Answer :
How many time constants must elapse before a capacitor is charged?
The equation of charging capacitor is:
[tex]Q=Q_{0} (1-e^{-\frac{t}{r} })[/tex]
Where,
[tex]Q_{0} =[/tex] equilibrium charge
[tex]t=[/tex] time constant
[tex]r=[/tex] time elapse
- First, we need to set the [tex]Q[/tex] to [tex]Q = 0.80Q_{0}[/tex], so the equation becomes,
- [tex]Q = 0.80Q_{0}[/tex] = [tex]Q_{0} (1-e^{-\frac{t}{r} } )[/tex]
- Then, we can eliminate the [tex]Q_{0}[/tex]
- 0.80 = [tex]1-e^{-\frac{t}{r} }[/tex]
- Substracting both sides by 1
- 0.80-1 = [tex]1-e^{-\frac{t}{r} }[/tex]
- -0.20 = [tex]-e^{-\frac{t}{r} }[/tex]
- [tex]e^{-\frac{t}{r} }[/tex] = 0.20
After that, we need to determine [tex]\frac{t}{r}[/tex] by taking natural log to both sides
- ㏑ [tex]e^{-\frac{t}{r} }[/tex] = ㏑ 0.20
- [tex]-\frac{t}{r}[/tex] = ㏑ 0.20
- [tex]t = -r[/tex]㏑ 0.20
- [tex]t = 1.609r[/tex]
The number of time constants must elapses are [tex]\frac{t}{r}=1.609[/tex]
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