Question

A capacitor of capacitance 500 μF is connected to a battery through a 10 kΩ resistor. The charge stored in the capacitor in the first 5 s is larger than the charge stored in the next.
(a) 5 s
(b) 50 s
(c) 500 s
(d) 500 s

Answer 1

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D) 500 s.............!!!!

Answer 2

The charge stored in the capacitor in the first 5 s is larger than the charge stored in the next 5,50,500 Sec.

Explanation:

Step 1:

The charge of the capacitor is given by $\mathrm{Q}=\mathrm{CV}\left(1-\mathrm{e}^{-\mathrm{t} / \mathrm{R}}\right)$ .

C = Capacitance of the given Capacitor = 500 μF

R = Resistance connected in the circuit = 10 kΩ

Step 2:

The charge stored in the capacitor in the first $5 s=\mathrm{Q}_{0}=\mathrm{CV}\left(1-\mathrm{e}^{-5 / 5}\right)=\mathrm{CV} \times 0.632$

(Where $\mathrm{RC}=10 \times 10^{3} \times 500 \times 10^{-6}=5 \mathrm{S}$)

Step 3:

Charge developed in next 500 S = Q1 = CV (1- 0.632) = 0.368 CV

Thus we can say that charge developed on the capacitor in first 5 SEC is greater than the charge developed by the next given seconds.