There was a question in my school physics exam which is as follows:

Consider a 100W bulb which operates at 50V dc. John wants to light this bulb by a 200V ac source. What component should he use, also give its specifications.

The answer to the question is an inductor, with which I have got no issues. But in the exam, I wrote my answer to be a step-down transformer, which I have myself used for such applications. In the explanation I wrote the following:

A step-down transformer shall be used for this purpose. The specifications would be : 200V/50V (rms). The 50V rms voltage will dissipate the same power as 50V dc in one cycle, so irrespective of the current required (assuming the transformer to be capable enough to handle 2A), supplying the required voltage will achieve the same results. The winding of the primary and the secondary coils will be in the ratio 4:1 (assuming lossless transmission)

My teacher didn't accept this answer and neither did she give any explanations for why my answer was wrong. Despite having used it myself for the same purpose (powering my ac motors which operate at lower voltages - 12V), my answer was marked incorrect. Why is it so, or is it incorrect at all (since the question was ambiguous, I chose to write transformer as my answer)?

  • $\begingroup$ Your answer sounds fine to me. But I don't like the question: it doesn't specify whether the bulb is a simple old-fashioned resistive type, or if it's an LED that requires a DC supply, etc. $\endgroup$ – PM 2Ring Oct 8 '16 at 8:27
  • $\begingroup$ It is a normal resistive bulb. LED would have been specified explicitly $\endgroup$ – Pranshu Malik Oct 8 '16 at 8:49
  • $\begingroup$ @PranshuMalik Do you mean that the operating voltage is 50 V ac rather than dc otherwise the given answer, an inductor, is not correct? However if it is a filament light bulb rather than a fluorescent lamp you can use a transformer. If it is a fluorescent lamp then a transformer can be used but it must be capable of limiting the current using the idea of leakage impedance. $\endgroup$ – Farcher Oct 8 '16 at 9:05
  • $\begingroup$ @Farcher No, the rating is 50V DC and not AC. However, with a resistive type bulb, it really doesn't matter until the power dissipation is the same which is the case with rms voltage $\endgroup$ – Pranshu Malik Oct 8 '16 at 9:34
  • $\begingroup$ @PranshuMalik Thank you for the clarification. In the modern world the question is ambiguous because of the availability of LED bulbs. $\endgroup$ – Farcher Oct 8 '16 at 9:50

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