Find average value of current - (1) 50a (2) 100a (3) 10a (4) 25a

General
1

find average value of current - (1) 50a (2) 100a (3) 10a (4) 25a

2

Find average value of current - (1) 50A (2) 100A (3) 10A (4) 25A

Answer:
From typical rectifier or pulse wave analysis in basic electrical engineering, the average value of a current waveform depends on its shape (square, sine, half-wave rectified sine, full-wave rectified sine, etc.) and conduction interval. Although the exact wave shape is not specified in your question, if the waveform is assumed to be a simple “square” pulse that is high (e.g., 50 A) for half of its period and zero for the remaining half, its average would be half of the peak value, which is:

I_{\text{avg}} = \frac{1}{2} \times I_{\text{peak}}

So, for a 50 A rectangular pulse conduction half the time, the average value would be:

I_{\text{avg}} = \frac{1}{2} \times 50 \text{ A} = 25 \text{ A}

Hence, under this common assumption, the correct choice from the given options would be 25A.

Note: For sine waves or other waveforms, the average value would differ (often involving factors such as 1/\pi, 2/\pi, etc.). However, most introductory questions referencing a “50 A” maximum and asking directly for an average “assuming no further details” typically imply a rectangular waveform used half the time.

Key Points Summary

  1. Waveform Shape Matters

    • A square or rectangular wave that is “on” for half its period has an average value = ½ Ă— peak.
    • For a sine wave or rectified sine wave, other formulas apply (e.g., I_{m}/\pi, 2 I_{m}/\pi, etc.).

  2. Given Options and Assumption

    • With no additional information, the common assumption is a 50 A square pulse with a 50% duty cycle.
    • Thus, the average current would be 25 A.

  3. Most Likely Correct Answer

    • 25 A is the likely average value from the provided choices under the simplest rectangular assumption.
Waveform Peak Current (A) Average Value Expression Example Result (If Peak = 50A)
Sine Wave (full cycle) ( I_m ) ( 0 ) 0 A (no net DC)
Half-Wave Rectified Sine ( I_m ) ( \frac{I_m}{\pi} \approx 15.9 A
Full-Wave Rectified Sine ( I_m ) ( \frac{2 I_m}{\pi} \approx 31.8 A
Square Wave (50% duty) ( I_m ) \frac{I_m}{2} 25 A

@LectureNotes

1 Like