)^(2)=(1)/(2)xx8xx10^(-6)xx(3)^(2)` `U_(a)36 muJ` |
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