When a transformer is connected to \(220\) V a.c. supply, then the output voltage is \(22\) V. A device rated as \(30\) V, \(60\) W is being run by the transformer. The current in primary coil of the transformer will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(0.2\) A
Note: The given answer is as per the official NTA Answer Key. However, none of the given options appears to be correct, as explained below.
For an ideal transformer, \( V_pI_p = V_sI_s \).
Working as per the NTA key
\( \displaystyle I_s = \frac{P}{V} = \frac{60\,\mathrm{W}}{30\,\mathrm{V}} = 2\,\mathrm{A} \)
\( \displaystyle I_p = \frac{V_sI_s}{V_p} = \frac{22\,\mathrm{V} \times 2\,\mathrm{A}}{220\,\mathrm{V}} = 0.2\,\mathrm{A} \)
Working as per the standard method
\( \displaystyle R = \frac{V^{2}}{P} = \frac{(30\,\mathrm{V})^{2}}{60\,\mathrm{W}} = 15\,\Omega \)
\( \displaystyle I_s = \frac{22\,\mathrm{V}}{15\,\Omega} \approx 1.47\,\mathrm{A} \)
\( \displaystyle I_p = \frac{22\,\mathrm{V} \times 1.47\,\mathrm{A}}{220\,\mathrm{V}} \approx 0.147\,\mathrm{A} \)
So the primary current is about 0.15 A, which is not among the options. NTA seems to have assumed that the device draws its rated current of 2 A at 22 V, which gives 0.2 A.