Questions on EMF and internal resistance
Q5. A battery of EMF
and internal resistance r is connected in series to a variable resistor R and an ammeter of negligible resistance. A voltmeter is connected across R, as shown in the diagram.
![](5.png)
(a)
(i) State what is meant by the EMF of the battery.
![](5a.png)
The EMF of the battery is the electrical energy produced per unit charge
[- it is equal to the potential difference/voltage across terminals of the battery when there is no current being drawn from it – or the reading on the voltmeter when connected on open circuit]
(ii) The reading on the voltmeter is less than the EMF. Explain why this is so.
The reading on the voltmeter gives the potential difference across the external circuit. When a current flows some of the EMF is used to drive that current through the battery
. This results in voltage being 'lost' across the internal resistance
as this is not measured on the instrument.
(3)
(b) A student wishes to measure
and r. Using the circuit shown above, the value of R is decreased in steps, and at each step the readings V and I on the voltmeter and ammeter respectively are recorded.
These are shown in the table below.
reading on voltmeter/V |
reading on ammeter/A |
8.3 |
0.07 |
6.8 |
0.17 |
4.6 |
0.33 |
2.9 |
0.44 |
0.3 |
0.63 |
(i) Give an expression relating V, I,
and r.
![](5a.png)
= V + Ir ![](../../../../graphics/symbols/nuclides/ticksmall.png)
(ii) Draw a graph of V (on the y-axis) against I (on the x-axis) on graph paper.
Marks were allocated as follows:
correctly labelled axes (with units)
correct plotting
best fit straight line ![](../../../../graphics/symbols/nuclides/ticksmall.png)
![](5b.png)
(iii) Determine the values of
and r from the graph, explaining your method.
= V + Ir can be rearranged to be the equation of a straight line ![](../../../../graphics/symbols/nuclides/ticksmall.png)
V = -rI +
if we compare this to the equation for a straight line
Y = mx + c we can see that
Y is V - the reading on the voltmeter (external potential difference)
x is I - the ammeter reading – the current
-r is m - the gradient of the graph ![](../../../../graphics/symbols/nuclides/ticksmall.png)
(one mark for the negative sign) = -14 ![](../../../../graphics/symbols/omega_red.png)
so r = 14 ![](../../../../graphics/symbols/omega_red.png)
![](../../../../graphics/symbols/nuclides/ticksmall.png)
is c - the intercept on the y axis
= 9.2 V ![](../../../../graphics/symbols/nuclides/ticksmall.png)
(9)
(Total 12 marks)