Physics › Practise › Questions › EMF and internal resistance
Questions on EMF and internal resistance
Electromotive force is the energy a source transfers to each coulomb passing through it, and a real cell carries an internal resistance r. The terminal pd is then V = ε − Ir, the lost volts growing with the current. Plot terminal pd against current: the intercept is ε and the gradient is −r.
20 original questions · 63 marks · the EMF and internal resistance notes · Electricity
To print: the questions (PDF, 84 KB) · the worked answers (PDF, 78 KB)
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening a mark scheme: the schemes award marks point by point, and the marks are easier to see when you have something of your own to compare against.
Define the electromotive force (emf) of a source.
Mark scheme
The emf is the energy transferred to each unit of charge by the source, ε = W/Q (1). It is measured in volts, or joules per coulomb (1).A battery of emf 6.0 V and internal resistance 1.0 Ω supplies a current of 0.50 A. Calculate its terminal potential difference.
Mark scheme
V = ε − Ir = 6.0 − 0.50 × 1.0 (1)
V = 5.5 V (1)State the equation linking emf, current, external resistance and internal resistance.
Mark scheme
ε = I(R + r) (1), where R is the external resistance and r the internal resistance (1).State the condition under which the terminal pd of a cell equals its emf.
Mark scheme
When no current flows (I = 0), since the lost volts Ir are then zero (1).A battery of emf 9.0 V has a terminal pd of 8.4 V while supplying a current. Calculate the lost volts, and state where the corresponding energy is transferred.
Mark scheme
Lost volts = ε − V = 9.0 − 8.4 = 0.6 V (1); the energy is dissipated in the internal resistance, warming the battery (1).Explain why a digital voltmeter of very high resistance connected directly across an isolated cell reads almost exactly the cell's emf.
Mark scheme
The voltmeter draws almost no current from the cell (1), so the lost volts Ir are negligible and the terminal pd is (almost) equal to ε (1).A cell of emf 12 V and internal resistance 1.0 Ω is connected to an external resistor of 5.0 Ω. Calculate the current in the circuit.
Mark scheme
I = ε/(R + r) (1)
I = 12/(5.0 + 1.0) (1)
I = 2.0 A (1)A battery of emf 9.0 V and internal resistance 0.50 Ω is connected to a 4.0 Ω resistor. Calculate (a) the current, (b) the terminal potential difference and (c) the power dissipated inside the battery.
Mark scheme
(a) I = 9.0/(4.0 + 0.50) = 2.0 A (1)
(b) V = IR = 2.0 × 4.0 (1)
V = 8.0 V (1)
(c) Pinternal = I2r = 2.02 × 0.50 = 2.0 W (1)A cell of emf 1.5 V delivers a current of 0.30 A, and its terminal potential difference falls to 1.35 V. Calculate the internal resistance of the cell.
Mark scheme
Lost volts = ε − V = 1.5 − 1.35 = 0.15 V (1)
r = (ε − V)/I = 0.15/0.30 (1)
r = 0.50 Ω (1)A cell of emf 6.0 V has an internal resistance of 0.20 Ω. Calculate the current if its terminals are short-circuited (R = 0).
Mark scheme
I = ε/r (1)
I = 6.0/0.20 = 30 A (1)A battery has a terminal pd of 5.7 V when supplying a current of 0.60 A, and 5.1 V when supplying 1.8 A. Determine the internal resistance and the emf of the battery.
Mark scheme
Write V = ε − Ir for each reading (1); subtracting: 5.7 − 5.1 = r × (1.8 − 0.60), so r = 0.50 Ω (1); substitute back: ε = 5.7 + 0.60 × 0.50 (1) = 6.0 V (1).A car battery of emf 12.6 V and internal resistance 25 mΩ supplies 160 A to the starter motor. Calculate the terminal pd of the battery, and determine the fraction of the battery's power that is delivered to the external circuit.
Mark scheme
r = 0.025 Ω (1); V = ε − Ir = 12.6 − 160 × 0.025 = 8.6 V (1); fraction = IV/Iε = V/ε (1) = 8.6/12.6 = 0.68 (1).A cell of emf 6.0 V and internal resistance 0.50 Ω is connected to two 5.0 Ω resistors in parallel. Calculate the current supplied by the cell and the cell's terminal pd.
Mark scheme
External resistance = 5.0/2 = 2.5 Ω (1); I = ε/(R + r) = 6.0/(2.5 + 0.50) = 2.0 A (1); V = IR = 2.0 × 2.5 = 5.0 V (1).A graph of terminal potential difference V against current I for a cell is a straight line with a V-axis intercept of 1.5 V and a gradient of −0.5 V A−1. State the emf and internal resistance, and calculate the terminal pd when the current is 1.0 A.
Mark scheme
From V = ε − Ir, the intercept is the emf, so ε = 1.5 V (1)
The gradient is −r, so r = 0.5 Ω (1)
At I = 1.0 A: V = 1.5 − 0.5 × 1.0 (1)
V = 1.0 V (1)A battery of emf 12 V and internal resistance 2.0 Ω is connected to a variable resistor. Calculate the resistance value that makes the terminal potential difference equal to 10 V.
Mark scheme
Lost volts = 12 − 10 = 2.0 V (1)
I = (ε − V)/r = 2.0/2.0 = 1.0 A (1)
R = V/I = 10/1.0 (1)
R = 10 Ω (1)Explain why the terminal potential difference of a battery falls as it supplies a larger current.
Mark scheme
Some energy per unit charge is used driving the current through the battery's own internal resistance, giving 'lost volts' Ir (1). As the current increases, Ir increases (1), so the terminal pd ε − Ir available to the external circuit falls (1).Describe an experiment to determine the emf and internal resistance of a cell using a variable resistor, an ammeter and a voltmeter. Explain how the results are processed, and why this method gives a better value for the emf than a single voltmeter reading across the cell.
Mark scheme
Connect the cell in series with the variable resistor, a switch and the ammeter, with the voltmeter across the cell's terminals (1). Vary the resistance to obtain a wide range of currents, recording V and I each time, opening the switch between readings to limit heating (1). Plot terminal pd V against current I; V = ε − Ir gives a straight line (1). The intercept on the V axis equals the emf (1) and the magnitude of the gradient equals the internal resistance r (1). The intercept extrapolates to I = 0, where there are no lost volts, whereas a voltmeter always draws a small current and so reads slightly below ε (1).A model boat's motor needs a pd of at least 4.4 V across it while drawing 2.0 A. Two batteries are available: battery P (emf 6.0 V, internal resistance 0.90 Ω) and battery Q (emf 5.0 V, internal resistance 0.20 Ω). Deduce which battery should be used.
Mark scheme
P: V = 6.0 − 2.0 × 0.90 = 4.2 V (1); Q: V = 5.0 − 2.0 × 0.20 = 4.6 V (1); only Q's terminal pd exceeds the 4.4 V required, while P falls short (1); choose Q: despite its lower emf, its much lower internal resistance leaves more pd for the motor (1).Two identical cells, each of emf 1.5 V and internal resistance 0.30 Ω, are connected in series to a 2.4 Ω resistor. Calculate the current in the circuit and the pd across the resistor.
Mark scheme
Total emf = 3.0 V and total internal resistance = 0.60 Ω, since both add in series (1); I = 3.0/(2.4 + 0.60) = 1.0 A (1); V = IR = 1.0 × 2.4 = 2.4 V (1).A student varies the external resistance in a circuit containing a single cell and plots the cell's terminal potential difference against the current it supplies, obtaining the graph shown. Determine the emf and the internal resistance of the cell, and hence calculate the current the cell would supply to a 1.6 Ω resistor.
Terminal potential difference against current for one cell, with the line of best fit extended back to meet the pd axis at I = 0. Mark scheme
Gradient taken between two well-separated points on the line of best fit, e.g. (1.40, 0.96) and (0.20, 1.44), giving (0.96 − 1.44)/(1.40 − 0.20) (1)
Gradient = −0.40 V A−1, so r = 0.40 Ω, quoted positive with the gradient described as negative (allow 0.38 Ω to 0.42 Ω) (1)
The intercept at I = 0 is the emf, ε = 1.52 V (allow 1.50 V to 1.54 V) (1)
ε = I(R + r) with the candidate's own values, I = 1.52/(1.6 + 0.40) (1)
I = 0.76 A (1)
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise EMF and internal resistance one question at a time
The player marks nothing for you. It shows one question, waits, then shows the scheme so you can mark yourself, and brings a question back sooner when it went badly.