EMF and Internal Resistance

Electricity Lesson 9

Learning Objectives

To know why the pd of a cell in use is less than its emf.

To know how to measure the internal resistance of a cell.

To know how to calculate how much power is wasted in a cell.

Question

What is the difference between emf and potential difference?

Answer

Emf is the total energy supplied to the circuit per unit charge by the cell.

pd is the energy per unit charge converted to other energies by the components. 

Electromotive Force The energy supplied to a circuit by a cell is given

by:

E is the energy in J, Q is the charge in C, ε is the emf in volts (NOT Newtons).

No circuit at all is 100 % efficient.  Some energy is dissipated in the wires, or even in the cell itself.

Q

E

Internal Resistance

The internal resistance of a source is the loss of potential difference per unit current in the source when current passes through the wire.

It is caused by the opposition to the flow of charge through the source.

We will use the symbol r to represent internal resistance.

A simple circuit… In this circuit the

voltmeter reads (very nearly) the emf.

(A perfect voltmeter has infinite resistance.  A digital multimeter has a very high resistance, so needs a tiny current; it is almost perfect.  )

Add a resistor… This time we find that

the terminal voltage goes down to V. 

Since V is less than E, this tells us that not all of the voltage is being transferred to the outside circuit; some is lost due to the internal resistance which heats the battery up. 

Emf = Useful volts  + Lost volts

Terminal pd

The electrical energy per unit charge delivered by the source when it is in a circuit.

The terminal pd, V, is less than the emf, ε, whenever current passes through the source.

The difference is the lost pd, v, due to the internal resistance of the source.

Including internal resistance

The resistors are connected in series.

So total resistance is:-

R + r

Current through the cell:-

rRI

Lost pd

So the cell

In other words:-

In energy terms the lost pd is the energy per coulomb dissipated or wasted inside the cell due to the internal resistance.

IrIRrRI )(

pd lost'' the pd terminaltheemf cell

Circuit Diagram

Lost pd

The terminal pd can be calculated using:-

The equation becomes:-

The lost pd can be calculated using:-

Irv IRV

vVIrIR

Worked Example

A battery of emf 12 volts and internal resistance 0.5 ohms is connected to a 10 ohm resistor.  What is the current and what is the terminal voltage of the battery under load?

Worked Example Diagram

Step 1

Treat the circuit as a perfect battery in series with an internal resistor.  The circuit becomes:

Worked Solution

Step 2:  Work out the total resistanceR tot = R1 + R2 = 10 ohms + 0.5 ohms

= 10.5 ohms Step 3: Now work out the current:I =

V/R = 12 ÷ 10.5 = 1.14 A

Step 4: work out the voltage across the internal resistor (lost voltage): v = Ir = 1.14 amps × 0.5 ohms = 0.57 volts

Worked Solution

Step 5: work out the terminal voltage: Terminal voltage = emf - lost voltage = 12 - 0.57 = 11.43 volts

We can of course work out the terminal voltage by working the voltage across the 10 ohm resistor, assuming there are no losses.

Power Wasted in a Cell

Recall that power can be calculated using:-

So we can multiply our equation by I to get:-

Or in words:-

IVP IrIR

rIRII 22 supplied,Power

cell in the edPower wastR todeliveredPower

cell by the suppliedPower

Measuring Internal Resistance

Measuring Internal Resistance

Terminal pd and current can be plotted on a graph.

This equation is in the form of a straight line

vV vV IrV

rIV cmxy