Peak current from rms voltage and resistance: If the rms voltage across a 15 kΩ resistor is 16 V, what is the peak (maximum) current through the resistor?

Introduction / Context:
Converting between rms and peak values is routine for sinusoidal steady-state analysis. Given an rms voltage across a resistor, find the rms current via Ohm’s law, then convert that rms current to its peak counterpart by multiplying by √2. This skill is frequently applied in amplitude ratings and current-limiting calculations.

Given Data / Assumptions:

• Vrms = 16 V.
• R = 15 kΩ = 15,000 Ω.
• Sinusoidal voltage across a purely resistive element.

Concept / Approach:
First compute Irms from Ohm’s law: Irms = Vrms / R. For a sinusoid, Ipeak = Irms * √2. Carefully handle kilo-ohm to ohm conversion to obtain milliamperes correctly.

Step-by-Step Solution:

Verification / Alternative check:
Back-check Vrms via Ipeak: Irms = Ipeak / √2 ≈ 1.5 / 1.414 ≈ 1.06 mA. Then Vrms = Irms * R ≈ 0.00106 * 15,000 ≈ 15.9 V (rounding to 16 V), confirming consistency with the given value.

Why Other Options Are Wrong:

• 15 mA and 10 mA: One or two orders of magnitude too large; they ignore the 15 kΩ resistance.
• 1 mA: Closer, but this is rms-like rather than peak; it does not match the √2 conversion from 1.0667 mA rms.

Common Pitfalls:

• Mishandling prefixes: using 15 Ω instead of 15 kΩ.
• Confusing rms with peak values or forgetting the √2 factor for sinusoids.

Final Answer:
1.5 mA