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

Answer:

$C = 2.48 \times 10^{-4} Farad$

Explanation:

As per the equation of voltage on capacitor we know that

$V = V_{max}(1 - e^{-\frac{t}{\tau}})$

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

so we will have

$0.80 V_{max} = V_{max}(1 - e^{-\frac{4}{\tau}})$

$0.20 = e^{-\frac{4}{\tau}}$

$-\frac{4}{\tau} = ln(0.20)$

$-\frac{4}{\tau} = -1.61$

$\tau = 2.48$

as we know that

$\tau = RC$

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

$C = 2.48 \times 10^{-4} Farad$