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  • In the circuit shown in figure,,<img alt="\mathrm{C}_{1}=1 \mu \mathrm{F}, \mathrm{C}_{2}=2 \mu \m

In the circuit shown in figure,\mathrm{R}_{1}=10 \Omega,\mathrm{C}_{1}=1 \mu \mathrm{F}, \mathrm{C}_{2}=2 \mu \mathrm{F} and \mathrm{\mathrm{E}=6 \mathrm{V}}.Then charge on capacitor 1 in steady state is

Option: 1

-2 \mu \mathrm{C}


Option: 2

2 \mu \mathrm{C}


Option: 3

5 \mu \mathrm{C}


Option: 4

12 \mu \mathrm{C}


Answers (1)

best_answer

In steady state no current flow through capacitors. Therefore charge on each capacitor remains constant. Let, in steady state, the circuit draw a current I from the battery and let chare on capacitors be \mathrm{q_{1}} and \mathrm{q_{2}} as shown in the figure.

Applying Kirchoff's voltage law on mesh ABCEFGHA,


\mathrm{\mathrm{IR}_{2}-\mathrm{E}+\mathrm{IR}_{1}=0}
\mathrm{or \, \mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=2 \mathrm{~A}}

Now applying KVL on the mesh ABJHA,
\mathrm{\frac{\mathrm{q}_{1}}{\mathrm{C}_{1}}+\mathrm{IR}_{1}=0 \: or \: \mathrm{q}_{1}=-2 \mu \mathrm{C}}

 

Posted by

vishal kumar

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