Transformer Coil Design: Turns Ratio Versus Core Saturation
A student winds a transformer for a school project. The primary has 1200 turns on a laminated iron core of cross-sectional area 4.0 cm², designed for 240 V, 50 Hz. The secondary is intended to supply 12 V to a 24 W lamp. With the lamp connected, the primary current is measured as 0.12 A, while the secondary current is 2.0 A at 12 V. The core becomes noticeably warm. The teacher remarks that the core may be operating near saturation because the turns were chosen from the turns-ratio formula alone. The student also claims that the transformer steps up current and therefore delivers more power at the secondary than the primary draws. Assume sinusoidal flux and neglect winding resistance unless needed.
1. Calculate the number of secondary turns, the ideal primary current and the efficiency. Explain the student's power claim.
Read the worked answer
Turns ratio: Ns/Np = Vs/Vp, so Ns = 1200 × 12/240 = 60 turns. The lamp draws Pout = VsIs = 12 × 2.0 = 24 W. In an ideal transformer the primary current would be Ipin = Pout/Vp = 24/240 = 0.10 A. The measured input power is Pin = VpIp = 240 × 0.12 = 28.8 W, so efficiency η = 24/28.8 = 0.83, about 83%. The secondary current is larger only because the secondary voltage is lower. Power is not amplified: the secondary power is always less than the primary power once losses are included. The student's claim confuses a larger current with a larger power.
2. Estimate the peak flux density in the core and judge whether saturation is a credible explanation for the heating. What design changes would reduce the risk?
Read the worked answer
For a sinusoidal supply, Vrms = 4.44 f N Φmax. Thus Φmax = 240/(4.44 × 50 × 1200) = 9.0 × 10⁻⁴ Wb. The peak flux density is Bmax = Φmax/A, with A = 4.0 cm² = 4.0 × 10⁻⁴ m²: Bmax = 9.0 × 10⁻⁴ / 4.0 × 10⁻⁴ = 2.25 T. Typical laminated transformer iron saturates around 1.5–1.8 T, so 2.25 T is at or beyond saturation. Once saturated, the magnetizing current rises sharply and core losses increase, which fits the warming and the 83% efficiency. Remedies include increasing the primary turns (and secondary turns in the same ratio), using a core of larger cross-sectional area, or operating at a lower primary voltage. Simply adding secondary turns without changing the primary would alter the output voltage.
3. Another student connects the 12 V secondary to a 12 V DC supply and finds that the lamp lights. Connecting the primary to 12 V DC produces no secondary voltage and the primary gets hot. Explain both observations.
Read the worked answer
The secondary is just a coil. Connected directly to 12 V DC, it behaves as a conductor and the lamp receives a steady current, so it lights. No transformer action is involved. Transformer action requires a changing magnetic flux linking the coils. A steady primary current produces a constant flux, so dΦ/dt = 0 and the induced secondary emf is zero. With no back-emf, the primary current is limited only by the small resistance of the winding, so it becomes large and the coil overheats. This is why a transformer must be operated on alternating current, where the flux changes continuously.