For the arrangement shown in the figure- the key is closed at t - 0- C2 is initially uncharged while C1 has a charge of 2- C-a Find the current coming out of the battery just after switch is closed-b Find the charge on the capacitors in the steady state condition-

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Question

For the arrangement shown in the figure, the key is closed at $t = 0$ . $C_{2}$ is initially uncharged while $C_{1}$ has a charge of $2 μ C$ .
(a) Find the current coming out of the battery just after switch is closed.
(b) Find the charge on the capacitors in the steady state condition.

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Solution

a) At t=0, the resistance of uncharged capacitor is zero we can treat charged capacitors $C_{1}$ as
EMF where EMF $= \frac{q}{C} = \frac{2}{1} = 2 V$
Net EMF of circuit $= 9 - 2 = 7 V$ or
Net EMF of circuit $= 9 + 2 = 11 V$
Net Resistance $= 30 + \frac{30 \times 60}{30 + 60} = 50 Ω$
Current at t=0 would be maximum and
Current $= \frac{N e t E M F}{N e t R e s i s t a n c e}$
Current $\frac{11}{50} A o r \frac{7}{50}$
b) In steady condition ,no current will flow through circuit , $C_{2}$ will get short circuited while potential difference across $C_{1}$ will be 9V
So $Q_{2} = 0, Q_{1} = C V = 1 \times 9 = 9 μ C$
As all resistance will be short circulated at steady state .

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For the arrangement shown in the figure, the key is closed at t=0. C2 is initially uncharged while C1 has a charge of 2 μC.(a) Find the current coming out of the battery just after switch is closed.(b) Find the charge on the capacitors in the steady state condition.

For the arrangement shown in the figure, the key is closed at $t = 0$ . $C_{2}$ is initially uncharged while $C_{1}$ has a charge of $2 μ C$ .
(a) Find the current coming out of the battery just after switch is closed.
(b) Find the charge on the capacitors in the steady state condition.