When you put a voltage across a capacitor, current will flow into the capacitor and the voltage across the capacitor will increase until the voltage across it reaches the value of the supply voltage. This is not a linear function. By that I mean that the voltage will increase quite rapidly at first, but the rate of increase will slow as time goes on.

To see how this works, let’s consider the RC time constant. The time constant of an RC circuit is equal to the resistance in the circuit times the capacitance, or simply R x C. For example, the time constant of a circuit having two 220-microfarad capacitors and two 1-megohm resistors, all in parallel is **220 seconds**. (E5B04)

The equivalent resistance of two 1 MΩ resistors in parallel is 500 kΩ. The equivalent capacitance of two 220 μF capacitors in parallel is 440 μF. The time constant is RxC = 440 x 10^{-6} x 500 x 10^{5} = 220 s.

**One time constant **is the term for the time required for the capacitor in an RC circuit to be charged to 63.2% of the applied voltage. (E5B01) Similarly, **one time constant **is the term for the time it takes for a charged capacitor in an RC circuit to discharge to 36.8% of its initial voltage. (E5B02)

The capacitor in an RC circuit is discharged to **13.5% **of the starting voltage after two time constants. (E5B03) Similarly, a capacitor charges to 86.5% of the applied voltage after two time constants. After three time constants, a capacitor is charged up to 95% of the applied voltage or discharged to 5% of the starting voltage.

You can use these percentages to answer the questions about how much time it takes for a capacitor to discharge. The key is to figure out what percentage the voltage given is of the starting voltage. In one case, the starting voltage is 20 V and you must figure out how much time it will take for the capacitor to discharge to 7.36 V.

Well, 7.36 V just happens to be 36.8% of 20 V, so the time required will be one time constant. One time constant is R x C, or in this case 0.01 x 10^{-6} x 2 x 10^{6}, or .02 s. So, it takes **0.02 seconds **for an initial charge of 20 V DC to decrease to 7.36 V DC in a 0.01-microfarad capacitor when a 2-megohm resistor is connected across it. (E5B05)

In the second case, the starting voltage is 800 V and you must calculate the time required for the voltage across the capacitor to drop to 294 V. Well, fortunately, 294 V / 800 V is again 36.8%, so the time required will be one time constant.

In this circuit, R = 1 MΩ and the capacitance 450 μF. R x C = 10^{6} x 450 x 10^{-6} = 450 s. So, it takes **450 seconds **for an initial charge of 800 V DC to decrease to 294 V DC in a 450-microfarad capacitor when a 1-megohm resistor is connected across it. (E5B06)

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