Question 3.11: A 10 μF capacitor is charged to a potential of 20 V and then...
A 10 μF capacitor is charged to a potential of 20 V and then discharged through a 47 kΩ resistor. Determine the time taken for the capacitor voltage to fall below 10 V.
The formula for exponential decay of voltage in the capacitor is:
v_{ c }=V_{ s } e^{-\frac{t}{C R}}where V_{ s } = 20 V and CR = 10 μF × 47 kΩ = 0.47 s.
We need to find t when V_{ C } = 10 V. Re-arranging the formula to make t the subject gives:
t=-C R \times \ln \left(\frac{v_{ C }}{V_{ S }}\right)thus t=-0.47 \times \ln \left(\frac{10}{20}\right)=-0.47 \times \ln (0.5)
or t = −0.47 × − = 0.693 0.325 s
In order to simplify the mathematics of exponential growth and decay, Table 3.1 provides an alternative tabular method that may be used to determine the voltage and current in a C–R circuit.
| t/CR or t/(L/R) | k (growth) | k (decay) |
| 0.0 | 0.0000 | 1.0000 |
| 0.1 | 0.0951 | 0.9048 |
| 0.2 | 0.1812 | 0.8187 (1) |
| 0.3 | 0.2591 | 0.7408 |
| 0.4 | 0.3296 | 0.6703 |
| 0.5 | 0.3935 | 0.6065 |
| 0.6 | 0.4511 | 0.5488 |
| 0.7 | 0.5034 | 0.4965 |
| 0.8 | 0.5506 | 0.4493 |
| 0.9 | 0.5934 | 0.4065 |
| 1.0 | 0.6321 | 0.3679 |
| 1.5 | 0.7769 | 0.2231 |
| 2.0 | 0.8647 (2) | 0.1353 |
| 2.5 | 0.9179 | 0.0821 |
| 3.0 | 0.9502 | 0.0498 |
| 3.5 | 0.9698 | 0.0302 |
| 4.0 | 0.9817 | 0.0183 |
| 4.5 | 0.9889 | 0.0111 |
| 5.0 | 0.9933 | 0.0067 |
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