Peak voltage from rms current and resistance: If a 4.7 kΩ resistor carries 4 mA rms of current, what is the peak (maximum) voltage drop across this resistor?

Introduction / Context:Determining peak values from rms measurements is routine when dealing with sinusoidal signals. For a purely resistive element, the rms voltage is I_rms * R, and the peak voltage equals the rms value multiplied by √2. This problem also checks clean handling of milli- and kilo- prefixes.

Given Data / Assumptions:

• I_rms = 4 mA = 0.004 A.
• R = 4.7 kΩ = 4700 Ω.
• Sinusoidal signal across a resistor (no reactive effects).

Concept / Approach:First compute the rms voltage drop using Ohm’s law. Then convert rms to peak via Vpeak = Vrms * √2. The result represents the maximum instantaneous magnitude of the sinusoidal voltage across the resistor.

Step-by-Step Solution:

Verification / Alternative check:Power consistency: P = I_rms^2 * R = (0.004)^2 * 4700 ≈ 0.0752 W. Using Vpeak, the instantaneous maximum power would be (Vpeak^2 / R) at peaks; the average value aligns with the rms-based power, indicating consistent calculations.

Why Other Options Are Wrong:

• 18.8 V: That is the rms drop, not the peak value.
• 4 V and 2.66 V: Off by factors from neglecting R magnitude or the √2 conversion.

Common Pitfalls:

• Mistaking rms for peak or vice versa.
• Dropping prefixes: treating 4.7 kΩ as 4.7 Ω produces wildly incorrect results.

Final Answer:26.6 V