Physics · Alternating Current · NEET
Because energy cannot be created. An ideal transformer only transfers power, it does not add any. Power = Voltage × Current. If power in and power out are equal (ideal case) and the voltage on the secondary is larger, then the current on the secondary must be smaller to keep the product the same. So a step-up transformer gives high voltage but low current, and a step-down transformer gives low voltage but high current.
Voltage is DIRECTLY proportional to turns: Vs/Vp = Ns/Np. Current is INVERSELY proportional to turns: Is/Ip = Np/Ns. Many students wrongly write Is/Ip = Ns/Np. Remember: the coil with more turns has more voltage but less current. Write all three ratios together so you never flip current: Vs/Vp = Ns/Np = Ip/Is.
Use power conservation directly, you do not even need the turns. For an ideal transformer, primary power = secondary power = the device's power rating. So Ip = P / Vp. Example: a 44 W lamp on 220 V mains gives Ip = 44/220 = 0.2 A. The secondary voltage (11 V) is a distractor; ignore it for the primary current.
Ideal means 100% efficient with three assumptions: (i) the primary winding has zero resistance, (ii) all the magnetic flux made by the primary links the secondary (no leakage), and (iii) the core has no eddy-current or hysteresis loss. Under these, power input = power output exactly. Real transformers reach about 95% efficiency; the small gap is covered in the next concept, energy losses in a transformer.
No. A transformer works by mutual induction driven by the changing flux of the same AC source. The secondary voltage oscillates at the SAME frequency as the primary (50 Hz in India). A transformer changes voltage and current magnitudes only, never the frequency. This is a common NEET trap.
A step down transformer connected to an ac mains supply of 220 V is made to operate at 11 V, 44 W lamp. Ignoring power losses in the transformer, what is the current in the primary circuit?
A 12 V, 60 W lamp is connected to secondary of step down transformer, whose primary is connected to ac mains of 220 V. Assuming the transformer to be ideal, what is the current in the primary winding?
In an ideal transformer, the turns ratio is N_p/N_s = 1/2. The ratio V_s : V_p is equal to (the symbols carry their usual meaning):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Vs/Vp = Ns/Np = Ip/Is, together with Vp × Ip = Vs × Is (power in = power out). This single chain lets you find any voltage, current, or turns value if the others are known.
An ideal transformer has no losses (no winding resistance, no flux leakage, no eddy or hysteresis loss), so energy is conserved. All the power delivered to the primary appears at the secondary: Pp = Ps.
Voltage increases (Vs > Vp because Ns > Np) and current decreases (Is < Ip) by the same factor. This is why high-voltage, low-current transmission reduces I²R power loss in cables.
Ip = P / Vp, where P is the device's power rating and Vp is the mains (primary) voltage. This works because ideal power input equals the device power. The secondary voltage is not needed.
No. A transformer needs a changing magnetic flux, which only steady-changing AC provides. Steady DC gives constant flux, so no emf is induced in the secondary. Transformers work only with AC.
