Given, f=84 Hz .: W=2πf =2πT x 84 = 527.7876 for Capacitor (xc) = 1 1 jwc j527.7876 x 2×15-6 =-3947.35082 for Inductor (x) = jwL = j527.7876 x 50 x×10 Now Redraw the circuit in =j26-38942 given frequency 3640 -1947.35082 th V + {2k-2 √x3j26.3894 + 44233 Vo 1VX Apply Nodal √x -36400+ Vx Vx Vx-Vo + + 2000 j26.3894 j947.3508 360° = x[ Vx [2000 4 + j26.3894 + 0947.3505 0947.3500 == Apply Nodal on Vo, 3610° = 10.036842 L-89.2224) Vx+ Vo 0947-3508 V₂ Vo-Vx + 0.1Vx+ 4000 947.3508 4000 j947. -]+ Vx [0.1+ 1947. 308 13508 / +1 (0.001085 £76.6757) №6 + ·(0. from quation ①and ② 1947.3508 =O • 100006 L_0.605°) Vx=0 =O Vx=372.4317 L9.174567° Volts 16=34334.5034 2111.894° VOHS LTspice XVII - [Draft2] File Edit Hierarchy View Simulate Tools Window Help Aa .op I1 AC 36 ← SINE(0 36 84 0.0119) .ac lin 1 383,2743 383,2743 This is ground. ○ Ara C1 2μ 0.1 Vx R1 Vx 2k 50m G1 + > 0 0 Vo 4k х 5 14:38 9.11.2024
Given, f=84 Hz .: W=2πf =2πT x 84 = 527.7876 for Capacitor (xc) = 1 1 jwc j527.7876 x 2×15-6 =-3947.35082 for Inductor (x) = jwL = j527.7876 x 50 x×10 Now Redraw the circuit in =j26-38942 given frequency 3640 -1947.35082 th V + {2k-2 √x3j26.3894 + 44233 Vo 1VX Apply Nodal √x -36400+ Vx Vx Vx-Vo + + 2000 j26.3894 j947.3508 360° = x[ Vx [2000 4 + j26.3894 + 0947.3505 0947.3500 == Apply Nodal on Vo, 3610° = 10.036842 L-89.2224) Vx+ Vo 0947-3508 V₂ Vo-Vx + 0.1Vx+ 4000 947.3508 4000 j947. -]+ Vx [0.1+ 1947. 308 13508 / +1 (0.001085 £76.6757) №6 + ·(0. from quation ①and ② 1947.3508 =O • 100006 L_0.605°) Vx=0 =O Vx=372.4317 L9.174567° Volts 16=34334.5034 2111.894° VOHS LTspice XVII - [Draft2] File Edit Hierarchy View Simulate Tools Window Help Aa .op I1 AC 36 ← SINE(0 36 84 0.0119) .ac lin 1 383,2743 383,2743 This is ground. ○ Ara C1 2μ 0.1 Vx R1 Vx 2k 50m G1 + > 0 0 Vo 4k х 5 14:38 9.11.2024
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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