Question

A transformer of $100%$ efficiency has 200 turns in the primary and 40000 turns in secondary. It is connected to a $220~\mathrm{V}$ main supply and secondary feeds to a $100\mathrm{k}\Omega$ resistance. The potential difference per turn is

• A. $1.1\mathrm{V}$
• B. $25\mathrm{V}$
• C. $18\mathrm{V}$
• D. $11\mathrm{V}$

Hint:

For an ideal transformer, $\frac{V_p}{N_p} = \frac{V_s}{N_s}$.

Answer 1

The Correct Option is A

To determine the potential difference per turn for the given transformer, we will break down the problem using transformer formulas and logic:

1. The transformer is described as having 100% efficiency. This means there is no loss of power during the transformation from the primary to the secondary coil.
2. The number of turns in the primary coil (\(N_p\)) is 200, and the number of turns in the secondary coil (\(N_s\)) is 40,000.
3. The primary voltage (\(V_p\)) is given as 220 V.
4. For an ideal transformer, the ratio of the voltages is equal to the ratio of the number of turns: \(\frac{V_s}{V_p} = \frac{N_s}{N_p}\)where \(V_s\) is the secondary voltage.
5. Substituting the given values: \(\frac{V_s}{220} = \frac{40,000}{200}\)
6. Simplifying the number of turns ratio: \(\frac{40,000}{200} = 200\)
7. Therefore: \(V_s = 220 \times 200 = 44,000 \, \mathrm{V}\)
8. The potential difference per turn can be calculated by dividing the secondary voltage by the number of turns in the secondary coil: \(\text{Potential difference per turn} = \frac{V_s}{N_s} = \frac{44,000}{40,000} \, \mathrm{V} = 1.1 \, \mathrm{V}\)

Thus, the potential difference per turn is 1.1 V, so the correct answer is

$1.1\mathrm{V}$