MYP Physics · topic MYP-13.02
Current as rate of charge flow
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.
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.