Question

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10. + -/2 points My Notes + Ask Your Teacher In this circuit, the sources have values Vsi(t) = 24 · cos(1900 · t) V and Vs2(t) = 12 · cos(1900 · t + 35°) V, while the components have values R = 3.2 k22, L1 = 4.2 H, and L2 = 3 H. Using the node voltage method in the phasor domain, determine the magnitude and phase angle of the inductor voltage phasor Vo. magnitude = angle = rad

Answer 1

w 1900 rad /sec X24 = W, = (1900) (4.2) = 7,980.2 Xby = Why = (1900) (3) = 5400 2 3.2k12 Vo V + ' sya 12 Las asi 2x 240 3.3798. 3j5700 Apply nodal analysis at node vo .. V-240 + vo tvo-12 35 3.2K so 17980 15700 0.00031251 V.-2410) -j0.0001253 V6 – 20.0001754/vo-12/35) = 0 Vo 10.0003125 – 30.0001253 - 30.0001754) = 0.007510 - 20.0001954 (12[35) r 10.000 31 25 –jo.0003) = 0.0075 (o – 0.002104125 Vo = 0.008875L-11.1969 0.000433 - 43.836

Vo = 20.496|32.634 - Magnitude = 20.496 v the angle = 32.634 so angle in radianę in degreep 32.62 x angle = 32.634 x Typ + angle = 0.569 rad.
​​​​​​agnitude = 20.496 V

Angle = 0.569 rad