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 :

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, $$Q = 40\mu C$$

Answer

40 $$\mu $$C

Answer

36 $$\mu $$C

Answer

20 $$\mu $$C

Answer

54 $$\mu $$C

Electrostatics & Capacitance

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