If a lamp has a resistance of 120 Ω when it operates at a power of 100 W, what current does the lamp carry?

Question

If a lamp has a resistance of 120 Ω when it operates at a power of 100 W, what current does the lamp carry?

BlackTom AI · Accepted Answer

To tackle this problem, I’ll first identify the fundamental relationship between power, resistance, and current for a resistor-like device. The lamp is described by P = I^2 R, so the current can be found from I = sqrt(P/R).
Option A: 2.10 A. If we plug this into P = I^2 R, P would be I^2 R = (2.10)^2 × 120 ≈ 4.41 × 120 ≈ 529 W, which is far from the given 100 W. This shows this value is too large for the specified power and resistance.
Option B: 1.25 A. Squaring this gives 1.5625; multiplying by 120 yields P ≈ 187.5 W, still not matching the stated 100 W. Therefore this current cannot be correct for the given conditions.
Option C: 0.913 A. Squaring 0.913 gives about 0.833; multiplying by 120 gives P ≈ 100 W, which aligns precisely with the specified power. This consistency confirms compatibility with P = I^2 R.
Option D: 0.830 A. Squaring 0.830 yields ≈ 0.689; times 120 gives P ≈ 82.7 W, which is noticeably short of the required 100 W. Hence this value does not satisfy the given data.
A quick cross-check can also be done by using I = sqrt(P/R): sqrt(100/120) ≈ sqrt(0.8333) ≈ 0.9129 A, which matches the 0.913 A result in Option C and helps validate the calculation. The other options fail to produce the specified power when combined with the given resistance.

If a lamp has a resistance of 120 Ω when it operates at a power of 100 W, what current does the lamp carry?单项选择题

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