Capacitor Tutorial 1 - Capacitance of a Capacitor

Capacitors are short term charge-stores, a bit like an electrical spring.  They are used widely in electronic circuits.  At its simplest it consists of two metal plates separated by a layer of insulating material called a dielectric. 

The picture shows some different capacitors:

The two components on the right are electrolytic capacitors.  The others are non-electrolytic capacitors.

The symbol for a capacitor is shown below:

There are two types of capacitor, electrolytic and non-electrolytic.  We wont worry at the moment what these terms mean, other than to say:

If we pump electrons onto the negative plate, electrons are repelled from the negative plate.  Since positives do not move, a positive charge is induced.  The higher the potential difference, the more charge is crowded onto the negative plate and the more electrons repelled from the positive plate.  Therefore charge is stored.  The plates have a certain capacitance.


Capacitance is defined as:


The charge required to cause unit potential difference in a conductor.


Capacitance is measured in units called farads (F) of which the definition is:


1 Farad is the capacitance of a conductor, which has potential difference of 1 volt when it carries a charge of 1 coulomb.


So we can write from this definition:


Capacitance (F) = Charge (C)

                                     Voltage (V)


In code, this is written:


[Q - charge in coulombs (C); C capacitance in farads (F); V - potential difference in volts (V)]  


A 1 farad capacitor is actually a very big capacitor indeed so instead we use microfarads (mF) where 1 mF = 1 10-6 F.  Smaller capacitors are measured in nanofarads (nF), 1 10-9 F, or picofarads (pF), 1 10-12 F.  A working voltage is also given.  If the capacitor exceeds this voltage, the insulating layer will break down and the component shorts out.  The working voltage can be as low as 16 volts, or as high as 1000 V.  


Question 1

Write down what is meant by the following terms: 

  • Dielectric

  • Farad

  • Working voltage



The voltage rises as we charge up a capacitor, and falls as the capacitor discharges.  The current falls from a high value as the capacitor charges up, and falls as it discharges.



If we connect a capacitor in series with a bulb:

We can say that a capacitor blocks d.c., but allows a.c. to flow.


The capacitor does NOT conduct electricity.  The "flow" of a.c. is due to the charge and discharge of the capacitor.


Question 2 

Why does a capacitor appear to allow ac to flow, but not dc? 

Question 3

What is the charge held by a 470 microfarad capacitor charged to a p.d. of 8.5 V?    



Measuring Capacitance

This circuit can be used to measure the value of a capacitor:




We can use I = Q/t to work out the charge going onto the plates.  We also know that f = 1/t, so we can combine the two relationships to give I = Qf, Q = I/f


Since C = Q/V, we can now write C = I/fV


Question 4

Why does the circuit only operate on the forward half-cycle?

Question 5

A capacitor is connected to a 12-volt power supply by a reed switch operating at 400 Hz.  The ammeter reads 45 mA.  What is the capacitance of the capacitor?




Energy in a Capacitor

When we charge up a capacitor, we make a certain amount of charge move through a certain voltage.  We are doing a job of work on the charge to build up the electric field in the capacitor.  Thus we can get the capacitor to do a job of useful work.

  We know that:

1.      Energy = charge voltage

2.      Q = CV

This second relationship tells us that the charge voltage graph is a straight line:

The capacitor is charged with charge Q to a voltage V.  Suppose we discharged the capacitor by a tiny amount of charge, dQ.  The resulting tiny energy loss (dW) can be worked out from the first equation:

dW = V dQ           

This is the same as the area of the pink rectangle on the graph.

If we discharge the capacitor completely, we can see that:

Energy loss = area of all the little rectangles

 = area of triangle below the graph

 = QV


By substitution of Q = CV, we can go on to write:  

E = CV2

Question 6 

What is the energy held by a 50 000 mF capacitor charged to 12.0 V?