Ohm’s law with AC rms values: If 8 mA rms of current flows through a 6.8 kΩ resistor, what is the corresponding rms voltage drop across the resistor?

Introduction / Context:
In a linear, purely resistive circuit, Ohm’s law applies to instantaneous, rms, or average quantities provided the matching forms are used. With AC sinusoidal signals, rms current multiplied by resistance gives rms voltage. This problem reinforces unit conversion (mA to A, kΩ to Ω) and precise multiplication.

Given Data / Assumptions:

• I_rms = 8 mA = 0.008 A.
• R = 6.8 kΩ = 6800 Ω.
• Purely resistive element, so V_rms = I_rms * R.

Concept / Approach:
Apply Ohm’s law directly with rms quantities. No phase angles or reactive components are involved, so the relationship is linear and scalar. Keep careful track of milli (10^-3) and kilo (10^3) prefixes during the calculation to avoid factor-of-1000 errors.

Step-by-Step Solution:

Verification / Alternative check:
Power check: P = I_rms^2 * R = (0.008)^2 * 6800 ≈ 0.435 W. Then V_rms = √(P * R) = √(0.435 * 6800) ≈ √2958 ≈ 54.4 V, confirming the result by a different route.

Why Other Options Are Wrong:

• 5.44 V: Misses a power-of-10 factor (likely using 680 Ω instead of 6800 Ω).
• 7.07 V and 8 V: Not consistent with the Ohm’s law product for the given values; these resemble unrelated common rms numbers.

Common Pitfalls:

• Mistaking 6.8 kΩ for 680 Ω or 68 kΩ.
• Leaving current in mA and resistance in kΩ without ensuring the product yields volts (careful with prefixes).

Final Answer:
54.4 V