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Question
A capacitor of capacitance 8.0 μF is connected to a battery of emf 6.0 V through a resistance of 24 Ω. Find the current in the circuit (a) just after the connections are made and (b) one time constant after the connections are made.
Solution
Given:-
Capacitance, C = 8 μF
Emf of the battery, V= 6 V
Resistance, R = 24
(a) Just after the connections are made, there will be no charge on the capacitor and, hence, it will act as a short circuit. Current through the circuit,
\[i = \frac{V}{R} = \frac{6}{24} = 0 . 25 A\]
(b) The charge growth on the capacitor,
\[q = Q \left( 1 - e^{- \frac{t}{RC}} \right)\]
One time constant = RC = 8 × 24 = 192 × 10-6 s
For t = RC, we have:-
\[q = Q . \left( 1 - e^\frac{- RC}{RC} \right)\]
\[ \Rightarrow q = CV\left( 1 - e^{- 1} \right)\]
\[ \Rightarrow q = 8 \times {10}^{- 6} \times 6 \times 0 . 632\]
\[ = 3 . 036 \times {10}^{- 5} C\]
\[V = \frac{Q}{C} = \frac{3 . 036 \times {10}^{- 5}}{8 \times {10}^{- 6}} = 3 . 792 V\]
Applying KVL in the circuit, we get:-
E = V + iR
⇒ 6 = 3.792 + 24i
⇒ i = 0.09 A
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