Answer is given. I just don't know how to get there. Thank you! Q3. 5 pointsUse the mesh-current method to find the pha-sor current I in the circuit shown.Solution:1Ωwww33.8/0°Vj202302202 0.75 V₂5-j5nLet Ia, Ib, and I be the three clockwise mesh currents going from left toright. Summing the voltages around meshes a and b gives33.8 = (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.aj5)(Ib − Ia) + 2(Ib − Ic).-

Answered: Image /qna-images/answer/66ea9127-b481-4ee7-b077-0fdc75e99761.jpg

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.

Delmar's Standard Textbook Of Electricity

7th Edition

ISBN:9781337900348

Author:Stephen L. Herman

Publisher:Stephen L. Herman

Chapter7: Parallel Circuits

Section: Chapter Questions

Problem 4PP: Using the rules for parallel circuits and Ohmslaw, solve for the missing values....

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