A capacitor of capacitance C = 1 μ F is suddenly connected to a battery of 100 volt through a resistance R = 100 Ω. The time taken for the capacitor to be charged to get 50 V is: [Take ln 2 = 0.69] (a) 1.44 × 10 - 4 s (b) 3.33 × 10 - 4 s (c) 0.69 × 10 - 4 s (d) 0.30 × 10 - 4 s

Answer: (c). V = V₀(1-e-t/RC); 50 = 100(1-e-t/RC) ⇒ e-t/RC = 1/2. t = RCln 2 = (100)(10⁻⁶)(0.69) t = 0.69×10⁻⁴ s, so the answer is (c).

A capacitor of capacitance C = 1 μ F is suddenly connected to a… | JEE Main 2021 (27 Jul, Shift 1)

Q30JEE Main 2021 · 27 JUL · SHIFT 1Medium

Capacitors in DC Circuits & RC

A capacitor of capacitance C = 1 $\mu$ F is suddenly connected to a battery of 100 volt through a resistance R = 100 $\Omega$ . The time taken for the capacitor to be charged to get 50 V is: [Take ln 2 = 0.69]

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Answer: (c)

$V = V_0(1-e^{-t/RC})$ ; $50 = 100(1-e^{-t/RC}) \Rightarrow e^{-t/RC} = \dfrac{1}{2}$ .
$t = RC\ln2 = (100)(10^{-6})(0.69)$
$t = 0.69\times10^{-4}\text{ s}$ , so the answer is (c).

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