# Simple Capacitor charging and discharging question

Below is a circuit consisting of a capacitor, voltage supply and resistors. There is a switch that lets the capacitor charge and discharge. I have also uploaded the given answers below;

I understand that for:

(i) Tc: you find the thevenin resistance and multiply it by the capacitor value.

(ii) Vc Max: the voltage across the parralel (Rz) resistor.

(iii): I am not sure that he has done there. Has he used the capacitor charge or discharge formula and rearranged it of some sort?

(iv): Not sure what hes done here?

.

• What is your specific question? – The Photon Apr 23 '17 at 15:00
• Despite the propaganda, there is nothing simple about charging and discharging capacitors. However that aside, what is your question? – Trevor_G Apr 23 '17 at 15:37
• Not sure how he has calculated III and IV part of the question – Michael Apr 23 '17 at 15:42
• Im a bit confused about how he has calculated it in the answers – Michael Apr 23 '17 at 15:45

I am not sure that he has done there. Has he used the capacitor charge or discharge formula and rearranged it of some sort?

The charging equation look like this:

$$V_C = V_{th}(1 - e^{\frac{-t}{RC}})$$

Now to solve for the "time"$(t)$ we need to rearranged this formula.

First we can divide by $V_{th}$

$$\frac{V_C}{V_{th}} = 1 - e^{\frac{-t}{RC}}$$

We subtract 1 from both sides and multiply both sides by -1

$$\frac{V_C}{V_{th}} - 1 = - e^{\frac{-t}{RC}}$$

$$1-\frac{V_C}{V_{th}} = e^{\frac{-t}{RC}}$$

take the ln() of both sides

$$ln(1-\frac{V_C}{V_{th}}) = {\frac{-t}{RC}}$$

Multiply both sides by -RC

$$-RC*ln(1-\frac{V_C}{V_{th}}) = t$$

Swap sides

$$t = -RC*ln(1-\frac{V_C}{V_{th}})$$

(iv): Not sure what hes done here?

$t_d$ is a discharge time constant $t_d = R*C$

Where $R$ is a resistance seen by the capacitor when discharging.

• It makes total sense. Tried it with a few other questions and it works like a charm. Cheers – Michael Apr 23 '17 at 18:39