IGCSE · 4 · extended response
Reducing grid losses while respecting transformer limits
Electricity and magnetism · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
Ideal transformer voltage ratio
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Secondary voltage / V
Horizontal axis (x): Secondary/primary turns ratio
Read the exact data as a table
| Secondary/primary turns ratio | Secondary voltage / V |
|---|---|
| 1 | 250 |
| 10 | 2500 |
| 20 | 5000 |
| 40 | 10000 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. An ideal transformer raises an AC supply from 250 V to 10.0 kV. It delivers 20.0 kW to a transmission line with total loop resistance 5.0 Ω. Compare this with delivering the same 20.0 kW into the line at 250 V. Treat 20.0 kW as power at the sending end and use rms values. The graph supplies transformer voltage against turns ratio. Supplied model: For an ideal transformer use Vs/Vp = Ns/Np and VpIp = VsIs; wire heating is I²R. Supplied data (Secondary/primary turns ratio; Secondary voltage / V): (1, 250), (10, 2500), (20, 5000), (40, 10000).
- (a)
Find turns ratio, secondary and primary currents, line loss and receiving-end power for the 10 kV case.
3 marks - (b)
Evaluate the direct 250 V comparison and explain why a transformer does not work continuously with steady DC.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Find turns ratio, secondary and primary currents, line loss and receiving-end power for the 10 kV case.
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
Vs/Vp = Ns/Np = 10000/250 = 40. At the sending end Is = P/Vs = 20000/10000 = 2.0 A.
- 2
For an ideal transformer input power equals output power, so Ip = 20000/250 = 80 A.
- 3
Line heating is I²R = 2²(5) = 20 W, leaving 20000−20 = 19980 W at the receiving end; the line voltage drop is IR = 10 V.
(b)
Evaluate the direct 250 V comparison and explain why a transformer does not work continuously with steady DC.
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
Using sending current P/V = 20000/250 = 80 A would give a 400 V line drop and I²R = 32000 W, exceeding the sending power.
- 2
This contradicts a passive receiving load at the assumed operating point; the calculation demonstrates why voltage drop and the load must be solved together, and why high-voltage transmission is useful.
- 3
A steady DC current produces steady flux after the switching transient, so it induces no sustained secondary emf. Real AC transformers also have core and winding losses and insulation limits.
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.
- Vs/Vp = Ns/Np = 10000/250 = 40. At the sending end Is = P/Vs = 20000/10000 = 2.0 A.
- For an ideal transformer input power equals output power, so Ip = 20000/250 = 80 A.
- Line heating is I²R = 2²(5) = 20 W, leaving 20000−20 = 19980 W at the receiving end; the line voltage drop is IR = 10 V.
- Using sending current P/V = 20000/250 = 80 A would give a 400 V line drop and I²R = 32000 W, exceeding the sending power.
- This contradicts a passive receiving load at the assumed operating point; the calculation demonstrates why voltage drop and the load must be solved together, and why high-voltage transmission is useful.
- A steady DC current produces steady flux after the switching transient, so it induces no sustained secondary emf. Real AC transformers also have core and winding losses and insulation limits.
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