*The power of an electrical device equals the amount of energy that it changes per second from electrical energy to other forms of energy*

For example, a 60W light bulb is a ‘60 joules per second’ light bulb, i.e. it changes 60J of electrical energy to light and heat each second.

If this bulb runs from a 240V mains supply, it carries a current of:

**Example**

What is the power rating of an electrical immersion heater that
runs from the mains voltage of 240V and carries a current of 15A?

This can be remembered as:

In metallic conductors and carbon all the electrical energy supplied is converted to heat - such conductors are called ‘passive resistors’.

**The kilowatt-hour (kWh)**

The kWh is the commercial unit of electrical

one kWh is the energy used by a device of power 1kW run for one hour

**Example**

A 200W bulb is run for 4 hours.

- what is the power of the bulb in kW?
- how many kWh are used?
- what is the cost, if electricity is charged at 10 pence per ‘unit’, i.e. per kWh?

b) Energy used = 0.2kW * 4h = 0.8 kWh

c) Cost = kWh * cost per kWh = 0.8 * 10 = 8 pence

The direction of conventional current flow is from + to - outside the battery.

However, *within* the battery the charges have to flow
from the - to the +, and the energy for this comes from
the chemical energy in the battery.

So, internally, the battery gives the charges electrical energy (in raising them from a lower to a higher potential). These charges then flow round the external circuit and their electrical energy is converted into light and heat by the lamp.

An analogy to this is raising a mass above the ground, and then releasing it. In raising it you are giving the mass potential energy. When the mass is released, this PE gets changed to KE as it falls, which is dissipated on impact.

The basic process in the above electrical circuit is:

In general, a source of electrical energy is a device (such as a battery) which converts one form of energy (e.g. chemical) into electrical energy*The electromotive force of a source is the energy converted to electrical energy when one coulomb of charge passes through it*

E.g. a car battery has an emf of 12 volts = 12 joules per coulomb. So, this means that each coulomb of charge that passes through it is given 12 joules of electrical energy, which is converted to other forms of energy in the circuit outside the battery.

**Internal resistance**

In (a), a high resistance voltmeter records the emf E of the cell.

In (b), where the cell maintains a current through resistor R, the voltmeter reading has fallen to V.

The current I flows, not just through R, but also through the cell, which has an internal resistance r. (The high resistance voltmeter is assumed to take no current).

For purposes of analysis, we can treat the internal resistance r as if it were outside the cell.

Now, regarding this diagram:

V is called the terminal p.d. of the cell, i.e. the p.d.
across its terminals.

Dividing top and bottom on the right by R, we get:

We see that as R gets bigger and bigger, r/R gets smaller and
smaller, so [1/(1+r/R)] gets nearer and nearer to 1 (one) and V gets nearer and
nearer to E.

On ‘open circuit’, R = infinity,

Thus, a high resistance voltmeter, which takes negligible or
zero current, can measure emf directly, as in the previous dia. (a).

**Example**

**Maximum power output**

The power dissipated by the resistor R is given by:

Assuming values for E and r (e.g. E = 1.5V and r = 0.5 ohms), we can calculate P for a range of R values (e.g. R = 0 to 10 ohms). Plotting P against R gives a graph like:

Note that the

**In series**

**In parallel**

Suppose that three identical cells (same E, same r) are arranged as follows:

The combined internal resistance r

This can be generalised - if n identical cells are in parallel, each with internal resistance r, then their combined resistance r

Three accumulators, each with emf 2.0 volts and internal resistance 0.02 ohms, are joined in series and used as a supply for a circuit.

- What is the total emf of the supply?
- What current would be driven through a resistance of 1.94 ohms ?

b) Total resistance in the circuit = 1.94 + 0.02 + 0.02 + 0.02 = 2.00 ohms

For any electrical circuit:

**Law 1**: The sum of currents entering a junction equals the sum of currents leaving the junction**Law 2**: The algebraic sum of all p.d.s and emfs around any closed loop is zero

One systematic way of applying the laws (see the example that follows) is:

1. If there are n junctions, apply Law (1) at (n -1) junctions. Label the currents on each branch as
I_{1}, I_{2}, I_{3}… etc. including arrows for directions.
Write down an equation in accordance with law (1) for each of the junctions. It
does not matter initially that the true directions of the currents may not be
known - if a current turns out to be negative, then
this means that it flows opposite to the direction of its arrow.

2. Draw in each loop to be used. Use enough loops to ensure that each component is in at least one loop. The idea is to produce an equation for each loop, each component producing one term in an equation. Starting anywhere on a loop, follow it all the way round back to the starting point. A p.d. across a resistor (= IR) is taken as positive if the loop points in the same direction as the current. An emf is taken as positive if the loop enters the positive terminal first. All the terms in a loop are added, and then the total set equal to zero when the loop is complete.

The following example will be solved using what has previously been covered about resistors, and then using Kirchoff’s laws. Clearly, they should produce the same results. The advantage to Kirchoff’s laws is that they can be applied to more complex circuits.

**Example**

Using Kirchoff’s laws:

There are two junctions, and the currents have been labelled at one of them (= 2 - 1, in accordance with (1) above).

By Law 1: I = I_{1} + I_{2}

Two loops have been drawn such that each component occurs in at least one.

Then, we have: I = I_{1} + I_{2}
= 1 + 0.4 = 1.4 A

In measuring I and V in dia. (a) we do not want to change their values. Therefore:

- the ammeter (dia. (b)) should have a low resistance compared to R (so I is not changed)
- the voltmeter (dia. (c)) should have a high resistance compared to R (so negligible current goes though it)

We can do this by putting a low value resistor in parallel with it, called a

The shunt and the meter are in parallel, so:

Thus, using a shunt resistor of value 0.1 ohm turns the
microammeter into an ammeter reading up to 1A.

Now, if R = resistance of the above set-up, then:

Thus, the combined resistance is low, just as is required for
an ammeter.

**Conversion of a microammeter into a voltmeter**

The same microammeter can be converted to a voltmeter reading
0-1V by putting a high value resistor in series with
it, called a *multiplier*.

The total resistance is (fairly) high (1000 + 9000 ohms), as is required for a voltmeter.

**BALANCED POTENTIALS: WHEATSTONE BRIDGE AND POTENTIOMETER **

**1. Wheatstone bridge**

P, Q and S are resistors whose values can be controlled. R is an unknown resistor.

One or more of P, Q and S are adjusted until galvanometer G shows no deflection, and the bridge is then said to be ‘

G is used as a ‘current detector’ rather than a ‘current
measurer’, so it needs to be *sensitive* to current rather than be well
calibrated, i.e. it should give a visible deflection for a very small
current.

At balance:

- no current flows along the vertical branch, so I
_{1}flows through R*and*Q and I_{2}flows through P*and*S - the p.d. across G is zero (otherwise there would be a current through it), so:

Thus, if P, Q and S are known, then the value of R can be found.

Note: A procedure in which conditions are varied so that an instrument reads zero is called a 'null method'. The Wheatstone bridge is a null method of determining resistance. An advantage of the method is that the accuracy of the result does not depend upon the accurate calibration of an instrument - though it does depend on its sensitivity.

**Metre bridge**

This is the simplest practical form of a Wheatstone bridge.

D is a moveable contact called a ‘jockey’. The resistors P and S consist of the bare wire AC of uniform cross section, one metre long, with a resistance of several ohms. The wire is usually attached to a metre ruler (not shown). Q is a standard resistance whose value is accurately known. R is an unknown resistance.

With the key closed, D is pressed against the wire to make contact. It is moved in steps until G reads zero, i.e. until the bridge is balanced.

Thus, we only need to determine the ratio of the balance lengths, and know the value of the standard resistor Q, to then determine the value of R.

Notes:

1. D is not *dragged* along the wire, since that would
spoil its uniformity, and make the above theory less valid (cancelling the As
assumes A is constant along the length of the wire)

2. It is a good idea to protect G from large currents that may occur when the jockey D is well away from the balance point.

The key is initially open until balance is nearly found, then it is closed. This prevents a large current passing through G when well away from balance but allows the sensitivity of G to be exploited when close to balance, to accurately locate the balance point.

3. The value of Q is chosen to give a balance point close to
the middle of the wire. This means that any errors in L_{1} and L_{2
}are about the same.

4. Having found the balance point and calculated R, then R and Q are reversed and R determined again, and its average value found. This helps to compensate for:

- end-errors (due to ‘contact resistances’, caused by dirt)
- error due to a non-uniform wire
- metre rule incorrectly placed

This consists of a wire of uniform cross-section, with several ohms resistance, through which a ‘driver’ cell passes a steady current.

The reading V on the voltmeter steadily increases as the jockey is moved from A to C. We observe that V is directly proportional to L, as expected, since if the length L has resistance R

**Calibration**

We say that the potentiometer is calibrated when we know the ‘potential gradient’ along the wire:

- potential gradient equals voltage across each cm of wire

To get a more accurate calibration, we use a standard cell, as below, whose emf is accurately known.

Notice that the positive terminals of both the driver cell V and the standard cell E

As with the Wheatstone bridge, we want to find a point along the wire at which G reads zero, the 'balance point', so this is also a 'null method'. Again, in practice G should be protected from large currents.

Initially, J should be touched to ends A and C in turn, to check that G gives opposite deflections, in which case there must be a balance point somewhere between A and C, at which G gives no deflection.

Suppose that E_{0} = 1.57 V, and balance length L_{0} = 84.0 cm

**Uses of potentiometer**

**a) Measuring an unknown emf**

Once the potentiometer is calibrated, i.e. once we know the potential gradient, the volts per cm, we can measure unknown emfs. If the standard cell in the previous diagram is replaced by a cell of unknown emf E and a balance length L is found, then:

The balance length(s) should be as long as
possible for maximum accuracy. An initial check should be made to determine
which of E and E_{0} produces
the greater balance length, and then R should be adjusted to make the longer one
as long as possible. Subsequently, R is unchanged, since that would alter the
potential gradient.

An advantage of using a potentiometer in measuring the emf of a
cell is that *no current* flows through the cell at balance and the value
of its emf can be found accurately (recall from earlier notes that when a
current flows through a cell, its internal resistance causes its terminal p.d.
to be is less than its emf).

Also, the accuracy of the method does not depend on the calibration of the galvanometer, as it is a null method (though it does depend upon its sensitivity).

**b) Measuring small emfs**

A thermocouple consists of two different metals with their bare ends twisted together to form junctions, as represented below. If the junctions are at different temperatures, a small emf is produced (typically a few mV).

A potentiometer can be used to measure small emfs. Consider the following:

The potentiometer wire AC is 1.00m long and its resistance is 10.0 ohms. The driver cell has negligible internal resistance. Between X and Y is a thermocouple with one junction in ice-water, and the other in boiling water. The balance point is found at J, with AJ = 0.75m.

The total resistance R_{T} of the 990 ohm resistor and the 10 ohm wire
= 1000 ohms. At balance, no current goes through G, so the current I though AC at balance is found
from:

This is the p.d. across AJ produced by the driver cell and, as the
potentiometer is balanced, it is exactly balanced by the emf produced by the
thermocouple, i.e. the emf of the thermocouple is 15 mV

With the 990 ohm resistor as shown, the p.d. across AC is only 20mV, and hence we get a long balance length with the thermocouple. Without the 990 ohm resistor in place the balance length would be very short, making the determination of the emf of the thermocouple inaccurate.

**© **

A Level Physics - Copyright © A
C Haynes 1999 & 2004