Question A8.18: A capacitor is charged to a potential of 100 V. It is then d...
A capacitor is charged to a potential of 100 V. It is then disconnected from its charging source and left to discharge through a resistor. If it takes 10 s for the voltage to fall to 50 V, how long will it take to fall to 10 V?
Here we are dealing with exponential decay. If we use y to represent the capacitor voltage at any instant, and since the initial voltage is 100 V, we can determine the value of x (i.e. X_{1}) when the voltage reaches 50 V by substituting into the formula:
y=Y_{\max } e^{-x}hence: 50=100 e^{-x_{1}}
from which: e^{-x_{1}}=\frac{50}{100}=0.5
taking logs (to the base e this time) of both sides:
\log _{e}\left(e^{-x_{1}}\right)=\log _{e}(0.5)thus: -x_{1}=-0.693
hence: x_{1}=0.693
We can use a similar process to find the value ofx (i.e. X_{2}) that corresponds to a capacitor voltageof 10 V:
y=Y_{\max } e^{-x}hence: 10=100 e^{-x_{2}}
from which: e^{-x_{2}}=\frac{10}{100}=0.1
taking logs (again to the base e) of both sides:
\log _{e}\left(e^{-x_{2}}\right)=\log _{e}(0.1)thus: -X_{2}=-2.3
hence: X_{2}=2.3
We can now apply direct proportionality to determine the time taken for the voltage to reach 10 V. In other words:
\frac{t_{50}}{x_{1}}=\frac{t_{10}}{x_{2}}from which: t_{10}=t_{50} \times \frac{X_{2}}{X_{1}}
thus: t_{10}=10 \times \frac{2.3}{0.693}=33.2 s
Fig. A8.36 illustrates this relationship.
