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A 10 mH inductor, a 1 uF capacitor, and a vari- able resistor are connected in parallel in the circuit shown. The resistor is adjusted so that the roots of the characteristic equation are -8000 + j6000 rad/s. The initial voltage on the capacitor is 10 V, and the initial current in the inductor is 80 mA. Find ic lo Answer: (a) 62.5 2: (b) -240,000 V/s: (c) B 10 V, B, = -80/3 V: (d) i(t) = 10e 8 cos 6000t + (82/3) sin 6000r] mA when t 2 0. a) R: b) dv(0")/dt; %3D c) B and Bz in the solution for v; and COS %3D

A 10 mH inductor, a 1 uF capacitor, and a vari- able resistor are connected in parallel in the circuit shown. The resistor is adjusted so that the roots of the characteristic equation are -8000 + j6000 rad/s. The initial voltage on the capacitor is 10 V, and the initial current in the inductor is 80 mA. Find ic lo Answer: (a) 62.5 2: (b) -240,000 V/s: (c) B 10 V, B, = -80/3 V: (d) i(t) = 10e 8 cos 6000t + (82/3) sin 6000r] mA when t 2 0. a) R: b) dv(0")/dt; %3D c) B and Bz in the solution for v; and COS %3D

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
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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A 10 mH inductor, a 1 uF capacitor, and a vari- able resistor are connected in parallel in the circuit shown. The resistor is adjusted so that the roots of the characteristic equation are -8000 + j6000 rad/s. The initial voltage on the capacitor is 10 V, and the initial current in the inductor is 80 mA. Find iR RE Vo Answer: (a) 62.5 0: a) R: (b) -240,000 V/s: (c) B= 10 V. By = -80/3 V: (d) i,(t) = 10e NOXorT8 cos 6000t + (82/3) sin 6000r] mA when t 2 0. b) dv(0")/dt: c) B and B, in the solution for v; and d) i(1). DELL AESICOFT CiT
Transcribed Image Text:A 10 mH inductor, a 1 uF capacitor, and a vari- able resistor are connected in parallel in the circuit shown. The resistor is adjusted so that the roots of the characteristic equation are -8000 + j6000 rad/s. The initial voltage on the capacitor is 10 V, and the initial current in the inductor is 80 mA. Find iR RE Vo Answer: (a) 62.5 0: a) R: (b) -240,000 V/s: (c) B= 10 V. By = -80/3 V: (d) i,(t) = 10e NOXorT8 cos 6000t + (82/3) sin 6000r] mA when t 2 0. b) dv(0")/dt: c) B and B, in the solution for v; and d) i(1). DELL AESICOFT CiT
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