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.

Answer 2

It would take 2.8 ms for the capacitor to reach 90% of its final charge

What is time constant?

The time constant is given by:

τ = Resistance * Capacitance

Resistance = 10000 Ω, Capacitance = 12 μF, hence:

τ =  10000 * 12 * 10⁻⁶ = 0.12

The time is given by:

$0.9Q=Q(1-e^\frac{-t}{\tau} )\\\\t=2.8\ ms$

It would take 2.8 ms for the capacitor to reach 90% of its final charge

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