**RC circuits** can be driven by DC voltage sources. Below is a diagram of a very simple RC circuit - it is a circuit with one resistor **R** and a capacitor **C** and an electromotive force **E** constant in time.

Imagine that at time t = 0 we close the key (switch). When the key is closed, the capacitor will be charged. The voltage across the capacitor, which is time dependent, can be found using Kirchhoff's second law (see equation 1 below).

Considering all the relationships in this circuit requires solving a differential equation (see equation 2).

I have provided ready-made solutions here (formulas: 3, 4 and 5). These are exponential functions (**e** is Euler's number - a mathematical constant approximately equal to 2.72).

Formula 3 **I(t)** describes the value of the current in the circuit, **U _{R}(t)** describes the value of the voltage on the resistance

In the circuit under consideration we have a maximum current value equal to **I _{0}** and zero voltage

The time needed for the current

Parameter τ is expressed in seconds if we express R in ohms and C in farads. The graph above was made for a circuit where

From formula 5, it is easy to calculate that after time

You already know enough about charging capacitors, so it's time for an experiment.

**How can you perform the experiment?**

• Use my online lab.

• Of course, such an experiment can also be performed in school laboratory - see Photo.

I assume that you have measured the voltage across the capacitor as a function of time - you have a table with the experimental results and plotted them on a graph **U _{C}(t)**. I also assume that you know the capacitance of capacitor

I will show you how I performed such an experiment - the graph of the **U _{C}(t)** function is above.

I transformed formula 5 as follows.

I got a linear relationship: **y = -t/τ** where: **y = ln( (E - U _{C}(t))/E )** and

Now I can use the same table of experimental results and plot the function y(t). I got the chart as below. In order to fit the straight line to the experimental points, I used the method of least squares.

We have success, the relationship expressed by formula 5 is confirmed. You can see from the graph and I got it with the least squares method that the slope of the straight line is equal to:

I used a 5uF capacitor in the experiment (5uF = 5•10^{-6}F). The capacitor was charged by resistance **R**. It is easy to calculate the value of this resistance because:

In other words, I got the resistance which was 1kΩ.

• Repeat this procedure for your experiment (use your measurement data).

• Try to extend this exercise and plot the function **I(t)** and **U _{R}(t)**.

• Do such an experiment for discharging a capacitor. Of course, also calculate the value of the resistor R through which the capacitor C discharges.

Good luck :)