OCR A Level Physics A: Series and parallel circuits, internal resistance and potential dividers (H556) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 4.3 (Electrical circuits) of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017: series and parallel circuits, internal resistance and potential dividers. Revise first with the study guide or the revision notes. Sources have negligible internal resistance unless a question says otherwise.

Questions

1. (a) Calculate the total resistance of 6.0 Ω, 12 Ω and 4.0 Ω resistors connected in parallel. [2] (b) This group is connected in series with a 3.0 Ω resistor across a 10 V source. Calculate the current from the source. [2]

2. State Kirchhoff’s first law and second law, and for each say which conserved quantity it expresses. [4]

3. Two cells of e.m.f. 9.0 V and 3.0 V are connected in opposition in a series loop with resistors of 4.0 Ω and 2.0 Ω.

(a) Calculate the current. [2] (b) Calculate the p.d. across the 4.0 Ω resistor. [1] (c) Calculate the power dissipated in the 2.0 Ω resistor. [1]

4. A 9.0 V source in series with a 3.0 Ω resistor is connected in parallel with an 8.0 V source in series with a 2.0 Ω resistor. A 3.0 Ω resistor R is connected across the pair of branches. Calculate the current in each branch and the p.d. across R. [6]

5. A cell of e.m.f. 1.6 V supplies a current of 0.50 A. Its terminal p.d. is then 1.35 V.

(a) Calculate the lost volts. [1] (b) Calculate the internal resistance. [1] (c) Calculate the resistance of the external circuit. [1] (d) Calculate the power wasted inside the cell. [1]

6. (a) A car battery’s terminal p.d. dips when the starter motor is switched on. Explain why. [3] (b) A 1.5 V cell has internal resistance 0.30 Ω. Calculate the current if its terminals are joined by a thick wire of negligible resistance. [2]

7. A student investigates a power supply. The table shows the readings.

I / A	0.20	0.40	0.60	0.80	1.00
V / V	5.52	5.04	4.56	4.08	3.60

(a) Describe the circuit and procedure used to obtain these readings. [3] (b) Use the data to determine the e.m.f. and internal resistance of the supply. [4] (c) Explain why the student opens the switch between readings. [1]

8. Two identical cells each have e.m.f. 1.5 V and internal resistance 0.40 Ω. They supply a 2.0 Ω resistor.

(a) Calculate the current when the cells are in series. [3] (b) Calculate the current when the cells are in parallel. [3]

9. A 5.0 V supply is connected across a 3.3 kΩ fixed resistor in series with an NTC thermistor. The output is taken across the fixed resistor. The thermistor resistance is 10 kΩ at 10 °C and 1.0 kΩ at 50 °C.

(a) Calculate Vout at 10 °C and at 50 °C. [3] (b) Explain why Vout changes in this way. [2] (c) State the thermistor resistance at which Vout is 2.5 V. [1]

10. A 9.0 V supply is connected across resistor R1 in series with a 4.0 kΩ resistor R2. Vout is across R2 and is required to be 2.0 V.

(a) Calculate R1. [3] (b) R1 is replaced by an LDR. State and explain how Vout changes when the light intensity increases. [1]

11. (a) A 12 V supply is connected across a uniform potentiometer track 80.0 cm long. Calculate the output between the slider and the zero end when the slider is 30.0 cm from that end. [2] (b) Two 10 kΩ resistors form a potential divider across the 12 V supply. Calculate Vout across the lower resistor, first unloaded, then with a 10 kΩ load connected across it. [4] (c) Explain why the loaded output is smaller. [2]

Answers

1. (a) 1/R = 1/6.0 + 1/12 + 1/4.0 = 2/12 + 1/12 + 3/12 = 6/12 [1], so R = 2.0 Ω [1] (b) Total R = 2.0 + 3.0 = 5.0 Ω [1]; I = 10/5.0 = 2.0 A [1] Examiner insight: Writing the reciprocal sum before inverting shows where an error arose; leaving 0.5 as the answer is a common slip.

2. First law: the total current arriving at a junction equals the total current leaving [1]; this expresses conservation of charge [1]. Second law: round any closed loop the sum of e.m.f. equals the sum of p.d. [1]; this expresses conservation of energy [1] Examiner insight: Name each law and the quantity it conserves; a bare list of formulae does not answer a “state” question.

3. (a) Net e.m.f. = 9.0 − 3.0 = 6.0 V [1]; I = 6.0/(4.0 + 2.0) = 1.0 A [1] (b) V = 1.0 × 4.0 = 4.0 V [1] (c) P = I²R = 1.0² × 2.0 = 2.0 W [1] Examiner insight: Subtracting the e.m.f.s is the point of the question; adding them gives 12 V and a different current.

4. Let I1 be the current from the 9.0 V branch and I2 from the 8.0 V branch, both towards R. Current in R = I1 + I2 (first law). [1] Loop with 9.0 V branch: 9.0 = 3.0 I1 + 3.0(I1 + I2), so 9.0 = 6.0 I1 + 3.0 I2. [1] Loop with 8.0 V branch: 8.0 = 2.0 I2 + 3.0(I1 + I2), so 8.0 = 3.0 I1 + 5.0 I2. [1] Double the second: 16 = 6.0 I1 + 10 I2. Subtract the first: 7.0 = 7.0 I2, so I2 = 1.0 A. [1] Then 9.0 = 6.0 I1 + 3.0, so I1 = 1.0 A. [1] Current in R = 2.0 A, so p.d. across R = 2.0 × 3.0 = 6.0 V. [1] Examiner insight: Define your currents and loop direction on the diagram before writing equations; a check that 9.0 − 3.0 − 6.0 = 0 on the first loop shows the working is consistent.

5. (a) Lost volts = E − V = 1.6 − 1.35 = 0.25 V [1] (b) r = 0.25/0.50 = 0.50 Ω [1] (c) R = 1.35/0.50 = 2.7 Ω [1] (d) P = I²r = 0.50² × 0.50 = 0.125, so 0.13 W [1] Examiner insight: Dividing lost volts by current, not terminal p.d. by current, is what gives r; keep each p.d. labelled.

6. (a) The starter motor draws a large current [1]. Lost volts Ir across the battery’s internal resistance increase [1]. Terminal p.d. V = E − Ir therefore falls [1] (b) I = E/r = 1.5/0.30 = 5.0 A [1] (with R = 0, E = I(R + r)) [1] Examiner insight: “The battery runs down” does not answer an explain question; link the current, the internal resistance and the equation.

7. (a) Put an ammeter and a rheostat in series with the supply, and wire a voltmeter straight to the supply’s two terminals [1]. Change the resistance in steps and record I and V each time [1]. Plot V against I (or use the table directly) [1] (b) V = E − Ir, so the vertical intercept is E and the gradient is −r [1]. Gradient from the first and last points = (3.60 − 5.52)/(1.00 − 0.20) = −2.4 V A⁻¹ [1]. So r = 2.4 Ω [1]. At I = 0, V = 5.52 + 0.20 × 2.4 = 6.0 V, so E = 6.0 V [1] (c) Current through the supply causes heating (and drains it), which could change r and E [1] Examiner insight: A gradient taken from widely spaced points, with the minus sign dealt with, is more convincing than one found from neighbouring points.

8. (a) In series: E = 3.0 V, r = 0.80 Ω [1]. Total resistance = 2.0 + 0.80 = 2.8 Ω [1]. I = 3.0/2.8 = 1.1 A [1] (b) In parallel: E = 1.5 V, r = 0.40/2 = 0.20 Ω [1]. Total resistance = 2.2 Ω [1]. I = 1.5/2.2 = 0.68 A [1] Examiner insight: State the combined e.m.f. and combined internal resistance first, so the method is clear to a reader.

9. (a) At 10 °C: Vout = 3.3/(3.3 + 10) × 5.0 = 1.2 V [1]. At 50 °C: Vout = 3.3/(3.3 + 1.0) × 5.0 = 3.8 V [1]. Resistances can stay in kΩ because the units cancel [1] (b) The thermistor’s resistance falls as temperature rises [1]. The thermistor then takes a smaller share of the 5.0 V, so the fixed resistor takes a larger share [1] (c) Equal shares of 2.5 V each need equal resistances: 3.3 kΩ [1] Examiner insight: Say which component the output is across; the same thermistor gives the opposite trend if Vout is taken across it.

10. (a) V1 = 9.0 − 2.0 = 7.0 V [1]. V1/V2 = R1/R2, so R1 = 4.0 × 7.0/2.0 [1] = 14 kΩ [1] (b) The LDR’s resistance falls, so its share falls and Vout across R2 rises [1] Examiner insight: Write the ratio form with the numbers in it before rearranging; a brief reason for the direction of change is more secure than a bare “increases”.

11. (a) Vout = (30.0/80.0) × 12 [1] = 4.5 V [1] (b) Unloaded: Vout = 6.0 V [1]. Load in parallel with the lower resistor: 10 kΩ and 10 kΩ give 5.0 kΩ [1]. Vout = 5.0/(10 + 5.0) × 12 [1] = 4.0 V [1] (c) The load is in parallel with the lower resistor, which lowers its effective resistance [1]. It then takes a smaller share of the supply p.d. [1] Examiner insight: Replace the lower resistor by the parallel combination before using the divider equation; using the unloaded resistance is the usual error.

Where marks are usually lost

• Leaving a parallel result as 1/R, or adding resistances when the resistors are in parallel.
• Adding opposing e.m.f.s instead of subtracting.
• Using E in place of the terminal p.d. when finding the load current or resistance.
• Dropping the minus sign when relating the gradient of a V-I graph to r.
• Using two neighbouring points to find a gradient, which gives a poor estimate.
• Not stating which component Vout is across before explaining a sensor circuit.
• Giving a bare “resistance changes” explanation without saying how the share of the supply p.d. changes.
• Forgetting that a load across R2 is in parallel with R2.

Next steps

• Recap with the revision notes or the study guide.
• Course pages: OCR A Level Physics hub and the printable checklist.
• Find your gaps with the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR Advanced GCE in Physics A (H556), Version 3.0 (March 2026), first assessment 2017, published by OCR (Oxford Cambridge and RSA). Section 4.3: Electrical circuits.