Determine the capacitor voltage in Figure 12-33 at a point in time 6 ms after the switch is closed. Draw the discharging curve.
Question 12.13: Determine the capacitor voltage in Figure 12-33 at a point i...
The discharge time constant is RC = (10 kΩ) (2.2 μF) = 22 ms. The initial capacitor voltage is 10 V. Notice that 6 ms is less than one time constant, so the capacitor will discharge less than 63%. Therefore, it will have a voltage greater than 37% of the initial voltage at 6 ms.
v_{C}= V_{i}e^{-t/RC}= (10 \ V)e^{-6 ms/22ms}= (10 \ V)e^{-0.27}= (10 \ V)(0.761)= 7.61 \ VThe discharging curve for the capacitor is shown in Figure 12-34.
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