- State than in an electric field a charged particle experiences a force.
- State that an electric field applied to a conductor causes the free electric charges in it to move.
- State that work W is done when a charge Q is moved in an electric field.
- State that the potential difference between two points is a measure of the work done in moving one coulomb of charge between the two points.
- State that if one joule of work is done moving one coulomb of charge between two points, the potential difference between the two points is one volt.
- Carry out calculations involving the relationship between potential difference, work and charge.
- State that the e.m.f. of a source is the electrical potential energy supplied to each coulomb of charge which passes through the source.
- State that an electrical source is equivalent to a source of e.m.f. with a resistor in series, the internal resistance.
- Describe the principles of a method for measuring the e.m.f. and internal resistance of a source.
- Explain why the e.m.f. of a source is equal to the open circuit p.d. across the terminals of the source..
- Explain how the conservation of energy leads to the sum of the e.m.f.s round a closed circuit being equal to the sum of the p.d.s round the circuit.
- Derive the expression for the total resistance of any number of resistors in series, by consideration of the conservation of energy.
- Derive the expression for the total resistance of any number of resistors in parallel by consideration of the conservation of charge..
- Carry out calculations involving the resistances in a balanced Wheatstone bridge.
- State that for an initially balanced Wheatstone bridge, as the value of one resistor is changed by a small amount, the out-of-balance p.d. is directly proportional to the change in resistance.
- Carry out calculations involving potential differences, currents and resistances in circuits containing resistors..
- Use the following terms correctly in context: charge, current, p.d., resistance, terminal p.d., load resistor, bridge circuit, e.m.f., lost volts, short circuit current.

Aids to Understanding

Fields in physics

In physics when we speak of a field, for example a gravitational field or magnetic field we are talking about a region of space in which a body experiences a force. A mass will experience an attractive force in a gravitational field and an iron based metal will experience a force in a magnetic field. The source of a gravitational field is a mass and the source of a magnetic field is either a permanent magnet or a current carrying conductor. To help us visualize electric fields we use arrows. The direction of the arrow is the direction a relevant body will experience a force. A gravitational field due to the Earth is shown below. The force experienced by a mass is in the direction of the field.

Energy matter

A free charge will experience a force and, if in a vaccum, will therefore
accelerate. The energy change is electrical energy to
moving energy. However if the electric field is applied to
a conductor, the free charges **do not** accelerate due
to the natural resistance of the conductor (ie the extent free
charges are available in the conductor and the extent they collide with the stationary atomic nuclei). In this
case the energy change is electric energy to heat energy.

Electric Current

Electric current, I, is defined
as Q/t where Q is measured in Coulombs and t is measured
in seconds. So if a current of 6 A is flowing in a wire
then 6 C of charge drifts past each point every second.

1 C = 1.6 X 10^{18} charges.
Most of you should know that the higher the resistance the lower
the current but it should be clear that the current is the same
at all points in a series circuit, including the wires and the
resistors.

Working against the gravitational field

When a force is doing work on a body it is transferring energy to that body.
For example an unbalanced force of 6N pulling a toy car for 2 m,
will have transfer 12 J of energy and the car will have acquired 12 J of kinetic energy (see 1.3).
If the force was balanced then 12 J of heat energy will be produced and there is no change
in kinetic energy. However, if a balanced force works on a body and no heat is produced, then the
energy transferred is stored in the body as potential energy. This is the situation when a force of 9.8 N
raises a mass of weight 9.8 N a height of 1 m. The balanced lifting force does 9.8 J of
work against the gravitational field and this is stored in the weight as gravitational potential energy.

Potential Energy and Potential Difference

You will have noticed that in this unit energy matters have arisen frequently, but now
we are suddenly speaking of potential difference that have units of Volts (not Joules). And when we numerically label an electric field arrangement,
the numbers have units of V. The diagram on the right is labelled 100 V to 260 V. Why not simply label it 100 J to 260 J, you may ask? Well the
point is when we label something it must be the same for everything relevant to it. For example a height of 6 m does not depend on what is
sitting at that height and all bodies sitting there are a height of 6 m, but the potential energy of a body at that height also depends on the mass of the
body. In an electric field context the Energy per
Coulomb (potential) is the same for all charges at a given point in the field (ie our dotted lines). For example suppose a 2 C and a 5 C charge are sitting at a potential of 15 V, what is the potential
energy of each charge? The answers are 30 J and 75 J. To get our answer we used E_{p} = QV = E_{w}. If both charges sat at zero potential
initially then 30 J and 75 J of Work had to be done to place the charges at point 15 V.

It should be clear that work done moving a charge between two
points equals change in potential energy of the charge and is proportional to the potential difference between the points. You should see how this is similar to the gravitational context.

Finding the speed of a particle

The speed of a charged particle is worked out by invoking the principle of conservation of energy.
When a charge falls through a pd it loses potential energy and gains kinetic energy. If the particle is initially at rest
we can say QV = 1/2mV^{2}.

**Example** an electron of mass 9.1 x 10^{-31} kg and charge 1.6 x 10^{-19} C falls through a pd of 1000.0V.

i) Calculate the work done by the field. Answer, QV = 1.6 x 10^{-19}. This is equal to increase in Ek.

ii) Calculate the electron's velocity.

Answer, 1.6 x 10^{-19} = 1/2mv^{2}, V = 5.3 x 10^{6}ms^{-1}

Note: the speed can never exceed 3.0 x 10^{8} so if you find you have an answer too big, you have probably forgotten
to take the square root.

Electro Motive Force and Internal resistance

EMF is defined in terms of joules
per coulomb and not in Newtons. The usefulness of defining a cell's action
in terms of Forces is that it reminds us of the forces inside
cell due to the cell's chemical actions. These forces displace charges through electric
field and give them electric potential energy. When these forces also work against the
cell's internal resistance then heat energy is also produced.

The energy change per coloumb is therefore Chemical to electric energy and heat, from this is
follows that the total energy supplied to each coulomb, E, is given by

E = V + Lost volts.

Convince yourself of the validity of this expression.

Straight line Mathematics and investigative ideas

The equation of a straight line
is y = mx + c. This means the y value will change if the
x value changes. The m and c are constant and represent
the gradient of the line and c is where the line cuts the
y-axis at x = 0. Now the expression for a real cell is given by

E = V_{tpd} + Ir or

V_{tpd} = -rI + E. This means if we change I the V_{tpd} will
change, -r will be the gradient of the line and the vertical axis intercept is the EMF.

EMF and Internal Resistance problem

The cell in the circuit below has an EMF
of 3.0 V. From circuit (a) find the value of the load resistance
R and internal resistance r. In circuit (b) and (c) the circuits
are modified with a resistor identical to resistor R in circuit (a)
Find the readings on the ammeters and voltmeters in both cases.

Resistors in Parallel and Series

Work out the value of resistor is equivalent or equal to the resistors in each network below and pro

The voltage divider

If we have two resistors in series
with a source of EMF, then

i) the potential drop across both resistors add up to the EMF.

ii)the current is each resistor is the same.

Use these facts to show that the ratio of potential drops across each resistor equals the ratio of the resistances.
ie

V_{1}/V_{2} = R_{1}/R_{2}

Description in Physics

Suppose you were given a resistor of resistance 200 W, for example, and were asked to explain what 200 W meant. To answer this properly you must quantify in terms of a unitary quantity, in this case our unitary quantity is Amps. You would write R = V/I and say 200 W means 200 V A^{-1}. In other words if we wish 1 A to flow in our 200 W resistor we need to apply 200 V across its terminals. If we have a 400 W resistor then we need 400 V before 1.0 A will
flow in it. When you are asked to write units after a quantity, you are writing the meaning of the quantity that you have calculated.

The green arrows show the electric field and the black vectors show the direction of the force on the particle. Note how the force on a positive charge is always in the direction of the field and its magnitude will get greater as the lines of force get closer. This follows for both radial and parallel fields. You should notice that for a uniform parallel field the field lines are at the same distant apart therefore the force on a charged particle is independent of where the charge is. You are invited to draw the same diagrams with a negative charge,

It may have occurred to you that the diagram above is simply a resistor between two wires. We will return to this argument when discussing voltage rules (see 11) below) for series and parallel circuits.

The force displacing the positive charge has done work, W, and energy has been stored in it as potential energy. When the charge is released the unbalanced force due to the field does work and it acquires kinetic energy as it travels along the field lines.

DV = V

Suppose an alpha particle resides at 140 V in the electric field below. If the particle is allowed to fall with the field, then..

(i) Calculate is potential energy as it passes through the 80 V and 40 V points

(ii) Calculate the speed of the particle at these points

Suppose a 1.0 kW resistor has a current of 2.0 A flowing in it and a potential difference (voltage) across it of 6.0 V.

(i) Calculate the heat energy produced each time 1 C of negative charge flows across this pd.

(ii) Calculate the heat energy produced per second.

You should now begin to see the important energy distinctions between charges flowing through vacuums and charges flowing through conductors. You should also be able to see the principle of conservation of energy in action in both cases.

(i) The sum of the work done on each coulomb by the electric fields across each resistor is equal to the total energy supplied to each coulomb by the cell's EMF. (Conservation of Energy)

(ii) The current is the same at all points including the cell and the resistors.(Conservation of charge)

We are assuming the cell is an ideal cell with no internal resistance and the connecting wires also have no resistance, or a resistance so much lower than the circuit resistances thaty it can be ignored. You may modify this argument for a non-ideal cell and wires with a non-zero resistance, if you wish. These points are illustrated below.

The red graph shows the potential (energy/coulomb) 'drop' across each resistor. The electric field (shown in green) does work on each charge as it travels through each resistance and maintain the constant current. The charge loses electrical energy as it falls to lower and lowere potentials. The energy change is electrical energy to heat energy. The resistors are drawn at different lengths to illustrate the fact that higher value resistors create larger potential drops and therefore have larger fields across them (See 2) above). It should be clear that the the conservation of energy was used to put the diagram together and sum of the potential drops is equal to the EMF of the cell.

If we have a number of resistors in series and we wish to replace it with one equivalent resistor that will allow the same current and potential drop then this resistor will be the the sum of the resistors it replaces. The conservation of charge leads to the constancy of the current in each resistor and the conservation of energy means the sum of the potential drops across each resistor must equal the EMF of the cell.

The conservation of charge leads to the constancy of the current in each resistor and the conservation of energy means the potential drop across each resistor must be equal and equal the EMF of the cell. Sometimes we say the equivalent resistance is the total resistance.

The