Q5: The circuit shown in the figure below was at steady state before the switch opens att = 0 sec. find the following:a) The value of the capacitance C to make the inductor current , (t) follows the criticaldamping case.b) After finding the capacitance C from part a, obtain the inductor current for 0 ≤ t ≤ 00c) Sketch the inductor current for 0 ≤ t ≤ co20 V(+IL (t) 0.5H10ΩsxSt=0C

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Answered: Q5: The circuit shown in the figure…

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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-In the circuit shown on the right the switch S has been open for a long time. What is the current flow through the inductor (0, 2 to 1, or 1 to 2) ? -In the circuit shown on the right the switch S has been open for a long time. At the instant the switch S is closed, the current flow through the inductor is (0, 1 to 2, or 2 to 1)? -In the circuit shown on the right the switch S has been open for a long time. The current flow through the inductor is (increasing, steady, or decreasing)? -In the circuit shown on the right the switch S has been open for a long time. Immediately after the switch S is closed, the current flow through the inductor is (increasing, steady, or deacreasing)? -In the circuit shown on the right the switch S has been closed for a long time. The current flow through the inductor is (1 to 2, 2 to 1, or 0)? -In the circuit shown on the right the switch S has been closed for a long time. At the instant the switch S is opened, the current flow through the inductor is (1…
About the circuit below: a) How long does it take for the current in the inductor to reach 40% of its maximum value? b) After 1 time constant in the closed position, the switch is opened at t = 0. Find the current and voltage in the inductor when t>0. c) Sketch the behavior (waveforms) of voltage and current in the inductor and resistor R2 in the storage and discharge (decay) phase of the inductor. R1 = R5 = 55k Ω R2 = R4 = 20k Ω R3 = R6 = 25k Ω V = 2500 v L1 = 50 mH
11. The problem is all about Capacitors and Inductors. Please provide FREE BODY DIAGRAM in the Solution.
USING FOL Differential Equation APPLICATION SOLVE THE FOLLOWING PROBELM AND SHOW COMPLETE SOLUTION AND FINAL ANSWER.A coil has a time constant of 1 second and an inductance of 8 H. If the coil is connected toa 100 V dc source, determine the rate of rise of current at the instant of switching.
The switch in the figure below is open for t < 0 and is then thrown closed at time t = 0. R 2R ww www www (a) Find the current in the inductor. (Use the following as necessary: L, R, t, and for E.) 12- (b) Find the current in the switch as functions of time thereafter. (Use the following as necessary: L, R, t, and & for 8.) switch
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All capacitors were initially discharged.at t = 0, S1 is placed at position 1 and S2 is closed.At t = 15 ms, S1 is placed at position 2 and S2 is kept closed. At t = 25 ms, S1 is kept at position 2 and S2 is opened.after 25 ms, the voltage vT will Question options: increase until it reaches the battery voltage at steady state exponentially decay to zero stay constant be a sine wave
5. An RC (DC) circuit consists of a Resistor with a resistance of 25000 SQ and Capacitor in series with a source potential of 30.0 V. The capacitor is initially uncharged and the circuit is completed at t= 0. a) If it is to take 45.0 seconds for the capacitor to be 88.0% fully charged, what must the be capacitance? b) At what time is the potential difference across the capacitor 17.0 Volts?
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In the circuit shown, suppose that the switch has been closed sufficiently long for the capacitor to become fully charged. a) Find the steady-state current through each resistor. b) Find the charge on the capacitor. c) The switch is now opened at t= 0. Write an equation for the current iR2 through R2 as a function of time. d) Find the time that it takes for the charge on the capacitor to fall to 1/5 of its initial value.

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