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Question
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 of the capacitor to become 12.6 μC. Find the resistance of the circuit.
Solution
Given:-
Capacitance of the capacitor, C = 10 μF = 10−5 F
Emf of the battery, emf= 2 V
Time taken to charge the capacitor completely, t = 50 ms
= 5 × 10−2 s
The charge growth across a capacitor,
`q = Q(1-e^(-t/(RC)))`
Q = CV = 10-5 × 2
q = 12.6 × 10-6F
= `12.6 xx10^(-6) = 2xx10^(-5) (1-e^((-5xx10^-2)/(Rxx10^-5)))`
= `(12.6xx10^-6)/(2xx10^-5) = 1-e^((-5xx10^-2)/(Rxx10^-5))`
= 1- 0.63 = `e^((-5xx10^3)/R)`
`= (-5000)/R` in 0.37
`= R = 5000/0.9942 = 5028Ω`
`=5028 xx 10^3Ω`
= 5kΩ.
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