This document discusses the internal resistance of batteries and power supplies. It explains that batteries and power supplies behave as if they have a series resistor inside them, called the internal resistance. When a battery is connected to a circuit, current flows through the internal resistance, causing a voltage drop and reducing the terminal voltage from the theoretical EMF voltage. The greater the current drawn from the battery, the greater the voltage drop due to the internal resistance.

1. Internal
Resistance
The hidden resistance of cells,
batteries and power supplies

2. What would happen if a low resistance copper wire
is connected across the terminal of a 1.5V torch
• Theoretically using
V=IR
V
– if R is small (as it would
be in a short thick wire)
the current should be
large, really large if R
is very small
battery?
R
I 

3. Lets try some figures
• For such a short and thick piece of copper
wire a resistance of 0.01 is not
unreasonable
• The battery has a voltage of 1.5 V
so theoretically;
1.5
0.01
V
I  
R
 150A
• This is serious amount of current the amount
needed to start your car
• Enough to heat the wire red hot and melt it
• That’s not what’s happening WHY??

4. What’s wrong and why?
• Must be the calculation
1.5
0.01
V
I  
R
150A 
• Either the value for V or R must be wrong,
how could we check?
– Measure R with a multimeter
– Measure V with a voltmeter

5. so;
• The battery has 1.5 V printed on it so let’s
measure R (R=0.001)
– That’s even worse and would make I 1500A ( )
• So what about the voltage of the battery?
1.5
0.001
I 
– When its connected to the wire the voltage dropped
– Why??

6. Internal Resistance of Batteries
and Power Supplies
• Batteries and power
supplies behave as if
they have a series
resistor inside them.

7. EMF
• When the battery or power
supply is not connected to a
circuit (no current drawn) the
voltage is at its theoretical
maximum
• This voltage is called the
EMF (electro motive force) of
the battery or power supply
(symbol εo ).
0 
V

8. Terminal Voltage
•As soon as the cell, battery or power
supply is attached to a circuit current
flows.
•This current passes through the
internal resistance and uses some of
the energy (voltage) provided by the
cell, battery or power supply (V=IR)
•So the voltage left for the circuit, the
terminal voltage (VT) is the EMF
minus the voltage loss across the
internal resistance
0 IntRes V V T  
or because V=IR
V IR T   0 
V
VT
V
VIntRes

9. Effect in Circuits
• The greater the current
drawn by the circuit the
lower the terminal voltage
(as more current flows it
passes through the internal
resistance and more energy
(voltage) is lost (V=IR))
• Older or flat batteries
have a higher internal
resistance than new or
fully charged batteries
(steeper line on graph)
Terminal Voltage of a Battery
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0 0.2 0.4 0.6 0.8 1 1.2
Current (A)
Voltage (V)
Older or flatter battery

10. Examples
1. When a 1.5V cell is connected to a 3.0Ω load resistor
the terminal voltage drops to 1.0 V. Calculate the
internal resistance of the cell.
1.5Ω
2. As a student starts her car the 12.0V emf of the car
battery drops to 8.0V. If the current drawn is 150A
calculate the internal resistance of the battery.
0.027Ω
3. A 9.0V battery is connected to a smoke alarm. The
alarm has a resistance of 630Ω and draws a current of
14mA. Find the internal resistance of battery.
13Ω

11. Concept Questions
• When a car has a flat battery why wont it start the car?
• What measurable differences are there between a flat
and fully charged battery?
• Dry cells and batteries have a relatively high internal
resistance. Explain why they are of limited use for low
resistance circuits that require a terminal voltage close to
the emf?
• What happens to the terminal voltage of a cell as it goes
flat
• What happens to the emf of a cell as it goes flat?
• How would you test a battery to see if it is flat?

12. Examples
• ESA, Activity 13B, Pg 210
• ABA Pg 136-138

13. Assignment

14. • That's an excellent question! Measuring internal resistance is a little more complicated. In general, the internal
resistance of a battery is not a fixed value. It varies over time as the battery loses energy and also varies
depending on the load, or how much current is drawn from the battery. Engineers refer to this as a non-linear
resistance. You might also find that if really load down a battery, the internal resistance changes dramatically. In
order to fully characterize the internal resistance, we need to take measurements over time and for different loads.
The result is a set of graphs, each plotting values of resistance as a function of load current. This can be an
involved process, depending on how much data you want. We can, however, get an estimate of the internal
resistance by taking only two measurements with a voltmeter. First, we need to measure the "open circuit" voltage
of the battery. This is simply the voltage at the battery terminals when no current is being drawn. Practically, it's
hard to measure voltage without drawing some current, but most voltmeters have a high enough input resistance
that it can be neglected. So we first measure the open circuit voltage, Vo. Let's say it's 0.9 volts, as in your case.
Now we need to load down the battery and measure the voltage at the battery terminals again, let's call that Vl.
Let's call the load resistance Rl. Given all those values, the equation for the internal resistance (Ri) of the battery
is: Equ 1: Ri = Rl * ((Vo/Vl) - 1) You can derive this equation from the equation for a voltage divider: Equ 2: Vl = Vo
* Rl / (Ri + Rl) This equation can be derived from application of Ohm law and Kirchoffs law. You can find that
information from any text on electric circuit theory. Now we need to choose a value of Rl that gives us good
numbers. From inspection of equ 2, we can observe that if Rl equals Ri, the load voltage will be exactly half of the
open circuit voltage. This trick is commonly used if you have a variable resistor within the range of internal
resistance. This is not always possible, so you might have to use several fixed resistors of different values. You
will probably find that typical alkaline cells tend to have relatively low internal resistances when they are new. It's
probably in the range of an ohm or so. You will probably need a resistor that is not more than 10 times that value
to make good measurements. If you can find a 10 ohm resistor that's OK. Of course, you should start with higher
values to see the effects. Start with 100 Ohms and see what you get. Then try lower values and run the
calculations again. You can enter your values into a spreadsheet and plot the results. A word of caution: be careful
when using low value resistors with good batteries, the current draw will be high enough the heat up the resistor
(and the battery!).