In the following figure,C_1=5 mu F,C_2=C_3=10 | vedclass

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(A) Given: $C_1=5 \mu F$,$C_2=C_3=10 \mu F$,and $\varepsilon=20 	ext{ V}$.When the switch $S$ is connected to point $A$,capacitor $C_1$ is charged by

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$40$

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(A) Given: $C_1=5 \mu F$,$C_2=C_3=10 \mu F$,and $\varepsilon=20 	ext{ V}$.When the switch $S$ is connected to point $A$,capacitor $C_1$ is charged by the battery of $EMF$ $\varepsilon=20 	ext{ V}$.The charge stored in $C_1$ is:$Q = C_1 \varepsilon = 5 \mu F 	imes 20 	ext{ V} = 100 \mu C$.When the switch $S$ is moved to point $B$,the charged capacitor $C_1$ is connected in parallel with the uncharged capacitors $C_2$ and $C_3$.Since they are connected in parallel,the charge $Q$ redistributes among the three capacitors until they reach a common potential difference $V$.The total charge remains conserved:$Q = (C_1 + C_2 + C_3) V$$100 \mu C = (5 \mu F + 10 \mu F + 10 \mu F) V$$100 \mu C = 25 \mu F 	imes V$$V = \frac{100}{25} 	ext{ V} = 4 	ext{ V}$.The charge on capacitor $C_3$ is:$q_3 = C_3 V = 10 \mu F 	imes 4 	ext{ V} = 40 \mu C$.

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