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a. If ix = 5A, find v₁ and iy; b. If v₁ = 3V, find ix and iy; c. What value of is will lead to v₁ # V₂?

a. If ix = 5A, find v₁ and iy; b. If v₁ = 3V, find ix and iy; c. What value of is will lead to v₁ # V₂?

Introductory Circuit Analysis (13th Edition)
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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Explain how the answer was derived in this problem (see the 2 attached image)

Problem 2. Referring to the circuit given in Figure 2, a. If ix = 5A, find v₁ and iy; b. If v₁ = 3V, find ix and iy; c. What value of is will lead to v₁ # V₂? is 10 Ω www + 21 10 Ω ww 10 Ω 10 Ω Figure 2. + V2 10 Ω
Transcribed Image Text:Problem 2. Referring to the circuit given in Figure 2, a. If ix = 5A, find v₁ and iy; b. If v₁ = 3V, find ix and iy; c. What value of is will lead to v₁ # V₂? is 10 Ω www + 21 10 Ω ww 10 Ω 10 Ω Figure 2. + V2 10 Ω
Problem 2 GIVEN: Aq is (↑ REQUIRED: EQUATION: "11 I = X R SOWTION: M + 1052 "▷ a.) V₁ = ?, ¡y = ? mtb.) ix = ? ly=? "DC.) is = ? a.) if ix=5A WD Ii V₁ W ix x 10 = 10N 10 +10 141/7 V₁= I₁ x 10 V₁ = 25 V 1x 110 ly = ¹x x 10 10 + 10 iy= 2₂5 A B = 2.5 × 10 = 25 V lx WWW 1012 Figure 2. 1052 b.) i √ = 1₁ + 1₂ it V₁ = 3 V current is 102 5x10 20 T₁= V K ₁² = 3/8 = 0.3A V 1₂= 11=3 R₂ 23 1. ix = 0+ 0.3=0.6A 1x = 0.6A =25A C V₂ 5×10=2.5A 10 +10 3/8 = 0.3A for iy= Diy= 1x xW 10+10 iy=0.3A Liy 10.12 0.6 xto 2x10
Transcribed Image Text:Problem 2 GIVEN: Aq is (↑ REQUIRED: EQUATION: "11 I = X R SOWTION: M + 1052 "▷ a.) V₁ = ?, ¡y = ? mtb.) ix = ? ly=? "DC.) is = ? a.) if ix=5A WD Ii V₁ W ix x 10 = 10N 10 +10 141/7 V₁= I₁ x 10 V₁ = 25 V 1x 110 ly = ¹x x 10 10 + 10 iy= 2₂5 A B = 2.5 × 10 = 25 V lx WWW 1012 Figure 2. 1052 b.) i √ = 1₁ + 1₂ it V₁ = 3 V current is 102 5x10 20 T₁= V K ₁² = 3/8 = 0.3A V 1₂= 11=3 R₂ 23 1. ix = 0+ 0.3=0.6A 1x = 0.6A =25A C V₂ 5×10=2.5A 10 +10 3/8 = 0.3A for iy= Diy= 1x xW 10+10 iy=0.3A Liy 10.12 0.6 xto 2x10
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