Transcript EMF and Internal Resistance

EMF and Internal Resistance
Electricity Lesson 8
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 source.

pd is the energy per unit charge converted to
other energies by the components.
Electromotive Force

The energy supplied to a circuit by a battery is given by:
E

Q

W 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 battery itself.
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.
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.
Circuit Diagram
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
Including internal resistance

The resistors are
connected in series.

So total resistance is:R+r


Current through the
cell:-
I

Rr
Lost pd

So the cell
  I ( R  r )  IR  Ir

In other words:-
cell emf  the terminal pd  the ' lost' pd

In energy terms the lost pd is the energy per coulomb
dissipated or wasted inside the cell due to the internal
resistance.
Lost pd

The terminal pd can be
calculated using:-
V  IR

The lost pd can be
calculated using:-
v  Ir
The equation becomes:-
  IR  Ir  V  v
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 resistance
R 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.