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|2 = 11/16/2 (Equation 1) Loop 2 (Right Loop): Apply KVL around Loop 2, which includes the resistances R2, R3, and the current source gm Vo. The current source gm V₁ = gm × Vo=9m × (I2R3), since V₁ = I2R3. The KVL equation for Loop 2 is: -R2(12-11)-R312 -9m Vo = 0 Substitute V₁ = I2R3 and the known values: -6000(12 - I₁) - 40001 – 7 × 10¯³ × (400012) = 0 Simplify: -600012 + 60001₁-400012-281₂ = 0 6000₁ = (1000012 +2812) 6000₁ =1002812 60001₁ =1002812 (Equation 2) Use loop analysis to find V in the circuit shown given that V₁ = = 2V, 9m = 7m¹, R₁ = = Vo= V 5kQ, R₂ = 6k and R3 = 4kQ. V₁ R₁ R₂ gm Vo R3 + Vo

|2 = 11/16/2 (Equation 1) Loop 2 (Right Loop): Apply KVL around Loop 2, which includes the resistances R2, R3, and the current source gm Vo. The current source gm V₁ = gm × Vo=9m × (I2R3), since V₁ = I2R3. The KVL equation for Loop 2 is: -R2(12-11)-R312 -9m Vo = 0 Substitute V₁ = I2R3 and the known values: -6000(12 - I₁) - 40001 – 7 × 10¯³ × (400012) = 0 Simplify: -600012 + 60001₁-400012-281₂ = 0 6000₁ = (1000012 +2812) 6000₁ =1002812 60001₁ =1002812 (Equation 2) Use loop analysis to find V in the circuit shown given that V₁ = = 2V, 9m = 7m¹, R₁ = = Vo= V 5kQ, R₂ = 6k and R3 = 4kQ. V₁ R₁ R₂ gm Vo R3 + Vo

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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For Vo I got 646V which is incorrect. 

|2 = 11/16/2 (Equation 1) Loop 2 (Right Loop): Apply KVL around Loop 2, which includes the resistances R2, R3, and the current source gm Vo. The current source gm V₁ = gm × Vo=9m × (I2R3), since V₁ = I2R3. The KVL equation for Loop 2 is: -R2(12-11)-R312 -9m Vo = 0 Substitute V₁ = I2R3 and the known values: -6000(12 - I₁) - 40001 – 7 × 10¯³ × (400012) = 0 Simplify: -600012 + 60001₁-400012-281₂ = 0 6000₁ = (1000012 +2812) 6000₁ =1002812 60001₁ =1002812 (Equation 2)
Transcribed Image Text:|2 = 11/16/2 (Equation 1) Loop 2 (Right Loop): Apply KVL around Loop 2, which includes the resistances R2, R3, and the current source gm Vo. The current source gm V₁ = gm × Vo=9m × (I2R3), since V₁ = I2R3. The KVL equation for Loop 2 is: -R2(12-11)-R312 -9m Vo = 0 Substitute V₁ = I2R3 and the known values: -6000(12 - I₁) - 40001 – 7 × 10¯³ × (400012) = 0 Simplify: -600012 + 60001₁-400012-281₂ = 0 6000₁ = (1000012 +2812) 6000₁ =1002812 60001₁ =1002812 (Equation 2)
Use loop analysis to find V in the circuit shown given that V₁ = = 2V, 9m = 7m¹, R₁ = = Vo= V 5kQ, R₂ = 6k and R3 = 4kQ. V₁ R₁ R₂ gm Vo R3 + Vo
Transcribed Image Text:Use loop analysis to find V in the circuit shown given that V₁ = = 2V, 9m = 7m¹, R₁ = = Vo= V 5kQ, R₂ = 6k and R3 = 4kQ. V₁ R₁ R₂ gm Vo R3 + Vo
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