Question

A capacitor is being charged from a battery and through a resistor of 10 kΩ. It is observed that the voltage on the capacitor rises to 80% of its maximal value in 4 seconds. Calculate the capacitor's capacitance.

Answer 1

`C = 2.48 * 10^(-4) Farad`

Step-by-step explanation:

As per the equation of voltage on capacitor we know that

`V = V_(max)(1 - e^{-(t)/(\tau)})`

now we know that voltage reached to its 80% of maximum value in 4 second time

so we will have

`0.80 V_(max) = V_(max)(1 - e^{-(4)/(\tau)})`

`0.20 = e^{-(4)/(\tau)}`

`-(4)/(\tau) = ln(0.20)`

`-(4)/(\tau) = -1.61`

`\tau = 2.48`

as we know that

`\tau = RC`

`(10 k ohm)(C) = 2.48`

`C = 2.48 * 10^(-4) Farad`