The output voltage of an ideal transformer connected to 240 V a.c. mains is 24 V. When this transformer is used to light a bulb with ratings 24 V, 24 W, the primary current would be-
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.1 A
The bulb works at its rated 24 V, so it draws its rated power of 24 W from the secondary. For an ideal transformer the input power equals the output power, \( V_pI_p = V_sI_s \).
\( \displaystyle I_s = \frac{P}{V_s} = \frac{24\,\mathrm{W}}{24\,\mathrm{V}} = 1\,\mathrm{A} \)
\( \displaystyle I_p = I_s\,\frac{V_s}{V_p} = 1\times\frac{24}{240}\,\mathrm{A} = 0.1\,\mathrm{A} \)
Shortcut: The primary must supply the same 24 W, so \( I_p = P/V_p = 24/240 = 0.1\,\mathrm{A} \), and the 1 A of Option (1) is the secondary current.