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Karnataka Board PUCPUC Science Class 11

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.

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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.

Sum
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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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Chapter 32: Electric Current in Conductors - Exercises [Page 202]

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HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 32 Electric Current in Conductors
Exercises | Q 63 | Page 202

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