Analyze the circuit given in Figure in the steady-state condition. Charge on the capacitor in this state is $q_{0} = 16 μ C$
a. Find the current in each branch.
b. Find the emf of the battery.
c. If in the beginning, the battery is removed and nodes
A and C are shortened, then find the duration in which charge on the capacitor becomes $5.92 μ C$ .

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Solution

For parts (a) and (b), we have
For BAD, $V_{B} + 2 (i - i_{1}) + 1 (i - i_{1}) - 4 i_{1} = V_{D}$
or $3 i - 7 i_{1} = - 4$
For BCD, $6 i - 10 i_{1} = 4$ or $i_{1} = 3 A, i = 17 / 3 A$
$C u r r e n t i n A C = i_{1} = 3 A . I n A B C, i - i_{1} = 8 / 3 A$
$E = 8 i_{1} = 24 V$
$i = 2 i_{3} = i_{1} + i_{2}$ (i)
$\frac{q}{C} = 2 i_{1} + i_{l} + 4 i_{3} = 3 i_{1} + 4 i_{3}$ (ii)
Also $\frac{q}{C} = 3 i_{2} + 3 i_{2} + 4 i_{3} = 6 i_{2} + 4 i_{3}$ (iii)
From Eqs. (ii) and (iii), we have $6 i_{2} = 3 i_{1}$ or $i_{1} = 2 i_{2}$
Solve
to get $i = \frac{q}{4 C}$ or $\frac{- d q}{d t} = \frac{q}{4 C}$ or $q = q_{0} e^{- t / 4 c}$
or $5.92 = 16 e^{t} \frac{}{4 \times 4 \times 10^{- 6}}$ or $t = 16 μ s$

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