The problems are on the transformers worksheet. Try them on your own first, then compare with the answers below.
Quick answer: An alternating current reverses direction regularly, and its rms value, V₀/√2 for a sine wave, gives the equivalent steady heating effect. These worked answers correct the 20 problems of the Physics 11-12 worksheet on transformers, from peak from rms to spot the errors: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the result with its unit. When a problem gives an rms value, convert carefully to peak only if the question needs it. Small rounding differences in the last digit are acceptable.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.7.5
Problem 2 Rms current
Irms = I₀ / √2 = 4.2 / 1.414 = 2.97 A (about 3.0 A).Problem 3 Period and frequency
T = 1/f. UK: 1/50 = 0.020 s = 20 ms. Texas: 1/60 = 0.01667 s = 16.7 ms.Problem 4 Stove ring current
1. I = P/V = 2300 / 230 = 10 A.2. R = V/I = 230 / 10 = 23 Ω.
Problem 5 Secondary voltage
Vs = Vp Ns/Np = 230 × 100 / 500 = 46 V.Problem 6 Meaning of rms
Over a full cycle the positive and negative halves cancel, so the mean is zero and says nothing about the supply. The rms value is the steady dc value that gives the same heating effect in a resistor.Problem 7 High-voltage line data
1. 11 kV = 11 000 V.2. P = VI = 11 000 × 120 = 1320000 W = 1.32 MW.
Problem 8 Currents in a transformer
1. Is = P/Vs = 36 / 12 = 3.0 A.2. For an ideal transformer Vp Ip = 36 W, so Ip = 36 / 230 = 0.157 A (0.16 A).
Problem 9 Efficiency
1. Efficiency = output / input = 72 / 80 = 90 %.2. Power wasted = 8 W, so E = Pt = 8 × 3600 = 28800 J (28.8 kJ).
Problem 10 Loss in a village cable
1. I = P/V = 100 000 / 2000 = 50 A; loss = I²R = 50² × 5.0 = 12500 W (12.5 kW).2. I = 100 000 / 20 000 = 5 A; loss = 5² × 5.0 = 125 W.
3. At 2.0 kV: 12.5 %; at 20 kV: 0.125 %.
Problem 11 Oscilloscope trace
1. Peak = 3.2 divisions × 5.0 V/div = 16 V.2. One cycle spans 4.0 divisions, so T = 4.0 × 5.0 ms = 20 ms and f = 1/0.020 = 50 Hz.
3. Vrms = 16 / √2 = 11.3 V.
Problem 12 Power in a resistor
1. Vrms = 30 / √2 = 21.2 V.2. P = Vrms² / R = 450 / 40 = 11.25 W.
3. Peak power = V₀²/R = 900 / 40 = 22.5 W. The power varies between zero and this value during each cycle, and its mean is exactly half the peak.
Problem 13 Why high voltage?
1. For the same power, a higher V means a smaller I. The cable loss is I²R, so it falls with the square of the current, and very little energy is wasted as heat.2. Generators and homes need modest, safe voltages, so transformers step the voltage up before transmission and down before use. They only work with ac, which is one reason mains is alternating.
Problem 14 Laminated core
1. A solid block allows large eddy currents to circulate, heating the core. Thin insulated sheets break up the eddy-current paths and so reduce the heating.2. Soft iron is easily magnetised and demagnetised, so the energy lost in each cycle (hysteresis) is small.
Problem 15 Ac and dc heaters
1. The same: 1.0 kW, since rms is defined by equal heating.2. V₀ = √2 × 230 = 325 V.
3. The voltage peaks at 325 V but is lower for most of the cycle, so the mean power equals that of a steady 230 V.
Problem 16 Village supply
1. I = P/V = 440 000 / 22 000 = 20 A.2. Loss = I²R = 20² × 8.0 = 3200 W (3.2 kW).
3. ΔV = IR = 20 × 8.0 = 160 V.
4. Power delivered = 440 000 − 3200 = 436 800 W, which is 99.3 % of the power sent.
Problem 17 Real transformer
1. Is = P/Vs = 60 / 24 = 2.5 A.2. Input power = output / efficiency = 60 / 0.92 = 65.2 W.
3. Ip = P/Vp = 65.2 / 230 = 0.28 A.
Problem 18 Instantaneous mains voltage
1. V₀ = 325 V; ω = 100π rad/s, so f = ω/2π = 50 Hz.2. V = 325 sin(100π × 0.0040) = 325 sin(1.257 rad) = 325 × 0.951 = 309 V.
3. sin(100πt) = 230/325 = 0.708, so 100πt = 0.786 rad and t = 2.50 ms.
Problem 19 National grid design
1. I = P/V = 200 × 10⁶ / 400 × 10³ = 500 A; loss = I²R = 500² × 12 = 3.0 MW.2. At 25 kV, I = 8000 A and the loss would be 8000² × 12 = 768 MW, which is more than three times the power station’s output. This is impossible, so high-voltage transmission is essential.
3. 3.0 / 200 = 1.5 %.
Problem 20 Spot the errors
1. 100/230 = 0.43 A is the rms current. The peak current is √2 × 0.435 = 0.61 A.2. The ratio is upside down. Ns = Np Vs/Vp = 920 × 12 / 230 = 48 turns.
3. A step-down transformer must have fewer secondary turns than primary turns, so 17 600 turns is impossible.
