Tutorial 7 B - Internal Resistance

Learning Objective

To understand the concept of internal resistance;

To use internal resistance of a cell;

Key Question

 

How do cells provide energy to a circuit?

How do we deal with internal resistance?

Energy and Cells

Batteries (or more strictly speaking cells) convert chemical energy into electrical energy.  Generators turn kinetic energy into electrical energy.  In doing so, they keep the negative terminal with an excess of electrons and the positive terminal with a deficiency of electrons.  A battery does a job of work in pumping the electrons around the circuit.

 

A battery is said to produce Emf which is defined as

the energy converted into electrical energy when unit charge passes through the source.

 

This is similar to the definition for potential difference, except that it describes the conversion to electrical energy, rather than the conversion from electrical energy.  It represents the total energy that can be supplied to a circuit.  No circuit at all is 100 % efficient.  Some energy is dissipated in the wires, or even in the battery itself.

 

 

 

The curly E (E) is the physics code for emf, and the units are Joules per coulomb (JC-1) or volts (V).

 

A more simple and practical way of remembering emf  is to say that it is the terminal voltage of a battery or generator in open circuit, i.e. when no current is being drawn.

 

Question 3

Why does a battery not pump protons?

Answer

Question 4

A battery converts 13 000 J of chemical energy into electrical energy.  It does so by giving a current of 0.5 A for 2 hours.  What is its emf?

Answer

Internal Resistance

All batteries and generators dissipate heat internally when giving out a current, due to internal resistance.  A perfect battery has no internal resistance, but unfortunately there is no such thing as a perfect battery.  Nickel-Cadmium and Lead-Acid batteries have very low internal resistance, and we can regard these as almost perfect.  These batteries can provide very high currents.

 

Suppose we connect a cell to a high resistance voltmeter.  (A perfect voltmeter has infinite resistance.  A digital multimeter has a very high resistance, so needs a tiny current; it is almost perfect.  An ordinary moving coil voltmeter has a relatively low resistance, so it takes a small but appreciable current.)

 

The voltmeter will read (very nearly) the emf.

Suppose we now add a load.  We will assume the wires have negligible resistance.

This time we find that the terminal voltage goes down to V.  Since V is less than , 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. 

Question 5

What is the difference between emf and potential difference?

Answer

Question 6

What is meant by a perfect battery?  Why are real batteries not perfect?

Answer

 

EMF and Terminal Voltage

 

Emf = Useful volts  + Lost volts

 

In code:

= V + v

We can represent the circuit with internal resistance as:

 

 

So our cell is now a:

perfect battery in series with an internal resistor, r

 

We can now treat this as a simple series circuit and we know that the current, I, will be the same throughout the circuit.  We also know the voltages in a series circuit add up to the battery voltage.

Emf = voltage across R + voltage across the internal resistance

                =         V                 +                    v

 

We also know from Ohmís Law that:

V = IR

and

v = Ir

so we can write:

 

= IR + Ir

 

Many students panic at the sight of internal resistance problems.  All you have to do is turn the cell with the internal resistance into a

perfect battery in series with its internal resistor,

and treat it as a simple series circuit.

In experiments to determine the internal resistance, we get a graph like this:

The graph is a straight line, of the form y = mx + c.  

We can make the equation for internal resistance:

V = -rI + E

There are three features on the graph that are useful:

  • The intercept on the y-axis tells as the emf.

  • The intercept on the x-axis tells us the maximum current the cell can deliver when the p.d. is zero, i.e. a dead short circuit.

  • The negative gradient tells us the internal resistance.

Question 7

Explain how this statement is consistent with Kirchhoff II

Answer

Question 8

A battery has an internal resistance of 0.5 ohms.  The battery has an emf of 1.52 V.  When it is connected to a resistor, the terminal voltage falls to 1.45 V.  What current is flowing?  What is the value of the resistor?

Answer

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Tut 7 A

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