Session 9

Calculating Capacitance Values

The purpose of this session is to teach you how to calculate the total capacitance of a network of capacitors in series and then in parallel.

By the end of this session the learner will be able to:

  • Calculate the total capacitance, total charge, voltage dropped and energy stored by a network of three capacitors connected in series
  • Calculate the total capacitance, total charge and energy stores by a network of three capacitors connected in parallel

BTEC Outcomes Covered

  • Unit 6 Pass 6

First things first, lets briefly review some terminology from session 3

Capacitance is measured in Farads and is a direct measure of the ability of a capacitor to store an electric charge. The higher the capacitance the more charge/energy can be stored

Electric charge (Q) is measured in Coulombs (C) and is a measure of the total number of electrons flowing past a point in a circuit in 1 unit of time (i.e. 1 second). Note that in calculations (Q) is used to describe charge even though the unit is the coulomb

Voltage each capacitor in a series circuit will have a portion of the supply voltage dropped across it. In a circuit that contains only capacitors, the total available voltage (the supply voltage) will be dropped across all of them.

The dropped voltage is proportional to the size of capacitor. The larger the capacitor the smaller the voltage dropped across it.

Capacitors in Series

This video explains the concepts and how you can calculate total capacitance, charge, voltage dropped and energy stored by a circuit containing 3 capacitors in series.

Paper Workthrough

Take the following circuit

Caps in series

We have 3 capacitors in series. The first job is to work out the total capacitance of the network.

The easiest way to do this is to use a tecnique known as product over sum. Basically we take the value of any 2 of the capacitors and multiply them together, then we divide the sum of those capacitors into the product like this

(C1xC2)/(C1+C2)

(1×10)/(1+10)

= 10/11

= 0.909 uF

Please note that this only works with 2 capacitors at a time. We simply take the result of product/summing C1 & C2 and apply the same tecnique with the remaining capacitor. In this way we can work out the total capacitance for any number of capacitors, it is extremely quick once you get used to it.

(0.909×100)/(0.909+100)

90.9/100.909

Total Capacitance = 0.9008 uF

Before we move on there are 2 things you must note:

  • In a series circuit, the total capacitance value must always be lower than the lowest value capacitor in the circuit. In this case 1uF > 0.9008uF so we are good to go
  • Before you can use this value in any further calculations you must divide it by 1,000,000

This is because u = 1 millionth. If you were working in milli then you would divide by 1000, if you were working in nano you would divide by 1,000,000,000. Its a good idea to convert all of your values to the same type, do the calculation and then divide the result at the end.

The relationship between charge, voltage and capacitance

A relationship exists between charge (Q), voltage(V) and capacitance(V) as follows:

Q = CV

C = Q/V

V = Q/C

Therefore if we know the total capacitance and we know the supply voltage (12 volts in this case) we can work out the total charge (Q)

Using:

Q = CV

Q = 0.0000009008 x 12

Total Charge (Q) = 0.0000108 Coulombs

Now this charge is applied equally to all of the capacitors in the circuit, meaning that each capacitor, regardless of its size, has 0.0000108 coulombs of charge.

We can use this fact to work out the voltage dropped across each capacitor.

Using:

V = Q/C

And remembering that in a series circuit, the charge is constant for all capacitors.

The voltage across C1

Vc1 = 0.0000108/0.000001

Vc1 = 10.8 Volts

Vc2 = 0.0000108/0.00001

Vc2 = 1.08 Volts

Vc3 = 0.0000108/0.0001

Vc3 = 0.108 Volts

Now if we add all 3 voltages together

10.8 + 1.08 + 0.108 = 11.988 Volts

Or 12 Volts rounded up

This is your check. If the summed voltages do not pretty much equal the supply voltage then you have made a mistake somewhere so go back and check your working.

Energy Stored

To work out the energy story by a capacitor or group of capacitors we use the equation.

Energy = .5C x V^2

Half the capacitance times the voltage squared

So for this system

0.00000045 x 144

Energy stored = 0.0000648 Joules

 Capacitors in Parallel

The main thing to remember about parallel circuits is that the voltage drop across each branch of the parallel circuit = the supply voltage as illustrated next.

Voltage across parallel

The key things to remember are, that the total voltage across each branch = the supply voltage. And that the voltage across each component in the branch is split according to the value of the component. In my example the capacitors in each separate branch are of the same value so the voltage dropped across each of them is equal.

Therefore in the circuit below

For parallel calculations

The voltage is a constant and the charge stored by each capacitor will depend upon the value of the capacitance.

Total capacitance is easy to find for this type of circuit. You simply add up all the capacitances and then (because they are micro Farads) you divide the sum by 1,000,000.

Armed with this information you should now be able to find:

  • The total capacitance
  • The total charge
  • The charge stored by each capacitor
  • The energy stored by the whole network

If you are stuck, here is a great video showing how to work all of this out. Don’t forget to like, subscribe and generally positively comment on it 😉

When you have completed these calculations and filled in your report you are done for this week. Congratulations!!

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