In Figure 12-31, determine the capacitor voltage 50 μs after the switch is closed if the capacitor is initially uncharged. Draw the charging curve.

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Accepted Answer

The time constant is [latex]RC[/latex] = (8.2 kΩ)(0.01 μF) = 82 μs is. The voltage to which the capacitor will fully charge is 50 V (this is [latex]V_{F}[/latex]). The initial voltage is zero. Notice that 50 μs is less than one time constant; so the capacitor will charge less than 63% of the full voltage in that time.

[latex]v_{C}= V_{F}(1-e^{-t/RC})= (50 \ V)(1-e^{-50\mu s/82\mu s})[/latex]

[latex]\ \ \ \ = (50 \ V)(1-e^{-0.61})= (50 \ V)(1- 0.543)= 22.8 \ V[/latex]

The charging curve for the capacitor is shown in Figure 12-32.

You can determine an exponential function on your calculator by using the [latex]e^{x}[/latex] key and entering the value of the exponent of e.

Question 12.12: In Figure 12-31, determine the capacitor voltage 50 μs after...

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