Electromotive Force and Internal Resistance

Before we start, the golden rules from GCSE:

The ? through any components connected in series is the same through each component.The ? across any components connected in parallel is the same across each component.

Current Potential Difference

The PD between the terminals of a cell is caused by a chemical reaction within the cell.The voltage produced is known as the Electromotive force (EMF.)

As soon as we draw current from the cell or battery, a p.d. is produced over any internal resistance in the cell. (The cell is made of metals which have resistance!)

This p.d. reduces the voltage we can measure across the terminals of the cell - the terminal p.d.

emf Internal Resistance

An ideal cell is easy to deal with so we invent a model of a real cell as being made up of a fixed resistor (representing the internal resistance (r)) and a cell with no internal resistance of emf E.

Typical value E for a cell is 1.5V. Terminal p.d. will be a little less.

emf Internal Resistance

rE

The Terminal p.d. (V) will be equal to the p.d. across the load.

EE

rr

VV

Try to explain the following equations:V=IRp.d. lost across internal resistance = IrTerminal p.d. = E - IrV = E - IrE = V +IrE = IR + IrE = I(R+r)You can learn these if you want to but it is much better to understand and be able to construct them yourself.

Questions:

1 A battery has an internal resistance of 2 and anemf of 6V. What is the current drawn and terminalp.d. when a load of 8 is applied to it?Solution:

Draw a diagram:

2

I / A

6V

8

2

I / A

6V

8

Questions:

1 A battery has an internal resistance of 2 andan emf of 6V. What is the current drawn andterminal p.d. when a load of 8 is applied to it?

Total Resistance = 2+8 = 10

I=V/R

I = 6/10 = 0.6 A

V = E - Ir

V = 6 - (0.6 x 2)

V = 4.8V

2 A battery has a terminal p.d. of 1.5V in open circuit and 1.4V with a load of 10. What is the internal resistance of the cell?

In open circuit R is infinite, so I is zero. That will make E = V - Ir becomeV=Eso the emf is 1.5V

In the second case with a load of 10 using E = V +Ir will give1.5 = 1.4 +Ir (1)But I = V/Rso I = 1.4 / 10 = 0.14A

?

I / A

1.5V

10

substituting back into (1) we get1.5 = 1.4 + (0.14 x r)1.5 - 1.4 = 0.14 x rr = 0.1 / 0.14r = 0.714

Determination of emf and rV = E - IrUse a variable load, measure I and V

V

AWe could writeV = -r x I + EWhere r and E are constants, V and I are measurable variables.

This is of the form y=mx+c such that if we plot V on the y axis (ordinate) and I on the x axis, it will produce a straight line of gradient -r and intercept E on the y axis.

Vary R to obtain sets of values of V and I Plot V against I Determine the y intercept - that will be equal to E Determine the gradient - that will be equal to -r

It is better to plot the graph rather than just substitute the valuesbecause:

it has the effect of averaging out any errors which arerandom

obvious mistakes are easily seen and not included.

Method Determination of emf and r

V

A

High internal resistance ?

the load could have less resistance than rso more power would be developed in the power supply than the loadnot very good for delivering powerbut what if the supply was a very high voltage?it would effectively limit the current and terminal p.d.

So it is good for a high voltage, low current supply.

Low Internal resistance ?

very little voltage is lost across the internal resistancemost of the power is delivered to the load

So it would be good for a car batterySo it would be good for a car battery

providing low voltage and high current toproviding low voltage and high current to

turn the starter motor.turn the starter motor.

How do you get the maximum power from a power supply?

The max power theorem says when r = R (internal resistance = load)The max power isdevelopedin the load

note:50% of the energy is wasted producing heat in the supplyyou get a lot of power

but waste a lot of the energy stored in the power supply

The total pd (V) across the pair of resistors is the sum of the p.d.’s across each one i.e.

V = V1 + V2

But V=IR

so IR = I1R1 + I2R2

Since they are in series, I = I1 = I2

so R = R1 + R2

R1 R2

Resistors in series

You could be asked to prove that R = R1 + R2 in an exam!

This would be typical of a four mark question. That is 4/60 - well worth understanding and learning.

Resistors in Parallel

We want to find the resistance R that can replace these and draw the same current I and have the same pd across it, V.Conservation of charge will show that

I = I1 + I2

But I = V/Rso V/R = V1/R1 + V2/R2

Since the resistors are in parallel, V = V1 = V2

so

R1

R2

I1

I

I2

21

111

RRR

Don’t be over confident. Many A level students can’t do the maths accurately.

To add fractions you must find a common denominatorTo solve for R you must eventually turn the equation upside down

Question

What is the combined resistance of a 5 and 4 resistance connected first in series, then in parallel?

In series

R = R1 + R2

5+4 = 9

In parallel

R = 20/9 = 2.2

21

111

RRR

20

9

45

54

4

1

5

11 xR