The Transformer and the 'Free Energy' Claim
A student builds a demonstration with an ideal transformer: 240 V AC input to a primary of 1200 turns, and a secondary of 60 turns connected to a 12 Ω resistor. The student measures the secondary current as 1.0 A and the primary current as 0.05 A. They calculate input power as 240 × 0.05 = 12 W and output power as 12 × 1.0 = 12 W, and conclude the transformer is 100% efficient. A classmate objects: 'The secondary voltage should be 12 V and the current 1.0 A, so the resistor dissipates 12 W-but the primary current you measured is too small to supply that, so your ammeter must be wrong.' The teacher asks the class to adjudicate.
1. Use the turns ratio to find the secondary voltage and current, and judge whether the measurements are self-consistent for an ideal transformer.
Read the worked answer
Turns ratio: Ns/Np = 60/1200 = 1/20. For an ideal transformer, Vs/Vp = Ns/Np, so Vs = 240/20 = 12 V. The resistor current is I = V/R = 12/12 = 1.0 A, matching the measured secondary current.
For an ideal transformer, power in equals power out: Vp Ip = Vs Is, so Ip = (12 × 1.0)/240 = 0.05 A. This matches the measured primary current.
The measurements are therefore fully self-consistent with an ideal transformer. The classmate's objection is incorrect: the primary current of 0.05 A is exactly what is required, because the primary voltage is 20 times the secondary voltage, so the primary current is 1/20 of the secondary current. The classmate has confused 'small current' with 'insufficient power'-power depends on the product VI, and 240 V × 0.05 A = 12 W.
2. The student now replaces the 12 Ω resistor with a 3 Ω resistor and finds the secondary current becomes 4.0 A. Predict the new primary current and explain what would happen to the primary current if the secondary were left open-circuit.
Read the worked answer
With a 3 Ω resistor, Vs is still 12 V (set by the turns ratio and the 240 V supply), so Is = 12/3 = 4.0 A. For an ideal transformer, Ip = (Vs Is)/Vp = (12 × 4.0)/240 = 0.20 A. The primary current has increased by a factor of 4, tracking the secondary current, because the transformer transfers power and the secondary load determines the current drawn.
If the secondary is open-circuited, no current flows in the secondary (Is = 0), so no power is transferred and the ideal primary current is zero. In a real transformer, a small magnetising current would still flow in the primary due to the finite inductance of the coil and core losses, but it would be much smaller than the loaded current. The key judgement is that the primary current is not fixed by the supply voltage alone-it is determined by the load on the secondary, which is why the transformer is described as transferring power rather than generating it.