Submitted Answers ANSWER 1: Deduction:-14.2857142857143% 300e 800t cos (600t) - 384e 800f sin(600t) mA 300e 800t cos (600t) 192e 800t sin(600t) mA 300e 800t cos (600t) - 192e 800f sin(600t) mA 600e 800t cos (600t) 192e 800f sin(600t) mA 600e 800t cos(600t) 384e 800f sin(600t) mA 600e 800t cos(600t) - 192e800f sin(600t) mA 600e 800t cos(600t)- 384e 800f sin(600t) mA 300e 800t cos (600t) 384e-800t sin(600t) mA ANSWER 2: Deduction:-14.2857142857143% 300e 800t cos (600t) - 384e 800f sin(600t) mA 300e 800t cos (600t) 192e 800t sin(600t) mA 300e 800t cos (600t) - 192e 800f sin(600t) mA 600e 800t cos (600t) 192e 800f sin(600t) mA 600e 800t cos(600t) 384e 800f sin(600t) mA 600e 800t cos(600t) - 192e800f sin(600t) mA 600e 800t cos(600t)- 384e 800f sin(600t) mA 300e 800t cos (600t) 384e-800t sin(600t) mA
The circuit elements in the circuit in (Figure 1) are R-125 Ω, C-5 uF, and L 200 mH. The initial inductor current is-0.3 A, and the initial capacitor voltage is 25 V Correct Part C Find iz (t) for20 O-300e 800t cos(600t) 384e800t sin (600t) mA O-300e 800t cos(600t) +192e 800t sin (600t) mA O-300e 800t cos(600t) -192e800t sin(600t)mA -600e 800t cos(600t) +192e 800t sin(600t) mA O -600e800t cos(600t)+384e 800t sin(600t) mA O -600e 800 cos(600t) 192e800sin(600t) mA O-600e 800t cos(600t) 384e800t sin (600t) mA O-300e 800f cos(600t)+384e 800n(60) mA Figure 1 of 1 CV Submit 0 XIncorrect: Try Again;4 attempts remaining Provide Feedback
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ANSWER 3: Deduction:-14.285714285714390 -300e 800t cos (600t) 384e-800f sin(600t) mA cos (600t) 192e-800f sin(600t) mA -300e 800t cos (600t) - 192e-800f sin(600t) mA -600e 800t cos (600t) 192e-800f s...
pls answer 7&8 10 ks2 10 kg vo Problem 7) The switch in the circuit has...
pls answer 7&8
10 ks2 10 kg vo Problem 7) The switch in the circuit has been open for a long time. Find the time constant and the voltage on the capacitor as t oo and sketch the voltage vc(t) on the capacitor for t > 0. 10v 1 mA 20 k223 40 uF - Velt t = 012 2012 We Problem 8) The switch in the circuit has been closed for a long time before it opens at t...
All missing parts please. (7%) Problem 6: A 16-Ω resistor, 45-uF capacitor, and 3.5-mH inductor are...
All missing parts please.
(7%) Problem 6: A 16-Ω resistor, 45-uF capacitor, and 3.5-mH inductor are connected in series with an AC source of amplitude 13 V and frequency 135 Hz. 8% Part (a) What is the impedance of the circuit, in ohms? 8% Part (b) What is the amplitude of the current in the circuit, in amperes? 8% Part (c) What is the phase constant of the current, in degrees? 8% Part (d) With a source voltage of Vsourc,-Vocos(2π...
The circuit in the figure to the right is represented by the equation below, where I...
The circuit in the figure to the right is represented by the equation below, where I is the current through the inductor Land Ve is the voltage drop across the capacitor C. Find formulas for I, and Vc when R=0.5 ohms, C = 12.5 farads, L= 2.5 henry, the initial current is o amperes, and the initial voltage is 12 volts. R Het с 1 RC Vc 000 L Choose the correct answer below. 4.(t) Vc(t) 0.100 OA 30 sin(-0.08t)...
8.23 why ic(0+)=45mA not 50mA? 8.21 Assume the underdamped voltage response of the ocircuit in Fig....
8.23 why ic(0+)=45mA not 50mA?
8.21 Assume the underdamped voltage response of the ocircuit in Fig. 8.1 is written as mit wv(t)= (A1 +A2)e cos wt +j(A - A2)e a sin wt en in moves -at -at 0. The initial value of the inductor current is Io, and the initial value of the capacitor voltage is Vo- Show that A2 is the conjugate of A1. (Hint: Use the same process as and A2.) outlined in the chapter to find A...
Circuit Analysis in the s-Domain 15.3. The initial voltage across the capacitor in the circuit shown in Figure P15.3 is v(0) 1 V, and the initial current through the inductor is i(0)0 mA Find th...
Circuit Analysis in the s-Domain 15.3. The initial voltage across the capacitor in the circuit shown in Figure P15.3 is v(0) 1 V, and the initial current through the inductor is i(0)0 mA Find the voltage vo (t) across the capacitor for t 2 0 Figure P15.3 50 mH 1 kS2 V. Volt) T 0.1 μF The circuit in the s-domain is shown below. R2 Va 1k 0.05s 1/(sC)-1e7/s Vo R1 2k V (0-ys 5/s 1/s 1 format long; 2...
Engineering circuit analysis by Hayt 8th edition question 27 and figure 9.43 I think 10u(1-t) means 10 (for t<1)...
Engineering circuit analysis by Hayt
8th edition
question 27 and figure 9.43
I think 10u(1-t) means 10 (for t<1) and 0
(for t>1)
then, I can't remove this current source because it
continuously make 10micro A
(at t=500ms, t=1.002ms)
I don't know what's wrong now..
366 26. For the circuit of Fig. 9,43, 1 30-) mA. (a) Select R, so th O)6 V (b) Compute e2 ms). (c) Determine the settling, time of t capacitor voltage. (d) Is the inductor...
PLEASE ANSWER ALL QUESTIONS Problem 3: Equivalent Impedance, KCL, Curret Division. Consider the circuit in Figure...
PLEASE ANSWER ALL QUESTIONS
Problem 3: Equivalent Impedance, KCL, Curret Division. Consider the circuit in Figure 2Figure 1 i(t) aC v(t) cos(2π100x1 03t) 2 iL (t) R2 = 4 kQ, L = 3 mH, and C = 1 nF. Figure 2. Circuit for Equivalent Impedance Question 1 Question 6 Question 11 Real part of Zab [Ohms Phase of vit) [deg it) (leads or lags) iLit) by positive 0 21 O 22.5 O 25.2 O 112 degrees Question7 Question 2 Amplitude...
pls answer 7&8
10 ks2 10 kg vo Problem 7) The switch in the circuit has been open for a long time. Find the time constant and the voltage on the capacitor as t oo and sketch the voltage vc(t) on the capacitor for t > 0. 10v 1 mA 20 k223 40 uF - Velt t = 012 2012 We Problem 8) The switch in the circuit has been closed for a long time before it opens at t...
All missing parts please.
(7%) Problem 6: A 16-Ω resistor, 45-uF capacitor, and 3.5-mH inductor are connected in series with an AC source of amplitude 13 V and frequency 135 Hz. 8% Part (a) What is the impedance of the circuit, in ohms? 8% Part (b) What is the amplitude of the current in the circuit, in amperes? 8% Part (c) What is the phase constant of the current, in degrees? 8% Part (d) With a source voltage of Vsourc,-Vocos(2π...
The circuit in the figure to the right is represented by the equation below, where I is the current through the inductor Land Ve is the voltage drop across the capacitor C. Find formulas for I, and Vc when R=0.5 ohms, C = 12.5 farads, L= 2.5 henry, the initial current is o amperes, and the initial voltage is 12 volts. R Het с 1 RC Vc 000 L Choose the correct answer below. 4.(t) Vc(t) 0.100 OA 30 sin(-0.08t)...
8.23 why ic(0+)=45mA not 50mA?
8.21 Assume the underdamped voltage response of the ocircuit in Fig. 8.1 is written as mit wv(t)= (A1 +A2)e cos wt +j(A - A2)e a sin wt en in moves -at -at 0. The initial value of the inductor current is Io, and the initial value of the capacitor voltage is Vo- Show that A2 is the conjugate of A1. (Hint: Use the same process as and A2.) outlined in the chapter to find A...
Circuit Analysis in the s-Domain 15.3. The initial voltage across the capacitor in the circuit shown in Figure P15.3 is v(0) 1 V, and the initial current through the inductor is i(0)0 mA Find the voltage vo (t) across the capacitor for t 2 0 Figure P15.3 50 mH 1 kS2 V. Volt) T 0.1 μF The circuit in the s-domain is shown below. R2 Va 1k 0.05s 1/(sC)-1e7/s Vo R1 2k V (0-ys 5/s 1/s 1 format long; 2...
Engineering circuit analysis by Hayt
8th edition
question 27 and figure 9.43
I think 10u(1-t) means 10 (for t<1) and 0
(for t>1)
then, I can't remove this current source because it
continuously make 10micro A
(at t=500ms, t=1.002ms)
I don't know what's wrong now..
366 26. For the circuit of Fig. 9,43, 1 30-) mA. (a) Select R, so th O)6 V (b) Compute e2 ms). (c) Determine the settling, time of t capacitor voltage. (d) Is the inductor...
PLEASE ANSWER ALL QUESTIONS
Problem 3: Equivalent Impedance, KCL, Curret Division. Consider the circuit in Figure 2Figure 1 i(t) aC v(t) cos(2π100x1 03t) 2 iL (t) R2 = 4 kQ, L = 3 mH, and C = 1 nF. Figure 2. Circuit for Equivalent Impedance Question 1 Question 6 Question 11 Real part of Zab [Ohms Phase of vit) [deg it) (leads or lags) iLit) by positive 0 21 O 22.5 O 25.2 O 112 degrees Question7 Question 2 Amplitude...
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