Transformer Efficiency and a Battery Supply
A student builds a transformer with 600 turns on the primary and 60 turns on the secondary. The primary is connected to a 240 V, 50 Hz supply. With the secondary open, the secondary voltage is 22.0 V. When a 12 Ω resistor is connected across the secondary, the secondary voltage falls to 20.0 V and the primary current is 0.21 A. The student calculates the secondary current as 20.0/12 = 1.67 A, the output power as 33.3 W, the input power as 240 × 0.21 = 50.4 W, and the efficiency as 66%. He expected nearly 100% from the ideal-transformer formula and wonders whether the missing 17 W is stored in the laminated core. He also plans to run the same transformer from a 12 V battery to obtain about 1.2 V across the secondary.
1. Calculate the ideal secondary voltage and judge whether the no-load reading of 22.0 V proves that the transformer is faulty.
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The ideal turns ratio gives Vs = 240 × (60/600) = 24 V. The no-load reading is 22.0 V, about 8% low. This is not automatically a fault: a real small transformer has flux leakage, finite core permeability and magnetising current, and primary winding resistance, so the secondary emf can be below the ideal value even at no load. The supply may also differ slightly from 240 V. However, an 8% discrepancy is large enough to warrant checking the actual number of turns, the quality of the core joints, and whether a shorted turn exists. The correct judgement is that the reading is plausible for a poorly designed or small transformer but should be investigated rather than accepted as ideal.
2. Critically examine the student’s efficiency calculation. Is 66% necessarily the true efficiency?
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The student has calculated the input apparent power, 240 × 0.21 = 50.4 VA, not necessarily the real input power. In an AC transformer the primary current has a magnetising component in quadrature with the supply voltage, so the real input power is Pin = VrmsIrms cos φ. The output power, 33.3 W, plus copper and core losses is the real input power. The difference from 50.4 VA is not stored in the core in steady state; some of it is reactive energy exchanged with the magnetising inductance, and the rest is dissipated as heat in the windings and core. Without a wattmeter or a measurement of power factor, 66% is only a lower bound based on apparent power. For example, if the true efficiency were 80%, the real input would be about 41.6 W and the power factor about 0.83, which is plausible.
3. What will happen when the student connects the primary to a 12 V battery instead of the 240 V AC supply?
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A battery provides direct current. After the switch is closed, the primary current rises from zero and there is a brief changing flux, so a short transient pulse may appear across the secondary. Once the current becomes steady, the flux in the core is constant and no emf is induced; the secondary voltage falls to zero. The transformer cannot step 12 V DC down to 1.2 V. It works only while the flux is changing, which is why an alternating supply is required. The student would see at most a momentary kick at make or break, not a steady 1.2 V.