Answered: Z1- 15 ohm, Z2- 8 - j12 are connected…

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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Question

Z1- 15 ohm, Z2- 8 - j12 are connected in parallel .Total power of the circuit is 2000 watt find the PI and P2.

Expert Solution

Step 1

Given Z₁and Z₂ are in parallel

Total real power P=2000 watt

$Z_{1} = 15 Ω R_{1} = 15 Ω Z_{2} = 8 - j 12 Ω R_{2} = 8 Ω X_{2} = 12 Ω$

Step 2

$Z_{e q} = \frac{Z_{1} Z_{2}}{Z_{1} + Z_{2}} Z_{e q} = \frac{15 \times (8 - j 12)}{15 + 8 - j 12} Z_{e q} = \frac{15 \times (8 - j 12)}{23 - j 12} Converting rectangular form to polar form Z_{e q} = \frac{15 \times 14.422 ∠ - 56.31}{25.942 ∠ - 27.55} Z_{e q} = \frac{216.33 ∠ - 56.31}{25.942 ∠ - 27.55} Z_{e q} = 8.338 ∠ - 28.76 Converting the polar form to rectangle form Z_{e q} = 7.31 - j 4.01 Z_{e q} = R_{e q} - j X_{e q} By comparing R_{e q} = 7.31 Ω X_{e q} = 4.01 Ω R e a l p o w e r P = \frac{V^{2}}{R_{e q}} Therefore sustitute the given values 2000 = \frac{V^{2}}{7.31} V^{2} = 2000 \times 7.31 V = \sqrt{14620} V = 120.91 v Real power taken by Z_{1} is P_{1} = \frac{V^{2}}{R_{1}} P_{1} = \frac{{(120.91)}^{2}}{15} P_{1} = \frac{14620}{15} P_{1} = 974.66 W a t t Real power taken by Z_{2} is P_{2} = \frac{V^{2}}{R_{2}} P_{2} = \frac{{(120.91)}^{2}}{8} P_{2} = \frac{14620}{8} P_{2} = 1827.5 W a t t$

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