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Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
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ElectricalBasic Electrical
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If maximum value of current is 100A, then average value is

A

50A

B

100A

C

0

D

63.7A

Correct Answer

Direct FormulaElectricalBasic Electrical
Option D

63.7A

Quick Summary: Given: Maximum value of alternating current I_m = 100A

📋 Given

Maximum value of alternating current ImI_{m}Im​ = 100A

🔢 Formula Used

Iavg=2ImπI_{avg} = \frac{2 I_m}{\pi}Iavg​=π2Im​​

📊 Diagram / Illustration
I(t)=Imsin⁡(ωt)I(t) = I_m \sin(\omega t)I(t)=Im​sin(ωt)
Half-cycle average: Iavg=0.637ImI_{avg} = 0.637 I_mIavg​=0.637Im​
🔢 Step-by-Step Solution
1

Identify given values

The maximum value of the sinusoidal alternating current is given as Im=100AI_m = 100AIm​=100A.

Im=100AI_m = 100AIm​=100A

2

Select the average value formula

For a full sinusoidal waveform over a half-cycle, the average value is calculated using the relationship Iavg=2ImπI_{avg} = \frac{2I_m}{\pi}Iavg​=π2Im​​.

Iavg=2ImπI_{avg} = \frac{2 I_m}{\pi}Iavg​=π2Im​​

3

Substitute and solve

Substitute Im=100I_m = 100Im​=100 and π≈3.14159\pi \approx 3.14159π≈3.14159 into the formula: Iavg=2×1003.14159=2003.14159≈63.66AI_{avg} = \frac{2 \times 100}{3.14159} = \frac{200}{3.14159} \approx 63.66AIavg​=3.141592×100​=3.14159200​≈63.66A.

Iavg≈63.7AI_{avg} \approx 63.7AIavg​≈63.7A

✅

D is correct because the average value of a sinusoidal current over a half-cycle is defined as 2π\frac{2}{\pi}π2​ times the maximum value, resulting in approximately 63.7A.

Core Concepts Used
Click any tag to open in AI Tutor
AC Fundamentals Average Current Sinusoidal Waveform
💡 EXAM TIP

Always distinguish between Average value (0.637Im0.637 I_m0.637Im​), RMS value (0.707Im0.707 I_m0.707Im​), and Peak-to-Peak value (2Im2 I_m2Im​) for AC circuits in electrical machines and power systems.

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