IL Q.12: The non-sinusoidal applied voltage for the circuit shown is: (e;=5+4sin1000t+2sin3000t) V. 1. Derive the equation of currents (i) & (ic). 2. Find the total power consumed in the circuit? (Ans.: iT=1+0.73sin(1000t-24°)A, ic=0.09sin(1000t+ 155.8°)A & P7=6.33W) eT :-J180 122

IL Q.12: The non-sinusoidal applied voltage for the circuit shown is: (e;=5+4sin1000t+2sin3000t) V. 1. Derive the equation of currents (i) & (ic). 2. Find the total power consumed in the circuit? (Ans.: iT=1+0.73sin(1000t-24°)A, ic=0.09sin(1000t+ 155.8°)A & P7=6.33W) eT :-J180 122

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Electrical Heating Unit

The conversion of electrical energy to heat energy is known as electric heating. The heating system that converts the electrical energy to heat energy for various application is called electrical heating unit. Electricity is employed as the firing mechanism in electric heating systems, which transforms electric current to heat. Electric heaters have heating elements in the form of electric resistors.

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1T
1L
Q.12: The non-sinusoidal applied voltage for the
circuit shown is: (e,=5+4sin1000t+2sin3000t) V.
1. Derive the equation of currents (i) & (ic).
2. Find the total power consumed in the circuit?
(Ans.: iT=1+0.73sin(1000t-24°) A, İc=0.09sin(1000t+155.8°)A & P;=6.33W)
eT
1C
-J180 J22

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1T 1L Q.12: The non-sinusoidal applied voltage for the circuit shown is: (e,=5+4sin1000t+2sin3000t) V. 1. Derive the equation of currents (i) & (ic). 2. Find the total power consumed in the circuit? (Ans.: iT=1+0.73sin(1000t-24°) A, İc=0.09sin(1000t+155.8°)A & P;=6.33W) eT 1C -J180 J22
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