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Example 4.1 12892 ww For the circuit in Fig. 4.2, find I, when vs = 12 V and v = 24 V. Solution: Applying KVL to the two loops, we obtain ΖΩ τι www> + Vx i₁ 4Ω 12i₁ = 4i₂ + v. = 0 (4.1.1) 4i, +16i2-3UxUs = 0 (4.1.2) 60 But vx = 211. Equation (4.1.2) becomes Ux -10i 16i2 - Us = 0 (4.1.3) Adding Eqs. (4.1.1) and (4.1.3) yields 2₁₁ + 122 = 0 Substituting this in Eq. (4.1.1), we get i₁ = −6i2 is -76i2 + vs = 0 When vs = 12 V, 12 I。 = i2 A 76 Us When vs 24 V, 24 I₁ = i₂ = i2 A 76 || wt - +1 1 + i Figure 4.2 For Example 4.1. 402 i2 + + 30x 6% √s +3√x = + 2 11 + 4 (11-12) + Us = 0 -4+1 +1612 Us 76 Vs showing that when the source value is doubled I doubles -V₂ =

Example 4.1 12892 ww For the circuit in Fig. 4.2, find I, when vs = 12 V and v = 24 V. Solution: Applying KVL to the two loops, we obtain ΖΩ τι www> + Vx i₁ 4Ω 12i₁ = 4i₂ + v. = 0 (4.1.1) 4i, +16i2-3UxUs = 0 (4.1.2) 60 But vx = 211. Equation (4.1.2) becomes Ux -10i 16i2 - Us = 0 (4.1.3) Adding Eqs. (4.1.1) and (4.1.3) yields 2₁₁ + 122 = 0 Substituting this in Eq. (4.1.1), we get i₁ = −6i2 is -76i2 + vs = 0 When vs = 12 V, 12 I。 = i2 A 76 Us When vs 24 V, 24 I₁ = i₂ = i2 A 76 || wt - +1 1 + i Figure 4.2 For Example 4.1. 402 i2 + + 30x 6% √s +3√x = + 2 11 + 4 (11-12) + Us = 0 -4+1 +1612 Us 76 Vs showing that when the source value is doubled I doubles -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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I dont understand about node 2 equation. Explain it

Example 4.1 12892 ww For the circuit in Fig. 4.2, find I, when vs = 12 V and v = 24 V. Solution: Applying KVL to the two loops, we obtain ΖΩ τι www> + Vx i₁ 4Ω 12i₁ = 4i₂ + v. = 0 (4.1.1) 4i, +16i2-3UxUs = 0 (4.1.2) 60 But vx = 211. Equation (4.1.2) becomes Ux -10i 16i2 - Us = 0 (4.1.3) Adding Eqs. (4.1.1) and (4.1.3) yields 2₁₁ + 122 = 0 Substituting this in Eq. (4.1.1), we get i₁ = −6i2 is -76i2 + vs = 0 When vs = 12 V, 12 I。 = i2 A 76 Us When vs 24 V, 24 I₁ = i₂ = i2 A 76 || wt - +1 1 + i Figure 4.2 For Example 4.1. 402 i2 + + 30x 6% √s +3√x = + 2 11 + 4 (11-12) + Us = 0 -4+1 +1612 Us 76 Vs showing that when the source value is doubled I doubles -V₂ =
Transcribed Image Text:Example 4.1 12892 ww For the circuit in Fig. 4.2, find I, when vs = 12 V and v = 24 V. Solution: Applying KVL to the two loops, we obtain ΖΩ τι www> + Vx i₁ 4Ω 12i₁ = 4i₂ + v. = 0 (4.1.1) 4i, +16i2-3UxUs = 0 (4.1.2) 60 But vx = 211. Equation (4.1.2) becomes Ux -10i 16i2 - Us = 0 (4.1.3) Adding Eqs. (4.1.1) and (4.1.3) yields 2₁₁ + 122 = 0 Substituting this in Eq. (4.1.1), we get i₁ = −6i2 is -76i2 + vs = 0 When vs = 12 V, 12 I。 = i2 A 76 Us When vs 24 V, 24 I₁ = i₂ = i2 A 76 || wt - +1 1 + i Figure 4.2 For Example 4.1. 402 i2 + + 30x 6% √s +3√x = + 2 11 + 4 (11-12) + Us = 0 -4+1 +1612 Us 76 Vs showing that when the source value is doubled I doubles -V₂ =
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