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A capacitor of capacitance 10 μF is connected to a battery of emf 2 V. It is found that it takes 50 ms for the charge on the capacitor to become 12.6 μC. Find the resistance of the circuit.

Radhika Batra , 10 Years ago
Grade 11
anser 1 Answers
Kevin Nash

Last Activity: 10 Years ago

Sol. C = 10 μF = 10^–5 F, emf = 2 V t = 50 ms = 5 * 10^–2 s, q = Q(1 – e^–t/RC) Q = CV = 10^–5 * 2 q = 12.6 * 10^–6 F ⇒ 12.6 * 10–6 = 2 * 10^–5 (1 e^-5 10 2 /R 10 5 ) ⇒12.6 * 10^-6/2 * 10^-5 = 1 – e^-5*10^-2 /R*10^-5 ⇒1 – 0.63 = -^-5*10^3 /R ⇒ -5000/R = In 0.37 ⇒ R = 5000/0.9942 = 5028 Ω = 5.028 * 10^3Ω = 5 KΩ.

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