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S5 Physics Paper: Electricity, Circuits and Electromagnetism (100 marks)
S5 Physics – electricity and magnetism
S5 Physics paper: six questions on current, resistance, series and parallel circuits, power and cost, magnetic force and induction, with full answers and marking notes.
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Paper details S5 Physics practice paper with six questions covering current and resistance, series and parallel circuits, electrical power and cost, magnetic force, transformers and electromagnetic induction. Complete answers with marking notes.
Section A: Current, Charge and Resistance (16 marks)
Current is the charge passing a cross-section per unit time, I equals Q over t. Ohm law V equals I times R holds for metal conductors at steady temperature, and the power P equals VI, or I squared times R, gives the rate of energy transfer.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A current of 2.5 A flows in a circuit for 4 minutes. Calculate the charge that passes through the circuit. | Answer: t equals 4 times 60, that is 240 s, so Q equals 2.5 times 240, which is 600 C. (Marking: 1 mark for the time conversion, 1 mark for the formula, 1 mark for the value with unit.) |
| 2 | Given that the charge on one electron is 1.6 x 10^-19 C, estimate how many electrons carry the charge in part (a). | Answer: N equals 600 divided by 1.6 x 10^-19, which is about 3.75 x 10^21 electrons. (Marking: 1 mark for the formula, 1 mark for substitution, 1 mark for the order of magnitude.) |
| 3 | If the potential difference across the circuit is 12 V, find the resistance of the resistor and the power it dissipates. | Answer: R equals 12 divided by 2.5, which is 4.8 ohm, and P equals 12 times 2.5, which is 30 W. (Marking: 2 marks for the resistance, 2 marks for the power, 1 mark for units.) |
| 4 | Find the electrical energy dissipated by the resistor in 4 minutes, in kJ. | Answer: E equals 30 times 240, that is 7200 J, or 7.2 kJ. (Marking: 2 marks for the formula, 1 mark substitution, 2 marks for the unit conversion.) |
Section B: Series and Parallel Circuits (16 marks)
The equivalent resistance of a parallel section is smaller than any single branch. Potential differences add in series, while branches in parallel share the same potential difference and their currents add up.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | Two 4 ohm resistors are connected in parallel, and the pair is in series with an 8 ohm resistor across a 20 V supply. Find the equivalent resistance of the whole combination. | Answer: for the parallel pair, one over R equals one quarter plus one quarter, so R equals 2 ohm, and the total is 2 plus 8, that is 10 ohm. (Marking: 2 marks for the parallel expression, 1 mark for its value, 1 mark for the total.) |
| 6 | Find the total current drawn from the supply. | Answer: I equals 20 divided by 10, which is 2 A. (Marking: 1 mark for the formula, 1 mark for the value, 1 mark for the unit.) |
| 7 | Find the potential difference across the parallel pair and across the 8 ohm resistor. | Answer: the parallel pair has 2 times 2, that is 4 V, and the 8 ohm resistor has 2 times 8, that is 16 V; the two add to 20 V. (Marking: 2 marks for the parallel value, 1 mark for the series value, 1 mark for the check.) |
| 8 | Find the current in each 4 ohm resistor and the power dissipated in the 8 ohm resistor. | Answer: each 4 ohm resistor carries 4 divided by 4, that is 1 A; the 8 ohm resistor dissipates 4 times 8, that is 32 W. (Marking: 3 marks for the branch currents, 1 mark for the power formula, 1 mark for the value with unit.) |
Section C: Power, Energy Consumption and Cost (16 marks)
The rated power of an appliance is its power at the rated voltage. The energy needed to heat water comes from mass times specific heat capacity times temperature rise, and electricity is billed in kilowatt hours, where one unit is 3.6 x 10^6 J.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A kettle is marked 220 V, 2000 W. Find the current it draws and the resistance of its heating element when working normally. | Answer: I equals 2000 divided by 220, which is 9.09 A, and R equals 220 squared divided by 2000, which is 24.2 ohm. (Marking: 2 marks for the current, 2 marks for the resistance expression, 1 mark for the value.) |
| 10 | Water of mass 1.5 kg is heated from 25 °C to 100 °C. Find the energy absorbed by the water, taking its specific heat capacity as 4200 J per kg per K. | Answer: the temperature rise is 75 K, so E equals 1.5 times 4200 times 75, that is 472500 J, or about 472.5 kJ. (Marking: 1 mark for the temperature rise, 2 marks for the formula, 1 mark for the value.) |
| 11 | Assuming no heat loss, find the time needed to bring that water to the boil. | Answer: t equals 472500 divided by 2000, which is about 236 s, roughly 3 minutes 56 seconds. (Marking: 2 marks for the formula, 1 mark for the value, 1 mark for the unit.) |
| 12 | If electricity costs 1.2 dollars per unit, find the cost of heating that water once, and state whether a 5 A or a 13 A fuse should be used. | Answer: the energy used is 472500 divided by 3.6 x 10^6, about 0.131 unit, so the cost is about 0.16 dollars; the working current of 9.09 A exceeds 5 A, so a 13 A fuse is required. (Marking: 1 mark for the unit count, 1 mark for the cost, 1 mark for the fuse choice.) |
Section D: Magnetic Force, Parallel Wires and Transformers (17 marks)
A current-carrying wire in a magnetic field experiences a force of B times I times l times the sine of the angle, its direction given by Fleming left-hand rule, and the force is zero when the wire is parallel to the field.
| # | Question | Answer and marking notes |
|---|---|---|
| 13 | A straight wire of length 0.25 m carrying 3.0 A is placed at right angles to a uniform magnetic field of flux density 0.40 T. Find the magnitude of the force on the wire and name the rule that gives its direction. | Answer: the force is 0.40 times 3.0 times 0.25, which is 0.30 N, and its direction is given by Fleming left-hand rule. (Marking: 2 marks for the formula, 2 marks for substitution and value, 1 mark for the rule.) |
| 14 | If the wire is turned so that it lies parallel to the field, what happens to the force? Explain. | Answer: the force becomes zero, because the sine of the angle is zero and the charges move along the field, so there is no perpendicular component of force. (Marking: 2 marks for the answer, 1 mark for the explanation.) |
| 15 | Two long parallel wires 0.05 m apart carry 3.0 A each in the same direction. Find the force per metre on each wire and state whether they attract or repel. | Answer: the force per metre is 4 pi times 10^-7 times 3.0 times 3.0 divided by 2 pi times 0.05, which is 3.6 x 10^-5 N per m; since the currents are in the same direction, the wires attract. (Marking: 2 marks for the formula, 2 marks substitution, 1 mark for the value with unit, 1 mark for attraction.) |
| 16 | An ideal transformer has 100 turns on its secondary and steps 220 V down to 11 V, delivering 44 W. Find the number of primary turns and the secondary current. | Answer: the primary has 100 times 220 divided by 11, that is 2000 turns, and the secondary current is 44 divided by 11, which is 4.0 A. (Marking: 2 marks for the turns, 1 mark for the current.) |
Section E: Electromagnetic Induction and Alternating Current (18 marks)
Faraday law states that the induced electromotive force equals the number of turns times the rate of change of magnetic flux, and Lenz law says the induced current opposes the change producing it. The root mean square value of an alternating supply is its peak divided by root two.
| # | Question | Answer and marking notes |
|---|---|---|
| 17 | A coil of 200 turns has an area of 0.015 m squared. The flux density through it rises uniformly from 0.20 T to 0.60 T in 0.40 s. Find the change in magnetic flux through the coil. | Answer: the change in flux density is 0.40 T, so the change in flux is 0.40 times 0.015, which is 6.0 x 10^-3 Wb. (Marking: 1 mark for the change in flux density, 2 marks for the formula, 1 mark for the value with unit.) |
| 18 | Find the magnitude of the induced electromotive force in the coil. | Answer: the electromotive force is 200 times 6.0 x 10^-3 divided by 0.40, which is 3.0 V. (Marking: 2 marks for the formula, 2 marks substitution, 1 mark for the value.) |
| 19 | If the same change in flux takes 0.80 s instead, what is the induced electromotive force? State how the direction of the induced current relates to the change in flux. | Answer: the electromotive force is 200 times 6.0 x 10^-3 divided by 0.80, which is 1.5 V; by Lenz law the induced current produces a field that opposes the increase in flux, so it circulates the opposite way. (Marking: 3 marks for the value, 2 marks for the Lenz law statement.) |
| 20 | The peak voltage of the local mains supply is 340 V. Find its root mean square value and explain why appliances are rated at the root mean square value. | Answer: the root mean square voltage is 340 divided by root two, which is about 240 V; that value is the steady direct voltage giving the same heating effect, so it is used for ratings. (Marking: 2 marks for the value, 2 marks for the explanation.) |
Section F: Electromotive Force, Internal Resistance and Terminal Voltage (17 marks)
The electromotive force of a cell is the energy converted per unit charge, and internal resistance makes the terminal voltage fall on load according to the relation terminal voltage equals emf minus current times internal resistance.
| # | Question | Answer and marking notes |
|---|---|---|
| 21 | A cell of electromotive force 6.0 V drives a current of 0.5 A through an 11 ohm resistor. Find the internal resistance of the cell. | Answer: the emf equals the current times the sum of the external and internal resistance, so 6.0 equals 0.5 times the sum, giving an internal resistance of 1.0 ohm. (Marking: 2 marks for the formula, 2 marks substitution, 1 mark for the value with unit.) |
| 22 | Find the terminal voltage of the cell under this load. | Answer: the terminal voltage is 6.0 minus 0.5, that is 5.5 V, which also equals 0.5 times 11. (Marking: 2 marks for the formula, 1 mark for the value, 1 mark for the check.) |
| 23 | If the 11 ohm resistor is replaced by a 2 ohm resistor, find the current in the circuit. | Answer: the current is 6.0 divided by the sum of 2 and 1, which is 2.0 A. (Marking: 2 marks for the formula, 2 marks for the value.) |
| 24 | Find the power dissipated in the 2 ohm resistor, and state where the remaining power goes. | Answer: the external power is 4 times 2, that is 8.0 W; the remaining 4.0 W is dissipated as heat inside the cell, and the total of 12 W equals the emf times the current. (Marking: 2 marks for the external power, 1 mark for the internal power, 1 mark for the energy statement.) |
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