In the circuit given above, the switch was opened at t=0 after being closed for a very long time. Find vL(t) for t>0 using s-domain techniques.
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In the circuit given above, the switch was opened at t=0 after being closed for a very long time. Find vL(t) for t>0 using s-domain techniques.
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- Inductive Circuits Fill in all the missing values. Refer to the following formulas: XL=2fLL=XL2ff=XL2L Inductance (H) Frequency (Hz) Inductive Reactance ( ) 1.2 60 0.085 213.628 1000 4712.389 0.65 600 3.6 678.584 25 411.459 0.5 60 0.85 6408.849 20 201.062 0.45 400 4.8 2412.743 1000 40.841The rms voltage across a 0.018 μF capacitor is 1.5 V at a frequency of 50 Hz. Part A Whar is the rms current through the capacitor? Express your answer using two significant figures. 15] ΑΣΦ Irms= 0.00000848 Submit Previous Answers Request Answer X Incorrect; Try Again; One attempt remaining ? μAFind T 0.5 ms 40 92 www www 892 160 92 20 mH www
- Circuits 1 HW 7 Q9D. Find h22. Express your answer in millisiemens to three significant figures. Enter your answer in rectangular form.in V1 R2 100k SINE(0 10 100) R1 100k D1 D V2 0 out D D2 V3 0 DC offset[V]: 0 Amplitude[V]: 10 Freq[Hz]: 100 T delay[s]: Theta[1/s]: Phi[deg]: Ncycles: anter 2019-2220 Figure 3-5 Clipper Circuit for LTSPICE Simulation C4. To specify the type of simulation, select "Simulate>>Edit Simulation Cmd" from the menu. Choose "Transient" and enter 30m for Stop time, 0 for Time to start saving data, and 1m for Maximum Timestep. Click OK. C5. Click Run to start the analysis. C6. Place two voltage probes (voltage probes appear when placed on a wire during simulation), one at the input side (at the top of V1) and another at the output side (at the top of D2) of the circuit. What difference do you observe between the input and the output waveforms?
- usig the purtiel fraction メcn for me thed forOscilloscope-XSC1 MAAA Channel B 643.751 mV 643.751 mV 0.000 V T1 T2 T2-T1 Time 379.198 ms 379.198 ms 0.000 s Timebase Scale: 500 us/Div X pos.(Div): 0 Y/T Add B/A A/B Discussion Channel A 1.086 V 1.086 V 0.000 V Channel A Scale: 1 V/Div Y pos.(Div): 0 AC 0 DC ● Channel B Scale: 1 V/Div Y pos.(Div): 0 AC 0 DC Reverse Save Trigger Ext. trigger Edge: fz B Ext Level: 0 V Single Normal Auto None ZA X Fig(6): The input and output waveforms for differentiator Operational Amplifier 1- If Op Amp is operating at 1000 Hz as input wave, what is the frequency of the output signal? 2- Why do we use Spike function? List the possible application of spike waveform.Question 3 Not yet answered The maximum value of 50 Hz an AC voltage is 100 V. Its value after 1/300 second will be O a. 66.8 V b. 68.6 V c. 88.6 V d. 86.2 V
- Te netuork show in ope eating in the tondy Thandy state wilh sinussidal vol voltage douce and V, 2 Sin2t delernine the volla fe Va(t) 2178 F Va 2178F 2173F IHThe sinusoidal voltage source in the circuit seen in (Figure 1) is operating at a frequency of 46 krad/s. The coefficient of coupling is adjusted until the peak amplitude of 1 is maximum. Figure 2002 100 Q 5 mH3 7502 300 £2 ww 20 mH 1 of 1 40 nF Part A What is the value of k? Express your answer using three significant figures. k = .085 Submit Part B VAΣ vec X Incorrect; Try Again; 28 attempts remaining i₁ = Submit Previous Answers Request Answer What is the peak amplitude of 1 if Ug = 399 cos(4.6 x 10¹) V? Express your answer to three significant figures and include the appropriate units. Provide Feedback μA Value Request Answer Units ? ?Activity 1: You are testing an electrical device which can be modelled by the figure shown below: 18 -j12 a b j6 -j10 it E1 12 13 E2 13 The above electric circuit can be characterized by the following equations: 1811 + j6(11 – 12) + 4(11 – 12) = 1020° -j1212 + 8(12 – I3) + 20490° + 4(I2 – 4) + j6(I2 – I1) = 0 -j1013 + 1313 – 20290° + 8(13 – 12) = 0 Find I1, I2, and I3 using the Gaussian elimination method.