An electric iron draws a current of 15 A when connected to a 120-V power source. Its resistance is

Question

BlackTom AI · Accepted Answer

Consider the electrical setup described: a device connected to a 120 V source and drawing a current of 15 A. The key relationship here is Ohm's law, which connects voltage, current, and resistance through V = I R.

Option 0.125 ohms: If R were 0.125 ohms, then the current would be I = V / R = 120 V / 0.125 Ω = 960 A, which is far larger than the given 15 A. This option is inconsistent with the reported current and can be discarded.

Option 8 ohms: Applying Ohm's law with V = 120 V and I = 15 A gives R = V / I = 120 / 15 = 8 Ω. This directly matches the mathematical relationship and the stated current, so this is consistent with the data.

Option 16 ohms: If R were 16 Ω, the current would be I = V / R = 120 V / 16 Ω = 7.5 A. Since the actual current is 15 A, this value would imply a larger current than observed, making this option incorrect.

Option 1800 ohms: With a resistance of 1800 Ω, the current would be I = V / R = 120 V / 1800 Ω ≈ 0.0667 A, vastly smaller than the stated 15 A. This option does not fit the given current and is thus invalid.

In summary, the only option that aligns with Ohm's law and the given voltage and current is the one yielding R = 8 Ω. The other values produce currents that conflict with the reported 15 A. The reasoning hinges on consistent application of V = I R across all options.

An electric iron draws a current of 15 A when connected to a 120-V power source. Its resistance is 单项选择题

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