DELMAR'S STANDARD TEXT OF ELECTRICITY
DELMAR'S STANDARD TEXT OF ELECTRICITY
6th Edition
ISBN: 9780357323380
Author: Herman
Publisher: CENGAGE C
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Chapter 19, Problem 2PP

Assume that the current flow through the resistor, I R , is 15 A; the current flow through the inductor, I L is 36 A; and the circuit has an apparent power of 10,803 VA. The frequency of the AC voltage is 60 Hz.

E T _ _ _ _ _ _ _ _ _ _ _ _ E R _ _ _ _ _ _ _ _ _ _ _ _ E L _ _ _ _ _ _ _ _ _ _ _ _
I T _ _ _ _ _ _ _ _ _ _ _ _ _ I R 15 A _ _ _ _ _ _ _ _ _ I L 36 A _ _ _ _ _ _ _ _ _
Z _ _ _ _ _ _ _ _ _ _ _ _ _ R _ _ _ _ _ _ _ _ _ _ _ _ _ X L _ _ _ _ _ _ _ _ _ _ _ _
V A 10 , 803 P _ _ _ _ _ _ _ _ _ _ _ _ _ V A R S L _ _ _ _ _ _ _ _ _
P F _ _ _ _ _ _ _ _ _ _ _ _ θ _ _ _ _ _ _ _ _ _ _ _ _ L _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
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I need help checking if its correct -E1 + VR1 + VR4 – E2 + VR3 = 0 -------> Loop 1 (a) R1(I1) + R4(I1 – I2) + R3(I1) = E1 + E2 ------> Loop 1 (b) R1(I1) + R4(I1) - R4(I2) + R3(I1) = E1 + E2 ------> Loop 1 (c) (R1 + R3 + R4) (I1)  - R4(I2)    = E1 + E2 ------> Loop 1 (d) Now that we have loop 1 equation will procced on finding the equation of I2 current loop. However, a reminder that because we are going in a clockwise direction, it goes against the direction of the current. As such we will get an equation for the matrix that will be:   E2 – VR4 – VR2 + E3 = 0 ------> Loop 2 (a) -R4(I2 – I1) -R2(I2) = -E2 – E3  ------> Loop 2 (b) -R4(I2) + R4(I1) - R2(I2) = -E2 – E3  -----> Loop 2 (c)                                     R4(I1) – (R4 + R2)(I2) = -E2 – E3  -----> Loop 2 (d) These two equations will be implemented to the matrix formula I = inv(A) * b       R11                        R12   (R1 + R3 + R4)     -R4     -R4     R4 + R2

Chapter 19 Solutions

DELMAR'S STANDARD TEXT OF ELECTRICITY

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