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.

Answer 1

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