QJEEM18Phdce2: JEE Main 2018 Physics - Class 12 JEEM18Phdce2 Solution: A capacitor C 1 = 1.0 F is charged up to a voltage V = 60 V by…

Given : C₁ = 1.0 $\mu$ F; C₂ = 3.0 $\mu$ F; C₃ = 6.0 $\mu$ F

Capacitance C₂ and C₃ are in series, their equivalent capacitance is

${1 \over C} = {1 \over {{C_2}}} + {1 \over {{C_3}}} \Rightarrow {1 \over C} = {{{C_2} + {C_3}} \over {{C_2}{C_3}}}$

$\Rightarrow C = {{{C_2}{C_3}} \over {{C_2} + {C_3}}} = {{3.0\mu F \times 6.0\mu F} \over {3.0\mu F + 6.0\mu F}} = 2\mu F$

Charge on C₁ when battery is connected to switch (1) is

$= {C_1} \times 60$ ..... (1)

Charge when battery is connected to switch (2) is

$= ({C_1} + C)V'$ ...... (2)

Equating Eq. (1) and (2), we get

$({C_1} + C)V' = 60{C_1} \Rightarrow V' = {{60{C_1}} \over {{C_1} + C}}$

$\Rightarrow V' = {{60 \times 1\mu F} \over {1\mu F + 2\mu F}} = {{60 \times 1\mu F} \over {3\mu F}} = 20\,V$

Final charge on C₂ and C₃ is $Q = CV' = 2\mu F \times 20 = 40\mu C$

Therefore, $Q = 40\mu C$