Match List-I with List-II with reference to step-up/step-down transformers.
| List-I | List-II | ||
|---|---|---|---|
| (A) | Primary: \( 500 \) turns; Secondary: \( 1000 \) turns, Input: \( 220\ \mathrm{V},\ 1.0\ \mathrm{A} \) | (I) | Output current \( = 0.50\ \mathrm{A} \) |
| (B) | Primary: \( 1000 \) turns; Secondary: \( 500 \) turns, Input : \( 220\ \mathrm{V} \) | (II) | Output voltage \( = 110\ \mathrm{V} \) |
| (C) | Primary: \( 200 \) turns; Secondary: \( 100 \) turns, Input current : \( 2\ \mathrm{A} \) | (III) | Output current \( = 4\ \mathrm{A} \) |
| (D) | Step-up transfer ratio \( 1:5 \), Input : \( 230\ \mathrm{V} \) | (IV) | Output voltage \( = 1150\ \mathrm{V} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
For an ideal transformer the voltages go as the turns and the currents go the other way, because input power equals output power:
\( \displaystyle \frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}} = \frac{I_{p}}{I_{s}} \)
(A) Primary: \( 500 \) turns; Secondary: \( 1000 \) turns, Input: \( 220\ \mathrm{V},\ 1.0\ \mathrm{A} \) — this matches (I) Output current \( = 0.50\ \mathrm{A} \). It is a step-up transformer (turns ratio 2), so the current halves.
\( \displaystyle I_{s} = I_{p}\frac{N_{p}}{N_{s}} = 1.0\,\mathrm{A}\times\frac{500}{1000} = 0.50\,\mathrm{A} \)
(B) Primary: \( 1000 \) turns; Secondary: \( 500 \) turns, Input : \( 220\ \mathrm{V} \) — this matches (II) Output voltage \( = 110\ \mathrm{V} \). It is a step-down transformer, so the voltage halves.
\( \displaystyle V_{s} = 220\,\mathrm{V}\times\frac{500}{1000} = 110\,\mathrm{V} \)
(C) Primary: \( 200 \) turns; Secondary: \( 100 \) turns, Input current : \( 2\ \mathrm{A} \) — this matches (III) Output current \( = 4\ \mathrm{A} \). A step-down transformer raises the current.
\( \displaystyle I_{s} = 2\,\mathrm{A}\times\frac{200}{100} = 4\,\mathrm{A} \)
(D) Step-up transfer ratio \( 1:5 \), Input : \( 230\ \mathrm{V} \) — this matches (IV) Output voltage \( = 1150\ \mathrm{V} \). Here \( N_{p}:N_{s} = 1:5 \), so the voltage is multiplied by 5.
\( V_{s} = 5\times230\,\mathrm{V} = 1150\,\mathrm{V} \)