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the earliest time (after t = 0) when the current is 0.707 A. (c) Using identities, trigonometric that v(t) = VR(t) + vc(t) = show 15 sin 1000 t – 5 cos 1000 t. (d) Write v1(t) obtained in part (c) in the form v (t) = M cos(1000 t + 0); in other words, find M and 0. - i(t) = 0.707 sin(1000 t + 45°) A VR { R= 10 N v(t) VL L = 10 mH %D =10 sin(1000 t) V - vc C= 50 µF Figure P6.22 RLC circuit for problem P6-22.

the earliest time (after t = 0) when the current is 0.707 A. (c) Using identities, trigonometric that v(t) = VR(t) + vc(t) = show 15 sin 1000 t – 5 cos 1000 t. (d) Write v1(t) obtained in part (c) in the form v (t) = M cos(1000 t + 0); in other words, find M and 0. - i(t) = 0.707 sin(1000 t + 45°) A VR { R= 10 N v(t) VL L = 10 mH %D =10 sin(1000 t) V - vc C= 50 µF Figure P6.22 RLC circuit for problem P6-22.

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
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the earliest time (after t = 0) when the current is 0.707 A. trigonometric that vi(t) = vR(t) + vc(t) = (c) Using identities, show %3D 15 sin 1000 t - 5 cos 1000 t. (d) Write v,(t) obtained in part (c) in the form v1 (t) = M cos(1000t + 0); in other words, find M and 0. i(t) = 0.707 sin(1000 t + 45°) A 3R = 10 2 VR | v(t) VL L = 10 mH =10 sin(1000 t) V VC C= 50 µF Figure P6.22 RLC circuit for problem P6-22. + I + +
Transcribed Image Text:the earliest time (after t = 0) when the current is 0.707 A. trigonometric that vi(t) = vR(t) + vc(t) = (c) Using identities, show %3D 15 sin 1000 t - 5 cos 1000 t. (d) Write v,(t) obtained in part (c) in the form v1 (t) = M cos(1000t + 0); in other words, find M and 0. i(t) = 0.707 sin(1000 t + 45°) A 3R = 10 2 VR | v(t) VL L = 10 mH =10 sin(1000 t) V VC C= 50 µF Figure P6.22 RLC circuit for problem P6-22. + I + +
6-22. A sinusoidal voltage v(t) = 10 sin(1000 t) V is applied to the RLC circuit shown in Fig. P6.22. The current i(t) = 0.707sin(1000 t + 45°) flowing through the circuit produces voltages across R, %3D L, and C of VR(t) = 7.07 sin(1000 t + 45°) V VL (t) = 7.07 sin(1000 t + 135°) V vc(t) = 14.14 sin(1000 t – 45°) V. (a) Write down the amplitude, frequency (in hertz), period (in seconds), phase ba shift (in radians), and time shift (in milliseconds) of the current i(t) = 0.707 sin(1000t +45°) A. (b) Plot one cycle of the current i(t) = 0.707 sin(1000t + 45°) A and indicate
Transcribed Image Text:6-22. A sinusoidal voltage v(t) = 10 sin(1000 t) V is applied to the RLC circuit shown in Fig. P6.22. The current i(t) = 0.707sin(1000 t + 45°) flowing through the circuit produces voltages across R, %3D L, and C of VR(t) = 7.07 sin(1000 t + 45°) V VL (t) = 7.07 sin(1000 t + 135°) V vc(t) = 14.14 sin(1000 t – 45°) V. (a) Write down the amplitude, frequency (in hertz), period (in seconds), phase ba shift (in radians), and time shift (in milliseconds) of the current i(t) = 0.707 sin(1000t +45°) A. (b) Plot one cycle of the current i(t) = 0.707 sin(1000t + 45°) A and indicate
Expert Solution
Step 1

v(t) = 10 sin (1000 t) V

i(t) = 0.707 sin (1000 t + 45o) A

R = 10 Ω

L = 10 mH

C = 50 μF

vR(t) = 7.07 sin (1000 t + 45o) V

vL(t) = 7.07 sin (1000 t + 135o) V

vC(t) = 14.14 sin (1000 t -45o) V

 

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