CBSE · Class 12 · Physics · Alternating CurrentExplain the principle, construction, and working of a Transformer with a labeled diagrammatic representation. Also, discuss four major energy losses that occur in a real transformer along with methods used to minimize them.

Transformer: Principle, Construction, Working, and Losses

1. Principle of a Transformer\nA transformer is a static electrical device used to step-up (increase) or step-down (decrease) alternating voltage without changing the frequency of the power. It works on the principle of Mutual Induction. When an alternating current flows through the primary coil, it produces a continuously changing magnetic flux in the core, which links with the secondary coil and induces an electromotive force (emf) in it.

2. Construction\nA transformer consists of two main parts:

• Laminated Soft Iron Core: Made of thin sheets (laminations) of silicon steel insulated from each other by a layer of varnish to reduce eddy currents.
• Two Coils (Windings):
  • Primary Coil ($N_p$ turns): Connected to the input AC supply voltage ($V_p$).
  • Secondary Coil ($N_s$ turns): Connected to the output load resistance ($V_s$).

3. Working and Mathematical Relation\nLet $\Phi$ be the magnetic flux linked with each turn of the core at any instant.\nAccording to Faraday's Law of Electromagnetic Induction:

• Induced EMF in primary coil: $E_p = -N_p \frac{d\Phi}{dt}$
• Induced EMF in secondary coil: $E_s = -N_s \frac{d\Phi}{dt}$ \nDividing the secondary EMF equation by the primary EMF equation: $$\frac{E_s}{E_p} = \frac{N_s}{N_p} = K$$\nWhere $K$ is called the transformation ratio.\nFor an ideal transformer (no power loss, Power Input = Power Output): $$V_p I_p = V_s I_s \implies \frac{V_s}{V_p} = \frac{I_p}{I_s} = \frac{N_s}{N_p}$$
• Step-up Transformer: $N_s > N_p \implies V_s > V_p$ and $I_s < I_p$.
• Step-down Transformer: $N_s < N_p \implies V_s < V_p$ and $I_s > I_p$.

4. Major Energy Losses in a Transformer and Remedies\nEven though transformers are highly efficient devices ($90%-98%$), energy losses occur in various forms:

1. Copper Loss (Joule Heating Loss):

   • Cause: Heat produced ($I^2 R$) in the copper windings due to electrical resistance.
   • Remedy: Use thick copper wires with low resistance for high-current windings.

2. Eddy Current Loss (Iron Loss):

   • Cause: Induced circulating currents (eddy currents) in the iron core produce unwanted heat.
   • Remedy: Use a laminated soft iron core coated with insulating varnish.

3. Hysteresis Loss:

   • Cause: Continuous magnetization and demagnetization of the magnetic core during each AC cycle causes energy loss as heat.
   • Remedy: Use a core made of high-permeability, low-hysteresis loop materials like Silicon Steel or soft iron.

4. Flux Leakage Loss:

   • Cause: Not all magnetic flux produced by the primary coil links with the secondary coil due to poor magnetic coupling.
   • Remedy: Wind primary and secondary coils over each other on the same soft iron core limb.