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Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
Back to Practice Questions
ElectricalBasic Electrical
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Average value of current is given by

A

ImI_mIm​

B

0.5Im0.5I_m0.5Im​

C

0.707Im0.707I_m0.707Im​

D

0.637Im0.637I_m0.637Im​

Correct Answer

Concept & PrincipleElectricalBasic Electrical
Option D

0.637Im0.637I_m0.637Im​

Quick Summary: The average value of a sinusoidal alternating current over a half-cycle is defined as the arithmetic mean of all instantaneous values. For a symmetrical sinusoidal waveform, this is calculated as $\frac{2I_m}{\pi}$, which evaluates to approximately $0.637I_m$.

💡 Explanation

The average value of a sinusoidal alternating current over a half-cycle is defined as the arithmetic mean of all instantaneous values. For a symmetrical sinusoidal waveform, this is calculated as 2Imπ\frac{2I_m}{\pi}π2Im​​, which evaluates to approximately 0.637Im0.637I_m0.637Im​.

🔢 Key Formulas

Iavg=2ImπI_{avg} = \frac{2I_m}{\pi}Iavg​=π2Im​​ — Average value for a half-cycle

Irms=Im2≈0.707ImI_{rms} = \frac{I_m}{\sqrt{2}} \approx 0.707I_mIrms​=2​Im​​≈0.707Im​ — Root Mean Square value

⚙️ Working Principle

The average value is derived by integrating the instantaneous current i=Imsin⁡(ωt)i = I_m \sin(\omega t)i=Im​sin(ωt) over the interval from 000 to π\piπ and dividing by the interval duration π\piπ. Mathematically, Iavg=1π∫0πImsin⁡(θ)dθ=Imπ[−cos⁡(θ)]0°π=2Imπ≈0.637ImI_{avg} = \frac{1}{\pi} \int_{0}^{\pi} I_m \sin(\theta) d\theta = \frac{I_m}{\pi} [-\cos(\theta)]_0°{\pi} = \frac{2I_m}{\pi} \approx 0.637I_mIavg​=π1​∫0π​Im​sin(θ)dθ=πIm​​[−cos(θ)]0​°π=π2Im​​≈0.637Im​. Over a full cycle, the average value of a pure sinusoidal wave is zero due to the cancellation of the positive and negative half-cycles.

📌 Key Points
  • ▸

    The average value is strictly defined for a half-cycle in AC circuits.

  • ▸

    Form factor is defined as the ratio of RMS value to Average value, which is ≈1.11\approx 1.11≈1.11 for a sine wave.

  • ▸

    The average value over a complete period for a symmetrical AC signal is zero.

✅ Advantages
  • ▸

    Useful in calculating DC equivalent output in half-wave and full-wave rectifiers.

  • ▸

    Simplifies analysis of pulsating DC components in electrical machines.

❌ Disadvantages / Limitations
  • ▸

    Not representative of the actual power-delivering capability of the AC signal.

  • ▸

    Mathematically zero over a full cycle, requiring half-cycle rectification for meaningful measurement.

🛠️ Applications / Uses
  • ▸

    Moving coil (PMMC) instrument calibration.

  • ▸

    Design and analysis of rectifier circuits.

📄 Additional Information
  • ▸

    Option B (0.5Im) is incorrect and holds no standard physical significance for sine waves.

  • ▸

    Option C (0.707Im) refers to the RMS (Root Mean Square) value, not the average value.

📊 Diagram / Illustration
Average Current Calculation
2Im2I_m2Im​
π\piπ
Result: ≈0.637Im\approx 0.637 I_m≈0.637Im​
✅

D is correct — The average value of a sinusoidal current over one half-cycle is mathematically derived as 2Im/π2I_m/\pi2Im​/π, resulting in 0.637Im0.637I_m0.637Im​.

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

Always distinguish between Average and RMS values; recall that Form Factor = RMS/Average = 1.11 for sinusoidal signals.

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