Specification: AQA A-level Physics (7408) 3.7.5.6 | Paper: 2 | Data and formulae booklet: given in the exam
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Transformer: an alternating current in the primary coil produces a changing magnetic flux in the iron core. This flux links the secondary coil and induces an alternating emf in it. A step-up transformer has Ns > Np; a step-down transformer has Ns < Np. It does not work on steady dc.
Efficiency: efficiency = output power Ć· input power = IsVs Ć· IpVp. An ideal (100% efficient) transformer has IsVs = IpVp.
Causes of energy loss: eddy currents induced in the core heat it (reduced by a laminated core of thin insulated sheets); the resistance of the coils heats them (reduced by thick, low-resistance copper wire); flux leakage, where not all the primary flux links the secondary (reduced by winding the coils on top of each other); and energy lost in repeatedly magnetising the core (reduced by a soft iron core).
The equation
turns ratio = voltage ratio Ns Ć· Np = Vs Ć· Vp and efficiency = IsVs Ć· IpVp
| Quantity | Symbol | Unit |
|---|---|---|
| number of turns on the secondary coil | Ns | no unit |
| number of turns on the primary coil | Np | no unit |
| secondary (output) voltage | Vs | volts (V) |
| primary (input) voltage | Vp | volts (V) |
| secondary current | Is | amperes (A) |
| primary current | Ip | amperes (A) |
| efficiency | ā | no unit (a ratio or a percentage) |
Rearranged: Vs = Vp Ć Ns Ć· Np | Ns = Np Ć Vs Ć· Vp | ideal transformer: Ip = IsVs Ć· Vp | power lost in a cable = IĀ²R | Convert kV to V and kW or MW to W before you substitute. Stepping the voltage up steps the current down by the same factor (for an ideal transformer).
Use FIFA for every answer
- F ā Formula: write the equation as it appears on the sheet.
- I ā Insert: put in the numbers (convert to standard units first).
- F ā Fix: rearrange to make the unknown the subject.
- A ā Answer: calculate and always give the unit.
Core: Q1ā4. Exam standard: Q5ā8. Stretch: Q9ā10.
Questions
Q1 (Core) A transformer has 2000 turns on its primary coil and 100 turns on its secondary coil. The primary is connected to the 230 V rms mains. Calculate the secondary voltage and state whether this is a step-up or step-down transformer.
Show solution
F: Ns Ć· Np = Vs Ć· Vp
I: 100 Ć· 2000 = Vs Ć· 230
F: Vs = 230 Ć 100 Ć· 2000
A: 11.5 V rms, a step-down transformer (fewer turns on the secondary)
Q2 (Core) A transformer steps 230 V down to 12 V. Its primary coil has 4600 turns. Calculate the number of turns on the secondary coil.
Show solution
F: Ns Ć· Np = Vs Ć· Vp
I: Ns Ć· 4600 = 12 Ć· 230
F: Ns = 4600 Ć 12 Ć· 230
A: 240 turns
Q3 (Core) An ideal transformer steps 230 V down to 12 V. The secondary current is 2.5 A. Calculate the primary current in mA, to 2 significant figures.
Show solution
F: IsVs = IpVp (100% efficient)
I: 2.5 Ć 12 = Ip Ć 230, so 30 = Ip Ć 230
F: Ip = 30 Ć· 230 = 0.1304ā¦ A
A: 130.4ā¦ mA = 130 mA (2 s.f.)
Watch out: a step-down transformer steps the current up. The primary current is smaller than the secondary current, not larger.
Q4 (Core) A transformer connected to the 230 V mains draws a primary current of 0.25 A. Its secondary delivers 4.5 A at 12 V. Calculate the efficiency of the transformer, to 2 significant figures, and the power wasted.
Show solution
Step 1 ā powers
input power IpVp = 0.25 Ć 230 = 57.5 W; output power IsVs = 4.5 Ć 12 = 54 W
Step 2 ā efficiency
F: efficiency = IsVs Ć· IpVp
I: efficiency = 54 Ć· 57.5
F: efficiency is already the subject
A: 0.939ā¦ = 94% (2 s.f.); power wasted = 57.5 ā 54 = 3.5 W
Q5 (Exam) A laptop charger contains a transformer that is 92% efficient. It delivers 3.4 A at 19 V to the laptop from the 230 V mains. (a) Calculate the primary current, to 2 significant figures. (b) Calculate the energy wasted in 2.0 hours of use, to 2 significant figures. (c) State where most of this energy goes.
Show solution
Convert first: 92% = 0.92; 2.0 hours = 7200 s
Step 1 ā (a) primary current
output power = 3.4 Ć 19 = 64.6 W; input power = 64.6 Ć· 0.92 = 70.21ā¦ W
Ip = 70.21ā¦ Ć· 230 = 0.3052ā¦ A
A: 0.31 A (2 s.f.)
Step 2 ā (b) energy wasted
power wasted = 70.21ā¦ ā 64.6 = 5.617ā¦ W; energy = 5.617ā¦ Ć 7200 = 40 445ā¦ J
A: 4.0 Ć 104 J (2 s.f.)
(c) it is dissipated as heat in the coils (IĀ²R heating in their resistance) and in the core (eddy currents), which is why the charger feels warm
Watch out: input power = output power Ć· efficiency. Multiplying by 0.92 instead gives an input power smaller than the output, which is impossible.
Q6 (Exam) A power station generator produces 2.0 kA at 25 kV. An ideal step-up transformer with 1500 turns on its primary coil raises this to 400 kV for the National Grid. Calculate (a) the number of turns on the secondary coil and (b) the current in the transmission cables.
Show solution
Convert first: 2.0 kA = 2000 A; 25 kV = 25 000 V; 400 kV = 400 000 V
Step 1 ā (a) turns
F: Ns Ć· Np = Vs Ć· Vp
I: Ns Ć· 1500 = 400 000 Ć· 25 000 = 16
F: Ns = 16 Ć 1500
A: 24 000 turns
Step 2 ā (b) current
IsVs = IpVp, so Is = 2000 Ć 25 000 Ć· 400 000
A: 125 A (the voltage rises 16 times, so the current falls 16 times)
Q7 (Exam) A step-down transformer has 800 turns on its primary coil and 40 turns on its secondary coil. The primary is connected to the 230 V rms mains and the secondary to a heater of resistance 5.0 Ī©. The transformer is 92% efficient. Calculate (a) the secondary voltage and current, (b) the output power, to 3 significant figures, and (c) the primary current, to 3 significant figures. (d) Explain why the core is laminated.
Show solution
Step 1 ā (a) secondary voltage and current
Vs = 230 Ć 40 Ć· 800 = 11.5 V; Is = Vs Ć· R = 11.5 Ć· 5.0 = 2.3 A
Step 2 ā (b) output power
IsVs = 2.3 Ć 11.5 = 26.45 W = 26.5 W (3 s.f.)
Step 3 ā (c) primary current
F: efficiency = IsVs Ć· IpVp
I: 0.92 = 26.45 Ć· (Ip Ć 230)
F: Ip = 26.45 Ć· (0.92 Ć 230) = 26.45 Ć· 211.6
A: 0.125 A (3 s.f.)
(d) the changing flux induces eddy currents in the core, which heat it. Thin sheets separated by insulation increase the resistance of the eddy current paths, so the eddy currents and the heating are much smaller.
Q8 (Exam) A small wind farm sends 1.0 MW of power along cables of total resistance 4.0 Ī©. The power is transmitted at 25 kV. Calculate (a) the current in the cables, (b) the power lost in the cables and (c) the percentage of the power lost, to 2 significant figures.
Show solution
Convert first: 1.0 MW = 1.0 Ć 106 W; 25 kV = 25 000 V
Step 1 ā (a) current
F: P = IV
I: 1.0 Ć 106 = I Ć 25 000
F: I = 1.0 Ć 106 Ć· 25 000
A: 40 A
Step 2 ā (b) power lost
Plost = IĀ²R = 40Ā² Ć 4.0 = 6400 W
Step 3 ā (c) percentage
6400 Ć· 1.0 Ć 106 Ć 100 = 0.64% (2 s.f.)
Watch out: do not use VĀ²/R with the 25 kV transmission voltage. That voltage is across the load and the cables together, not across the cables alone.
Q9 (Stretch) A power station delivers 20 MW along transmission cables with a total resistance of 8.0 Ī©. Compare the power lost in the cables when the power is transmitted at 25 kV and at 400 kV, giving each loss as a percentage of the power transmitted, and explain the result.
Show solution
Convert first: 20 MW = 2.0 Ć 107 W; 25 kV = 25 000 V; 400 kV = 400 000 V
Step 1 ā at 25 kV
I = P Ć· V = 2.0 Ć 107 Ć· 25 000 = 800 A
Plost = IĀ²R = 800Ā² Ć 8.0 = 5.12 Ć 106 W, which is 26% of the power (2 s.f.)
Step 2 ā at 400 kV
I = 2.0 Ć 107 Ć· 400 000 = 50 A
Plost = 50Ā² Ć 8.0 = 20 000 W, which is 0.10% of the power
Step 3 ā compare
5.12 Ć 106 Ć· 20 000 = 256, which equals (400 Ć· 25)Ā² = 16Ā²
A: raising the voltage 16 times cuts the current 16 times, and because Plost = IĀ²R, the power lost falls by a factor of 256
Watch out: the losses depend on the current squared. Raising the voltage by a factor of 16 reduces the losses by 16Ā², not by 16.
Q10 (Stretch) A generator produces 120 kW at 690 V. It is connected to a village by cables of total resistance 0.50 Ī©. (a) Calculate the power lost if the generator is connected directly to the cables, to 2 significant figures, and give it as a percentage, to 2 significant figures. (b) A step-up transformer with a turns ratio Ns Ć· Np = 20 and an efficiency of 98% is now placed between the generator and the cables. Calculate the transmission voltage, the current in the cables, to 3 significant figures, and the total power lost (in the transformer and the cables), to 2 significant figures.
Show solution
Convert first: 120 kW = 120 000 W
Step 1 ā (a) direct connection
I = 120 000 Ć· 690 = 173.9ā¦ A; Plost = IĀ²R = (173.9ā¦)Ā² Ć 0.50 = 15 122ā¦ W
A: 15 kW (2 s.f.), which is 15 122ā¦ Ć· 120 000 = 13% (2 s.f.)
Step 2 ā (b) transmission voltage
Vs = 20 Ć 690 = 13 800 V (13.8 kV)
Step 3 ā current in the cables
output power = 0.98 Ć 120 000 = 117 600 W; Is = 117 600 Ć· 13 800 = 8.521ā¦ A = 8.52 A (3 s.f.)
Step 4 ā total power lost
in the transformer: 120 000 ā 117 600 = 2400 W; in the cables: (8.521ā¦)Ā² Ć 0.50 = 36.31ā¦ W
total = 2400 + 36.31ā¦ = 2436.3ā¦ W
A: 2.4 kW (2 s.f.), about a sixth of the loss without the transformer
Watch out: remember the transformer’s own 2% loss. Even so, it is far smaller than the IĀ²R loss it saves in the cables.
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