The periodic triangular current having a peak value of 180 [mA] shown below is passing through a 5[kQ] resistor. Find the average power that this current delivers to the resistor.etc.T/2 –T/4T/4 T2 3T/4T

Answered: The periodic triangular current having…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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The periodic triangular current having a peak value of 180 [mA] shown below is passing through a 5
[kQ] resistor. Find the average power that this current delivers to the resistor.
etc.
T/2 –T/4
T/4 T2 3T/4
T

Expert Solution

Step 1

Given

Peak to peak value of current I =180 mA

Peak to peak value is defined as difference of the maximum value in positive half and maximum value in negative half cycle over a complete time period.

Value of resistance R = 5000 Ohm

Rms value of current

Area under the curve will be $= \frac{I_{m} t}{h a l f c y l e} = \frac{I_{m}}{\frac{T}{2}}$

WE have calculated for half cycle as the wave is symmetrical for both the half cycles.

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