Questions ??-??. Consider an ideal RLC series circuit shown in the figurebelow. The current in the circuit is given by i(t) = I cos[(9450 rad/s)t], whereI is the amplitude current. The voltage amplitude of the AC source is 121 V.In addition to this, R = 378 , C = 0.802 µF, and L = 67.2 mH.9 of 15wwwMRL Find the equation for the voltage across the capacitor with time.(A) vc (t) = (25.4 V) cos[(9450 rad/s)t - rad](B) vc(t) = (25.4 V) cos[(9450 rad/s)t + rad](C) vc (t) = (25.4 V) cos[(9450 rad/s)t](D) vc (t) = (121 V) cos[(9450 rad/s)t - rad](E) vc(t) = (121 V) cos[(9450 rad/s)t - 0.926 rad]

Answered: Image /qna-images/answer/dafe28c9-7d6c-4a88-a563-036f1d45ddb4.jpg

Questions ??-??. Consider an ideal RLC series circuit shown in the figure below. The current in the circuit is given by i(t) = I cos[(9450 rad/s)t], where I is the amplitude current. The voltage amplitude of the AC source is 121 V. In addition to this, R = 378 , C = 0.802 µF, and L = 67.2 mH. 9 of 15 M R L

Questions ??-??. Consider an ideal RLC series circuit shown in the figure below. The current in the circuit is given by i(t) = I cos[(9450 rad/s)t], where I is the amplitude current. The voltage amplitude of the AC source is 121 V. In addition to this, R = 378 , C = 0.802 µF, and L = 67.2 mH. 9 of 15 M R L

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 an RLC series circuit, the voltage amplitude and frequency of the source are 100 V and 500 Hz, respectively, an R = 500Ω, L = 0.20 H, and C = 2.0μF. (a) What is the impedance of the circuit? (b) What is the amplitude of the current from the source? (c) If the emf of the source is given by v(t) = (100 V) sin 1000πt , how does the current vary with time? (d) Repeat the calculations with C changed to 0.20μF.
In the circuit in the figure, Vcc= 18 V, Rs1 = 0.9 kohm, Rohm = 0.5 kQ, R1 = 75 kohm, R2 = 15 kohm, RC = 2.2 kohm, RE = 1 kohm and B = 120. It is also known that Cwi = 6 pF, Cwo = 10 pF, Cbc = Cu = 7 pF, Cbe = C = 44 pF and Cce = 11 PF. Accordingly, considering the frequency response of the circuit, what would the high cutoff frequency (fH) be? NOTE-1: The output impedance of the transistor is ro = 20 k and will be taken into account in the calculations. NOTE-2: high frequency dependence of hfe, B. will be taken into account.
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ı just need fınal aswer In the circuit in the figure, Vcc = 18 V, Rs1 = 0.9 kΩ, Rs2 = 0.5 kΩ, R1 = 75 kΩ, R2 = 15 kΩ, RC = 2.2 kΩ, RE = 1 kΩ and β =120. It is also known that Cwi = 6 pF, Cwo = 10 pF, Cbc = Cu = 7 pF, Cbe = Cπ = 58 pF and Cce = 11 pF. Accordingly, considering the frequency response of the circuit, what would be the high cutoff frequency (fH)? NOTE-1: The output impedance of the transistor is r0 = 20 kΩ and will be considered in the calculations. NOTE-2: The high frequency dependence of hfe, that is β, will be taken into account. a. 189,95 kHz b. 617,33 kHz c. 569,84 kHz d. 474,87 kHz e. 332,41 kHz f. 47,49 kHz g. 142,46 kHz h. 712,30 kHz
Now, how (as what equivalent value) does this inductor transform into the frequency domain?
In the circuit in the figure, Vcc = 18 V, Rs1 = 0.9 kΩ, Rs2 = 0.5 kΩ, R1 = 75 kΩ, R2 = 15 kΩ, RC = 2.2 kΩ, RE = 1 kΩ and β =120. It is also known that Cwi = 6 pF, Cwo = 10 pF, Cbc = Cu = 7 pF, Cbe = Cπ = 51 pF and Cce = 11 pF. Accordingly, considering the frequency response of the circuit, what would be the high cutoff frequency (fH)? NOTE-1: The output impedance of the transistor is r0 = 20 kΩ and will be taken into account in the calculations. NOTE-2: The high frequency dependence of hfe, β, will not be taken into account.
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In the circuit in the figure, Vcc = 18 V, Rs1 = 0.9 kΩ, Rs2 = 0.5 kΩ, R1 = 75 kΩ, R2 = 15 kΩ, RC = 2.2 kΩ, RE = 1 kΩ and β =120. It is also known that Cwi = 6 pF, Cwo = 10 pF, Cbc = Cu = 7 pF, Cbe = Cπ = 48 pF and Cce = 11 pF. Accordingly, considering the frequency response of the circuit, what would be the high cutoff frequency (fH)? NOTE-1: The output impedance of the transistor is r0 = 20 kΩ and will be taken into account in the calculations. NOTE-2: The high frequency dependence of hfe, β, will be taken into account. a. 719,95 kHz b. 143,99 kHz c. 48,00 kHz d. 191,99 kHz e. 575,96 kHz f. 479,97 kHz g. 335,98 kHz h. 623,96 kHz