The diagrambelow shows two circuits A and B that were used by a student to test a battery of four identical cells. In circuit A, there was no load resistor and in circuit B a load resistor was connected. Assume that the meters in the circuits are ideal.
(a)
Explain why there is a difference in voltages recorded in the two circuits.
In circuit B, the resistance of the load resistor R is altered so that a series of values on the voltmeter and the corresponding values of the current on the ammeter are obtained.
(c)
(i)
On the axes above, sketch the graph you would expect to obtain as R is changed.
[2]
(ii)
Outline how the values of ε and r can be obtained from the graph.
A cell is connected to an external resistor and the terminal voltage across the cell monitored. The graph shows the discharge time for one cell with a current of 0.5 A.
(d)
Determine the terminal voltage of the single cell. Show your working clearly.
The diagram shows a cell of e.m.f. ε, and internal resistance, r, is connected to a variable resistor R. The current through the cell and the terminal p.d. of the cell are measured as R is decreased.
The graph below shows the results from the experiment.
(a)
State the relationship between the terminal p.d. and current and explain why this relationship occurs.
Draw a line on the graph above to show the results obtained from a cell with the half the e.m.f. but double the internal resistance of the first cell. Label your graph A.
The circuit diagram shows that the 9.0 Ω resistor is now connected in parallel with an unknown resistor, R. The battery now supplies a current of 3.0 A and has the same internal resistance r as the previous circuit.
A battery of e.m.f. 11.5 V and internal resistance r is connected in a circuit with three identical 13 Ω resistors. A current of 0.40 A flows through the battery.
(b)
Calculate the potential difference between points X and Y in the circuit.