CBSE · Class 12 · Physics · Alternating CurrentExplain the principle, construction, and working of a transformer. Differentiate between a step-up and a step-down transformer. Numerical Problem: A step-down transformer converts $2200\text{ V}$ to $220\text{ V}$. The primary coil has $5000$ turns. Calculate: The number of turns in the secondary coil. The output power and secondary current if the efficiency of the transformer is $90\%$ and the primary current is $2\text{ A}$.

Transformer: Principle, Construction, and Working

1. Principle:\nA transformer is a static electrical device that transfers electrical energy from one AC circuit to another at the same frequency through the principle of Mutual Induction.

2. Construction:

• Soft Iron Core: Made of thin, insulated soft-iron laminated sheets to reduce eddy current losses.
• Coils: Two insulated copper coils wrapped on the same core:
  • Primary Coil ($N_p$ turns): Connected to the input AC voltage source.
  • Secondary Coil ($N_s$ turns): Connected to the output load circuit.

3. Working:\nWhen an alternating voltage is applied across the primary coil, an alternating current flows, creating a continuously changing magnetic flux in the core. This changing flux links with the secondary coil, inducing an alternating electromotive force (emf) in it according to Faraday's Law of Electromagnetic Induction:

$$\frac{e_s}{e_p} = \frac{N_s}{N_p} = K \text{ (Transformation Ratio)}$$

Difference Between Step-Up and Step-Down Transformer

Numerical Solution

Given Data:

• Primary Voltage ($V_p$) = $2200\text{ V}$
• Secondary Voltage ($V_s$) = $220\text{ V}$
• Primary Turns ($N_p$) = $5000$
• Primary Current ($I_p$) = $2\text{ A}$
• Efficiency ($\eta$) = $90% = 0.90$

Step 1: Number of turns in secondary coil ($N_s$)\nFormula:

$$\frac{V_s}{V_p} = \frac{N_s}{N_p}$$

$$N_s = N_p \times \left(\frac{V_s}{V_p}\right) = 5000 \times \left(\frac{220}{2200}\right)$$ $$N_s = 5000 \times 0.1 = 500\text{ turns}$$

Step 2: Output Power ($P_{out}$)\nFirst, calculate Input Power ($P_{in}$):

$$P_{in} = V_p \times I_p = 2200\text{ V} \times 2\text{ A} = 4400\text{ W}$$ \nUsing efficiency formula $\eta = \frac{P_{out}}{P_{in}}$: $$P_{out} = \eta \times P_{in} = 0.90 \times 4400\text{ W} = 3960\text{ W}$$

Step 3: Secondary Current ($I_s$)\nSince $P_{out} = V_s \times I_s$:

$$I_s = \frac{P_{out}}{V_s} = \frac{3960\text{ W}}{220\text{ V}} = 18\text{ A}$$

Final Answers:

1. Turns in secondary coil ($N_s$) = $500$ turns
2. Output Power ($P_{out}$) = $3960\text{ W}$
3. Secondary Current ($I_s$) = $18\text{ A}$