*12. For the network in Fig. 71:a. Find currents 15, 12, and 16.b. Find voltages V₁ and V5.c. Find the power delivered to the 3 kn resistor.E = 28 V4₁₁V₁ R₁12 ΚΩR₂1212 ΚΩR323 ΚΩR49 ΚΩR₂www6 ΚΩ+ V₁ -R610.4 ΚΩ

Answered: Image /qna-images/answer/b2058a5a-4d97-4e13-b528-9a3c7b07e220.jpg

*12. For the network in Fig. 71: a. Find currents Is, Iz, and 16. b. Find voltages V₁ and V5. c. Find the power delivered to the 3 kΩ resistor. E = 28 V 12 ΚΩ R₂ 12 ΚΩ R₂ www • 3 ΚΩ R₁ Μ 9 ΚΩ Μ R₂ ΚΩ + Vs – R6 :10.4 ΚΩ

*12. For the network in Fig. 71: a. Find currents Is, Iz, and 16. b. Find voltages V₁ and V5. c. Find the power delivered to the 3 kΩ resistor. E = 28 V 12 ΚΩ R₂ 12 ΚΩ R₂ www • 3 ΚΩ R₁ Μ 9 ΚΩ Μ R₂ ΚΩ + Vs – R6 :10.4 ΚΩ

Delmar's Standard Textbook Of Electricity

7th Edition

ISBN:9781337900348

Author:Stephen L. Herman

Publisher:Stephen L. Herman

Chapter18: Resistive-inductive Parallel Circuits

Section: Chapter Questions

Problem 10PP: In an R-L parallel circuit, ET=240 volts, R=560, and XL=330. Find apparent power.

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Concept explainers

Norton Theorem

In the current circuit law, the Norton theorem (also known as the Mayer-Norton theorem) is a simplification that can be applied to networks generated by a constant line of opposition, power sources, and current assets. In two network terminals, it can be switched on with a current supply and single resistor respectively.

Question

*12. For the network in Fig. 71:
a. Find currents 15, 12, and 16.
b. Find voltages V₁ and V5.
c. Find the power delivered to the 3 kn resistor.
E = 28 V
4₁₁
V₁ R₁
12 ΚΩ
R₂
12
12 ΚΩ
R3
23 ΚΩ
R4
9 ΚΩ
R₂
www
6 ΚΩ
+ V₁ -
R6
10.4 ΚΩ

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Step 1: Determine said variables from given circuit:

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Step 2: Combine parallel resistors to reduce the circuit:

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Step 3: Find Current Is:

VIEW

Step 4: Find current I6:

VIEW

Step 5: Find current I2 and voltage V1 , V5 :

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Step 6: Find the power delivered to 3kΩ:

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