Current and Voltage in a Circuit Containing a Charging Capacitor
The purpose of this session is to teach you how to calculate the current and voltage levels in a circuit containing capacitor that is charging and discharging.
By the end of this session the learner will be able to:
- Carry out and experiment to determine the relationship between voltage and current for a charging and discharging capacitor
- Calculate the time constant for any circuit containing a resistor and capacitor
- Calculate the instantaneous voltage at any point in time for a circuit containing a resistor and capacitor in a dc circuit
BTEC Outcomes Covered
Unit 6 Pass 5
In the previous 2 sessions we learned quite a lot about capacitors. Lets just briefly review what we know:
- Capacitors are constructed using 2 metal plates separated by a material known as a dielectric
- Capacitors store an electric charge or energy in an electric field
- The distance between the plates, area of the plates and the dielectric all affect how much charge the capacitor can store
- Electric charge(Q) is measured in coulombs
- Capacitance is measured in Farads
- Capacitors connected in series, always have a total capacitance which is lower than the value of the smallest capacitor in the network
- Capacitors connected in parallel have their values added together and therefore can store a lot more charge
There is one more thing that we should really know about capacitors at this stage of our studies. We are going to conduct a simple experiment to demonstrate it.
Build the circuit shown below. Have a watch or some method of recording time ready
- Setup the power supply to 12 Volts dc
- Connect the power supply
- Run the circuit for about 1 minute
- Describe what happens to the current flowing in the circuit during this time
- Describe what happens to the voltage dropped across the resistor during this time
- Describe what happens to the voltage across the capacitor in this time
- Now switch off the circuit and observe what happens for 1 minute
- Describe the behaviour of the current and voltages during this time
- Is there anything about the construction of a capacitor that explains the behaviour of the current and voltages in this experiment?
- Can you identify this type of circuit on the virtual cat schematic?
Here is a video showing how to construct and test the circuit
Remember the dielectric is an insulator. An insulator does not allow current to flow through it. Therefore when we switch on the circuit, the current we measure is simply the flow of electrons charging up the plates of the capacitor. The plate connected to the positive terminal of the power supply charges to the positive potential and the terminal connected to the negative (or ground) charges to the negative potential
We have already seen that as the charge stored by a capacitor rises the voltage dropped across the capacitor increases. When the capacitor is fully charged:
All of the circuit voltage is dropped across the capacitor
And because of this no current can be flowing in the circuit. Ohms law tells us:
Current flowing = Voltage/Resistance
I = V/R
Because all of the voltage is dropped across the capacitor and the capacitor is an insulator, no current can flow through it. Therefore no current can flow through the circuit.
So for dc voltage a capacitor charges up until the potential across its plates is equal to the supply voltage and then no further current flows.
Whenever we have a resistor and capacitor in series with each other they form something known as a time constant.
To find the time constant of any circuit containing a resistor and capacitor we simply
Multiply the capacitance by the resistance
Time Constant = C x R
So for this circuit
Time Constant = 0.0001 x 100,000
Time Constant = 10 seconds
But what is the point of a time constant?
The time constant is a measure of how long the capacitor will take to reach full charge and for the current to stop flowing in the circuit.
It takes 5 time constants for any circuit containing a capacitor and resistor in series to reach full charge
Therefore our circuit should take 50 seconds for the capacitor to reach full charge.
How does this figure compare with your observations?
What factors do you think could make the reading higher or lower than 50 seconds?
Using the time constant
The capacitor does not charge at a constant rate. Its charging rate can be described by an exponential equation. The difference between linear growth and exponential type growth is illustrated below.
From previous information and the illustration above we can state:
“The voltage across a capacitor rises from its minimum value (0 volts) to its maximum value Vsupply in 5 time constants”
Now the challenge is to prove it mathematically
We are going to calculate the voltage across the capacitor at each of the 5 time constants it takes to fully charge
Before we do the paper proof, this video demonstrates the maths involved in carrying out the calculations. Please like, subscribe and generally positively comment 😉
The voltage across a charging capacitor at any point in time after a circuit has been switched on can be predicted using the following calculation.Click on the diagram if it is not clear.
- Where Vmax is the maximum value of the power supply
- Vinst (short for Vinstantaneous)is the value you are calculating i.e. the voltage across the capacitor
- C is the value of the capacitor
- R is the value of the resistor
And e can be found on the calculator by pressing shift then ln:
The first step when doing this calculation is to find the time constant of the circuit. The time constant is found by multiplying the capacitance by the resistance:
C x R
= 0.0001 x 100,000
= 10 seconds
So this value plugs into the CR part of the equation as shown:
We know that Vmax = 5 volts so that can be plugged into the equation as well
If we want to find the value of the voltage after 1 time constant then:
-t = CR
And so for each of the 5 time constants the power to e increases by 1, like so:
You should get this value > 0.3678
To find the value of Vinst we simply multiply out the brackets:
Find the percentage of the maximum voltage that this value represents:
3.16/5 x 100 = 63.2%
From theory we know that at t = 0 the voltage across the capacitor = 0 volts. You should use the above technique to calculate the voltage at the other 4 time constants and plot your results on a graph. The best way to do this is to use a spreadsheet in excel.
Finally we need to work out the current flowing in the circuit for each of the 5 time constants
We use this equation where I stands for current.
The only major difference is that we need to know Imax:
Theory tells us that at the very instant we switch the circuit on, the capacitor acts as a short circuit and no voltage is dropped across it. Therefore all of the circuit voltage must be dropped across the resistor.
It is at this time that the current flow is at its maximum. Therefore from Ohm’s law:
Imax = 5volts/100,000 Ohms
= 0.00005 Amps
Now use this information to calculate the current at t=0 and for the following 5 time constants. Plot the information on the same graph as you plotted the voltage and describe the relationship between current and voltage in a circuit containing a charging capacitor based on what you see.
This concludes all of the work required for the session. Well done!!