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The voltage across a 106 capacitor is 200 sin(3771-30°) V. Determine the sinusoidal expression of the current passing through the capacitor. 1.89 sin(377t + 60") A 1.89 sin(377t +90) A 1.89 sin(377t 120") A 1.89 sin(377t-90) A

The voltage across a 106 capacitor is 200 sin(3771-30°) V. Determine the sinusoidal expression of the current passing through the capacitor. 1.89 sin(377t + 60") A 1.89 sin(377t +90) A 1.89 sin(377t 120") A 1.89 sin(377t-90) 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
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The impedance of the AC circuit shown below is Z=120 + 190 2. The phasor of the current i(t) is 120° A. i(t) 120 Ω www + e(t)= 200 sin(3141) V VR(t) Determine the supply voltage. Refer to the figure below. 140 Ω -(0000) VL(t) R = 40 Q www Va Determine the voltage V₁. ve(t) XL = 360 (0000) Ve 50 Ω Xc = 120 Refer to the figure below. e(t)= 200 sin(3141) V 0 60131401 | R = 400 ww X=360 Determine the current 7 in the circuit. 1.89 sin(377t + 60°) A 1.89 sin(377t + 90°) A 1.89 sin(377t 120°) A 1.89 sin(377t - 90°) A Ve X = 120 The voltage across a 106 Q capacitor is 200 sin(377t-30°) V. Determine the sinusoidal expression of the current passing through the capacitor.
Transcribed Image Text:The impedance of the AC circuit shown below is Z=120 + 190 2. The phasor of the current i(t) is 120° A. i(t) 120 Ω www + e(t)= 200 sin(3141) V VR(t) Determine the supply voltage. Refer to the figure below. 140 Ω -(0000) VL(t) R = 40 Q www Va Determine the voltage V₁. ve(t) XL = 360 (0000) Ve 50 Ω Xc = 120 Refer to the figure below. e(t)= 200 sin(3141) V 0 60131401 | R = 400 ww X=360 Determine the current 7 in the circuit. 1.89 sin(377t + 60°) A 1.89 sin(377t + 90°) A 1.89 sin(377t 120°) A 1.89 sin(377t - 90°) A Ve X = 120 The voltage across a 106 Q capacitor is 200 sin(377t-30°) V. Determine the sinusoidal expression of the current passing through the capacitor.
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