A-Level · AS · extended response
A resistance experiment with an energy check
AS Level foundations · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 2
- Answer
- Complete
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.
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
Ready to self-mark?Reveal the detailed answer guide
(a)
Find resistance, charge transferred and electrical energy transferred in the 120 s run. Verify energy using two different expressions.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
The graph's slope is I/V = 0.050 A V⁻¹, the reciprocal of resistance, so R = 20 Ω.
- 2
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.
- 3
A second check is E = I²Rt = (0.30)²(20)(120) = 216 J. Current describes charge flow; it is not consumed by the resistor.
(b)
Describe the circuit and evaluate whether a straight line proves the resistor will remain ohmic at all voltages.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
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.
- 2
Collect several positive and negative voltages, limit heating and repeat at controlled temperature. A line through the origin supports I∝V over that range.
- 3
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.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- 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.
Private on this device