An ammeter reads 20 mA for a sinusoidal AC current (assume the meter indicates RMS). What is the corresponding peak-to-peak current of this sine wave?

Introduction:Many meters report RMS values, but design and analysis often require peak or peak-to-peak amplitudes. Converting between RMS and peak values for a pure sine wave is a common task in AC analysis.

Given Data / Assumptions:

• Measured current Irms = 20 mA.
• Waveform is a pure sinusoid.
• Relationship for sine: Ipeak = Irms * sqrt(2), Ipp = 2 * Ipeak.

Concept / Approach:For sinusoidal signals, RMS and peak quantities are linked by sqrt(2). Once peak is found, peak-to-peak is simply twice the peak. This stems from the trigonometric RMS of sinusoids over a full cycle.

Step-by-Step Solution:1) Compute Ipeak = Irms * sqrt(2) = 20 mA * 1.414 ≈ 28.28 mA.2) Compute Ipp = 2 * Ipeak ≈ 2 * 28.28 mA ≈ 56.57 mA.3) Round to the nearest option: approximately 57 mA.4) Report Ipp ≈ 57 mA.

Verification / Alternative check:Reverse-check: If Ipp ≈ 57 mA, then Ipeak ≈ 28.5 mA and Irms ≈ 28.5 / 1.414 ≈ 20.1 mA, consistent with the given meter reading.

Why Other Options Are Wrong:14 mA and 28 mA: correspond to peak or RMS misinterpretations, not peak-to-peak.

40 mA: equals 2 * Irms, but peak-to-peak requires 2 * Ipeak.

63 mA: would imply Irms ≈ 22.3 mA; not consistent with 20 mA reading.

Common Pitfalls:Confusing peak with peak-to-peak, or forgetting the sqrt(2) factor for sinusoids.

Final Answer:57 mA