Cambridge IGCSE Physics · topic 4.2
Electrical quantities
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
An electrostatic sprayer and an unintended spark
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A conducting sprayer nozzle behaves as a 100 pF capacitor relative to its surroundings. The supplied charge–potential data describe it. It operates at +20 kV and transfers 2.0 nC to each paint droplet. Assume identical droplets. A grounding connection can discharge the nozzle; no calculation here establishes a safe ignition threshold. Supplied model: Use Q = CV and stored electrostatic energy U = ½QV. Supplied data (Potential / kV; Charge / μC): (0, 0), (10, 1), (20, 2). (a) Find charge stored, stored energy and the number of charged droplets corresponding to that stored charge. [3 marks] (b) Explain why charged paint is attracted to a grounded conductor and how grounding differs from insulation as risk control. [3 marks]
Nozzle charge versus potential
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Charge / μC
Horizontal axis (x): Potential / kV
Read the exact data as a table
| Potential / kV | Charge / μC |
|---|---|
| 0 | 0 |
| 10 | 1 |
| 20 | 2 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find charge stored, stored energy and the number of charged droplets corresponding to that stored charge.
Reveal the worked solution
- The graph has C = Q/V = (1.0×10⁻⁶)/(1.0×10⁴) = 100 pF; at 20 kV charge is 2.0×10⁻⁶ C.
- Stored energy is ½QV = ½(2.0×10⁻⁶)(20000) = 0.020 J. The factor ½ accounts for the potential rising during charging.
- Q/qdrop = (2.0×10⁻⁶)/(2.0×10⁻⁹) = 1000 droplets if that charge were transferred with no replenishing supply.
- A positive droplet induces negative surface charge on the nearby conducting workpiece. Grounding lets electrons flow from Earth, maintaining a suitable potential and sustained attraction.
- Conductors allow mobile charge to redistribute; an insulator can retain local deposited charge. Static charge is not necessarily a continuous electric current.
- Bonding and grounding reduce charge accumulation and potential differences. Insulation can prevent contact current but may retain static charge, so its role must match the actual discharge or ignition risk.
Answer (a) Q = 2.0 μC; stored energy = 0.020 J; equivalent droplet count = 1000. (b) Charge redistribution produces attraction; grounding provides a discharge path while insulation restricts charge flow.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A resistance experiment with an energy check
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A fixed resistor is connected to a variable low-voltage supply. The supplied voltage–current data are collected with an ammeter in series and a voltmeter directly across the resistor. The resistor is then held at 6.0 V for 120 s. Treat meter loading as negligible unless considering limitations. Supplied data (Voltage / V; Current / A): (0, 0), (2, 0.1), (4, 0.2), (6, 0.3). (a) Find resistance, charge transferred and electrical energy transferred in the 120 s run. Verify energy using two different expressions. [3 marks] (b) Describe the circuit and evaluate whether a straight line proves the resistor will remain ohmic at all voltages. [3 marks]
Resistor current–voltage data
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Current / A
Horizontal axis (x): Voltage / V
Read the exact data as a table
| Voltage / V | Current / A |
|---|---|
| 0 | 0 |
| 2 | 0.1 |
| 4 | 0.2 |
| 6 | 0.3 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Find resistance, charge transferred and electrical energy transferred in the 120 s run. Verify energy using two different expressions.
Reveal the worked solution
- The graph's slope is I/V = 0.050 A V⁻¹, the reciprocal of resistance, so R = 20 Ω.
- At 6.0 V, current is 0.30 A; Q = It = 0.30(120) = 36 C. Voltage is energy transferred per unit charge, giving E = VQ = 216 J.
- A second check is E = I²Rt = (0.30)²(20)(120) = 216 J. Current describes charge flow; it is not consumed by the resistor.
- The ammeter must be in the resistor's current path; the voltmeter connects across its two terminals. Placing an ammeter across the supply can create a low-resistance fault.
- Collect several positive and negative voltages, limit heating and repeat at controlled temperature. A line through the origin supports I∝V over that range.
- At higher power the resistor may warm and change resistance or fail. Real voltmeters draw current and ammeters have resistance; these effects require a loading assessment when comparable to the component values.
Answer (a) R = 20 Ω; Q = 36 C; E = 216 J. (b) The evidence supports constant resistance over the measured range only, under the measurement conditions.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A battery whose terminal voltage depends on the load
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A battery supplies an external load. Its measured terminal-voltage versus total-current data are supplied. A 6.0 Ω resistor and a 3.0 Ω resistor are connected in parallel across its terminals. Model the battery as a fixed emf in series with internal resistance. Supplied data (Current / A; Terminal voltage / V): (0, 12), (1, 11), (2, 10), (3, 9). (a) Infer emf and internal resistance, then find total current, terminal voltage and both branch currents. [3 marks] (b) Calculate useful load power and battery internal heating, then predict the load voltage if the 3 Ω branch is disconnected. [3 marks]
Battery load test
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Terminal voltage / V
Horizontal axis (x): Current / A
Read the exact data as a table
| Current / A | Terminal voltage / V |
|---|---|
| 0 | 12 |
| 1 | 11 |
| 2 | 10 |
| 3 | 9 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Infer emf and internal resistance, then find total current, terminal voltage and both branch currents.
Reveal the worked solution
- V = ε−Ir has intercept ε = 12 V and gradient −r = −1.0 V A⁻¹, so internal resistance is 1.0 Ω.
- The parallel load is (1/6+1/3)⁻¹ = 2.0 Ω. Total current is ε/(Rload+r) = 12/3 = 4.0 A, giving terminal voltage 12−4 = 8.0 V.
- Both branches have 8.0 V across them, so I6 = 8/6 = 4/3 A and I3 = 8/3 A. Their sum checks the junction current rule.
- Power delivered externally is VI = 8(4) = 32 W. Internal heating is I²r = 16 W, and εI = 48 W checks energy conservation.
- With only the 6 Ω branch, total current is 12/(6+1) = 12/7 A.
- Terminal voltage is I(6) = 72/7 V, greater than before because the internal voltage drop is smaller. The emf is unchanged in this model.
Answer (a) emf = 12 V; r = 1.0 Ω; total current = 4.0 A; terminal voltage = 8.0 V; branch currents = 4/3 A and 8/3 A. (b) Load power = 32 W; internal heating = 16 W; disconnected-case terminal voltage = 72/7 ≈10.3 V.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.