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spin [16.1K]
3 years ago
13

A 2.0 µF capacitor is charged through a 50,000 ohm resistor. How long does it take for the capacitor to reach 90% of full charge

?
Physics
1 answer:
Nesterboy [21]3 years ago
4 0

Answer:

0.23 s

Explanation:

First of all, let's find the time constant of the circuit:

\tau=RC

where

R=50,000 \Omega is the resistance

C=2.0\mu F=2.0\cdot 10^{-6}F is the capacitance

Substituting,

\tau=(50,000 \Omega)(2.0\cdot 10^{-6}F)=0.1 s

The charge on a charging capacitor is given by

Q(t)=Q_0 (1-e^{-t/\tau} ) (1)

where

Q_0 is the full charge

we want to find the time t at which the capacitor reaches 90% of the full charge, so the time t at which

Q(t)=0.90 Q_0

Substituting this into eq.(1) we find

0.90 Q_0 = Q_0 (1-e^{-t/\tau})\\0.90=1-e^{-t/\tau}\\e^{-t/\tau}=1-0.90=0.10\\-\frac{t}{\tau}=ln(0.10)\\t=-\tau ln(0.10)=(0.1 s)ln(0.10)=0.23 s

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