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Flag question Marked out of 7.00 Not yet answered Question 13 For the system of capacitors shown in the figure C1 = 2F.C2 = 4
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Answer #1

Given that,

\Delta V=12V

C_{1}=2F

C_{2}=4F

C_{3}=3F

Net capacitance on C1 and C2 is,

C12 = C1 + C2 = 2 +4= 6

Total capacitance in a circuit is given as ,

C_{T}=\frac{C_{12}\times C_{3}}{C_{12}+C_{3}}

\Rightarrow C_{T}=\frac{6\times 3}{6+3}=2F

Total charge is given as,

Q_{T}=C_{T}\times \Delta V=2\times 12=24C

Voltage on C3 is gven as,

V_{3}=\frac{Q_{T}}{C_{3}}=\frac{24}{3}=8V

Voltage on C1 = Voltage on C2 ( Because of parallel path)

\Rightarrow V_{1}=V_{2}=\Delta V-V_{3}=12-8=4V

\Rightarrow V_{1}=V_{2}=4V

Now charge on C1 is given as,

Q_{1}=C_{1}\times V_{1}=2\times 4=8C

\Rightarrow Q_{1}=8C

Hence (a) 8C is correct option

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