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Q3. 5 points Use the mesh-current method to find the pha- sor current I in the circuit shown. Solution: 192 ww I 33.8/0 V j2 02 V. {302 202 0.75 V, -j5n Let Ia, Ib, and Ic be the three clockwise mesh currents going from left to right. Summing the voltages around meshes a and b gives 33.8 = (1 + j2) I₂ + (3 − j5) (Ia - Ib) and 0= (3- j5) (Ib — Ia) + 2(Ib - Ic). But V₂ = -j5(la - Ib), therefore Ic = -0.75[-j5(la - Ib)]. Solving for I = I₁ = 29+ j2 = 29.07/3.95° A.

Q3. 5 points Use the mesh-current method to find the pha- sor current I in the circuit shown. Solution: 192 ww I 33.8/0 V j2 02 V. {302 202 0.75 V, -j5n Let Ia, Ib, and Ic be the three clockwise mesh currents going from left to right. Summing the voltages around meshes a and b gives 33.8 = (1 + j2) I₂ + (3 − j5) (Ia - Ib) and 0= (3- j5) (Ib — Ia) + 2(Ib - Ic). But V₂ = -j5(la - Ib), therefore Ic = -0.75[-j5(la - Ib)]. Solving for I = I₁ = 29+ j2 = 29.07/3.95° A.

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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Q3. 5 points Use the mesh-current method to find the pha- sor current I in the circuit shown. Solution: 1Ω www 33.8/0° V j202 302 202 0.75 V₂ 5-j5n Let Ia, Ib, and I be the three clockwise mesh currents going from left to right. Summing the voltages around meshes a and b gives 33.8 = (1 + j2) Ia + (3-j5) (Ia - Ib) and 0 = = (3 − - But V₂ = -j5(Ia - Ib), therefore Ic = -0.75[-j5(Ia — Ib)]. Solving for I = I₂ = 29+ j2 = 29.07/3.95° A. a j5)(Ib − Ia) + 2(Ib − Ic). -
Transcribed Image Text:Q3. 5 points Use the mesh-current method to find the pha- sor current I in the circuit shown. Solution: 1Ω www 33.8/0° V j202 302 202 0.75 V₂ 5-j5n Let Ia, Ib, and I be the three clockwise mesh currents going from left to right. Summing the voltages around meshes a and b gives 33.8 = (1 + j2) Ia + (3-j5) (Ia - Ib) and 0 = = (3 − - But V₂ = -j5(Ia - Ib), therefore Ic = -0.75[-j5(Ia — Ib)]. Solving for I = I₂ = 29+ j2 = 29.07/3.95° A. a j5)(Ib − Ia) + 2(Ib − Ic). -
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