Amplitude relationships: A sinusoidal current has an RMS value of 14 mA. What is its peak-to-peak current (Ipp)?

Introduction / Context:Converting between RMS, peak, and peak-to-peak values is routine in AC analysis. RMS reflects equivalent DC heating, while peak and peak-to-peak describe waveform extrema. This problem converts a given RMS current to peak-to-peak.

Given Data / Assumptions:

• RMS current Irms = 14 mA.
• Pure sinusoid with zero DC offset.

Concept / Approach:

For a sine wave: Ip = Irms * √2, and Ipp = 2 * Ip. Combine to get Ipp = 2 * √2 * Irms.

Step-by-Step Solution:

Verification / Alternative check:

Reverse check: Irms = Ipp / (2√2) ≈ 39.6 / 2.828 ≈ 14.0 mA, confirming consistency.

Why Other Options Are Wrong:

16 mA confuses peak with RMS; 22.6 mA corresponds to √2 * 16 mA, not applicable. 45.12 mA would arise from using √(3) or other incorrect factor.

Common Pitfalls:

Using √3 or 2 instead of √2; mixing up RMS and average-rectified values.

Final Answer:

39.6 mA