Consider the RLC series circuit shown in the figure below.The current in the circuit is given by i(t) = I cos[(9580 rad/s)t], where I is theamplitude current. The voltage amplitude of the AC source is 112 V.R = 335 NC = 0.625 µFL = 75.8 mHi. Find the equation for the voltage across the inductor with time.A. vL(t) = (125 V) cos[(9580 rad/s)t + a rad]B. vµ(t) = (112 V) cos[(9580 rad/s)t|C. vĽ(t) = (125 V) cos[(9580 rad/s)t + «/2 rad]%3D%3DD. vµ(t) = (112 V) cos[(9580 rad/s)t + 1.03 rad]E. v1(t) = (125 V) cos[(9580 rad/s)t – a/2 rad]ii. Find the equation for the voltage across the capacitor with time.A. vo(t) = (28.7 V) cos[(9580 rad/s)t – 7/2 rad]B. vc(t) = (28.7 V) cos[(9580 rad/s)t + x/2 rad]C. vo(t) = (28.7 V) cos[(9580 rad/s)t + ™ rad]D. vďt) = (112 V) cos[(9580 rad/s)t + 2.11 rad]E. vo(t) = (215 V) cos[(9580 rad/s)t – a/2 rad]%3D

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Consider the RLC series circuit shown in the figure below. The current in the circuit is given by i(t) = I cos[(9580 rad/s){], where I is amplitude current. The voltage amplitude of the AC source is 112 V. R = 335 N C = 0.625 µF L = 75.8 mH i. Find the equation for the voltage across the inductor with time. A. vL(t) = (125 V) cos[(9580 rad/s)t + a rad] B. vL(t) = (112 V) cos[(9580 rad/s)t| %3D C. vL(t) = (125 V) cos[(9580 rad/s)t + 1/2 rad] D. vL(t) = (112 V) cos[(9580 rad/s) t + 1.03 rad] E. v1(t) = (125 V) cos[(9580 rad/s)t – 1/2 rad] %3D %3D ii. Find the equation for the voltage across the capacitor with time. A. vo(t) = (28.7 V) cos[(9580 rad/s)t – 1/2 rad] B. vo(t) = (28.7 V) cos[(9580 rad/s)t + 1/2 rad] %3D %3D C. vo(t) = (28.7 V) cos[(9580 rad/s) t + a rad] D. vďt) = (112 V) cos[(9580 rad/s)t + 2.11 rad] E. vo(t) = (215 V) cos[(9580 rad/s)t – a/2 rad] %3D %3D %3D

Consider the RLC series circuit shown in the figure below. The current in the circuit is given by i(t) = I cos[(9580 rad/s){], where I is amplitude current. The voltage amplitude of the AC source is 112 V. R = 335 N C = 0.625 µF L = 75.8 mH i. Find the equation for the voltage across the inductor with time. A. vL(t) = (125 V) cos[(9580 rad/s)t + a rad] B. vL(t) = (112 V) cos[(9580 rad/s)t| %3D C. vL(t) = (125 V) cos[(9580 rad/s)t + 1/2 rad] D. vL(t) = (112 V) cos[(9580 rad/s) t + 1.03 rad] E. v1(t) = (125 V) cos[(9580 rad/s)t – 1/2 rad] %3D %3D ii. Find the equation for the voltage across the capacitor with time. A. vo(t) = (28.7 V) cos[(9580 rad/s)t – 1/2 rad] B. vo(t) = (28.7 V) cos[(9580 rad/s)t + 1/2 rad] %3D %3D C. vo(t) = (28.7 V) cos[(9580 rad/s) t + a rad] D. vďt) = (112 V) cos[(9580 rad/s)t + 2.11 rad] E. vo(t) = (215 V) cos[(9580 rad/s)t – a/2 rad] %3D %3D %3D

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter33: Alternating Current Circuits

Section: Chapter Questions

Problem 33.1OQ: An inductor and a resistor are connected in series across an AC source as in Figure OQ33.1....

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