Question

A 0.12-mu or micro FF capacitor, initially uncharged, is connected in series with a 10-kohm resistor and a 12-V battery of negligible internal resistance. Approximately how long does it take the capacitor to reach 90% of its final charge?

Answer 1

Answer:

The time is 2.8 ms.

Explanation:

Given that,

Capacitor = 0.12 μF

Resistance = 10 kohm

Voltage = 12 V

Charge Q = 0.9 Q₀

We need to calculate the time constant

Using formula of time constant

$T=RC$

Put the value into the formula

$T=10\times10^{3}\times0.12\times10^{-6}$

$T=0.0012\ sec$

We need to calculate the time

Using formula of time

$Q=Q_{0}(1-e^{\frac{-t}{T}})$

Put the value into the formula

$0.9Q_{0}=Q_{0}(1-e^{\frac{-t}{0.0012}})$

$ln (0.1)=\dfrac{-t}{0.0012}$

$t=0.00276\ sec$

$t=2.8\ ms$

Hence, The time is 2.8 ms.