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

Explanation:

It is given in the question,

Capacitance, C = 8 μF

Battery’s Emf, V= 6 V

Resistance, R = 24

(a) The capacitor will have no charge just after the connections are done, henceforth, the capacitor will act as a short circuit. Through the circuit, the current is

i = VR = 624 = 0.25 A

(b) The capacitor’s charge growth,

q = Q 1-e-tRC

One time constant = RC = 192 $\times$ 10-6 s

For t = RC,

q = Q. 1-e-RCRC

⇒q = 8 $\times$ 10-6 $\times$ 6 $\times$ 0.632 = 3.036 $\times$ 10-5 C

V = 3.036 $\times$ 10-6 $\times$ 10-58 = 3.792 V

In the circuit, when KVL is applied, we obtain:

E =  iR + V

⇒ 6 = 24i + 3.792  

⇒ i = 0.09 A