Question 363: Electrostatics & Capacitance

Question

A capacitor C
1
= 1.0 $\mu $F is charged up to a voltage V = 60 V by connecting it to battery B through switch (1). Now C
1
is disconnected from battery and connected to a circuit consisting of two uncharged capacitors ${C_2} = 3.0\mu F$ and C
3
= 6.0 $\mu $F through switch (2), as shown in the figure. The sum of final charges on C
2
and C
3
is :
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l7wzkitg/81b9dba5-f89f-43e1-aba4-24b868f75d11/50a19040-319f-11ed-b3ea-45214d807530/file-1l7wzkith.png?format=png)

Accepted Answer

Given : C
1
= 1.0 $\mu$F; C
2
= 3.0 $\mu$F; C
3
= 6.0 $\mu$F
Capacitance C
2
and C
3
are in series, their equivalent capacitance is
${1 \over C} = {1 \over {{C_2}}} + {1 \over {{C_3}}} \Rightarrow {1 \over C} = {{{C_2} + {C_3}} \over {{C_2}{C_3}}}$
$ \Rightarrow C = {{{C_2}{C_3}} \over {{C_2} + {C_3}}} = {{3.0\mu F 	imes 6.0\mu F} \over {3.0\mu F + 6.0\mu F}} = 2\mu F$
Charge on C
1
when battery is connected to switch (1) is
$ = {C_1} 	imes 60$ ..... (1)
Charge when battery is connected to switch (2) is
$ = ({C_1} + C)V'$ ...... (2)
Equating Eq. (1) and (2), we get
$({C_1} + C)V' = 60{C_1} \Rightarrow V' = {{60{C_1}} \over {{C_1} + C}}$
$ \Rightarrow V' = {{60 	imes 1\mu F} \over {1\mu F + 2\mu F}} = {{60 	imes 1\mu F} \over {3\mu F}} = 20\,V$
Final charge on C
2
and C
3
is $Q = CV' = 2\mu F 	imes 20 = 40\mu C$
Therefore, $Q = 40\mu C$
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l32mogyk/82aecc69-5618-41fb-adb1-736104bc25e8/86c161c0-d1bc-11ec-9218-efb2cf12c71b/file-1l32mogyl.png?format=png)

Answer

40 $\mu $C

Answer

36 $\mu $C

Answer

20 $\mu $C

Answer

54 $\mu $C

Electrostatics & Capacitance

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