Download Higher Physics EMF and Internal Resistance Test – Marking

Higher Physics EMF and Internal Resistance Test – Marking Scheme
This test should be completed in 1 period under proper exam conditions. No formula sheet is required. Lined
and graph paper should be handed out. The completed test should be passed to Mr Watt for marking.
1. What do the letters EMF stand for?
ELECTRO MOTIVE FORCE
(1)
2. A certain battery has an EMF of 4v. What is meant by this statement?
(1)
4 JOULES OF ENERGY IS GIVEN TO EACH COULOMB OF CHARGE LEAVING THE BATTERY
3. Describe the term ‘lost volts’.
(1)
VOLTAGE DROP ACROSS THE INTERNAL RESISTANCE OF THE BATTERY
4. A battery has an EMF of 6v and an internal resistance of 0.5 Ohms. Calculate the terminal
potential difference (tpd) if a current of 2 Amps is drawn from the battery.
(2)
Tpd = EMF – lost volts = 6 – (2 x 0.5) = 5v
5. The following circuit is set up with switch s open:
(a) Calculate the tpd of the battery
If a current of 0.75A flows.
(2)
Tpd = voltage across 10 Ohm resistor
= IR = 0.75 x 10 = 7.5v
(b) Calculate the lost volts in this case. (1)
Lost volts = EMF – tpd = 9 – 7.5 = 1.5v
(c) Calculate the internal resistance of the battery. (2)
Int R = lost volts/current = 1.5/0.75 = 2 Ohms
(d) Switch s is now closed. Calculate the lost volts. (3)
When switch is closed, external resistance = 5 Ohms,
Total circuit resistance = 5 + 2 = 7 ohms, and circuit current changes to I = EMF/total R = 9/7 = 1.29A
Lost volts = IRint = 1.29 x 2 = 2.58v
6. A circuit was set up to determine the characteristics of a battery. Readings of voltage and
current were taken and are shown below:
Voltage (volts) Current (Amps)
2.4
0.1
1.8
0.35
1.2
0.6
0.6
0.8
(a) Use these results to create a line graph of voltage (y-axis) against current.
½ MARK FOR APPROPRIATE SCALES
½ MARK FOR LABELLED AXES
½ MARK FOR PLOTTED POINTS
½ MARK FOR BEST FIR STRAIGHT LINE
(b) Use your graph to determine:
(i)
The EMF of the cell.
This is the y-axis crossing = 2.6v (about)
(ii)
The internal resistance of the cell
(2)
(1)
(2)
(**) Internal resistance of battery = - gradient of graph
= -[(y2 – y1)/(x2 – x1)]
Make it clear on your graph which two points you
are using, and make sure you actually use (y2 – y1)
and (x2 – x1).
You will then get a negative x a negative, which
gives a positive.
Without the very first line (**) in your answer you
get zero marks.
(Total marks available = 18)