You have the following circuit in sinusoidal steady-state. Use phasor circuit analysis to find the time...
2- A circuit across the terminals of a sinusoidal voltage source, as shown in Figure 2. The steady-state expression for the source voltage is v;=50.cos(1000t+20). (40 points) 12 mH 100 MF 10 Figure 2 a) Construct the frequency-domain equivalent circuit. b) Calculate the steady-state current i by the phasor method.
8–31 A voltage vs(t) = 50 cos (5000t) V is applied to the circuit in Figure P8–31. (a) Convert the circuit into the phasor domain. (b) Find the phasor current flowing through the circuit and the phasor voltages across the inductor and the resistor. (c) Plot all three phasors from (b) on a phasor diagram. Describe if the current leads or lags the inductor voltage. i(t) 50 22 25 mH 00 + VL(t) - + Vr(t)- vs(t) (+) FIGURE P8-31
(a) Calculate the sinusoidal steady state expression of the current io(t) using nodal analysis as shown in Figure 1(a). [10 Marks) C R L -0000 10mH 200 10(1) 50 4F + vs(t) + 10 cos 1000+ V R22022 410(1) R3=300 Figure 1(a) (b) Calculate the sinusoidal steady state expression of the current is(t) using superposition principle as shown in Figure 1(b). [15 Marks) с ist) v.1) 1012 R: 1012 0.1 F 5 cos 2t VR L 0000 4H 2 sin...
(a) Calculate the sinusoidal steady state expression of the current io(t) using nodal analysis as shown in Figure 1(a). [10 Marks) C R L -0000 10mH 200 10(1) 50 4F + vs(t) + 10 cos 1000+ V R22022 410(1) R3=300 Figure 1(a) (b) Calculate the sinusoidal steady state expression of the current is(t) using superposition principle as shown in Figure 1(b). [15 Marks) с ist) v.1) 1012 R: 1012 0.1 F 5 cos 2t VR L 0000 4H 2 sin...
3. For the circuit shown in Figure 1, + VR - R = 3k 494 ET V 4.7 uF E = 8V sin(20,000t +60° -VL + Z = ? m L = 0.5 H Figure 1 i. ii. iv. v. vi. vii. Find the total impedance Zr in polar form. Draw the impedance diagram Find the current and the voltages VR, Vi and Vc in phasor form. Draw the phasor diagram of the voltages E, VR, V., and Vc and...
2502 In the adjoining circuit schematic, in steady-state, the current flowing through the loop causes a voltage drop across the resistor, having the waveform vr(t) = 15 cos (75 t) and a voltage drop across the capacitor given by ve(t) = 20 cos (75 t +90°) (a) Express the above two voltages in phasor form. (b) Find the source voltage shown in the circuit schematic, expressed in phasor form. (c) Express the source voltage v(t) as a function of time....
In the adjoining circuit schematic, in steady-state, the current flowing through the loop causes a voltage drop across the resistor, having the waveform vR(t) = 15 cos (75 t) and a voltage drop across the capacitor given by vC(t) = 20 cos (75 t + 90⁰) (a) Express the above two voltages in phasor form. (b) Find the source voltage shown in the circuit schematic, expressed in phasor form. (c) Express the source voltage v(t) as a function of time....
Use the node voltage method to find the steady-state expression for io in the circuit seen in (Figure 1) if ig 4 cos 2500t A and v, 16 cos(2500t + 90° ) V Write the steady-state expression for io(t) as to = L cos(wt + φ), where-180° <φ < 180° Figure く 1of1 100 μF 50 uF 12Ω View "31.6 mH 30
120 Problem 1, Use the node-voltage method to find the steady state expression for v () in the circuit shown. The sinusoidal sources are v,-35cos 50 t V'and i 20 sin 50 1 A 20 Ω 0 Problem 2 120) Use the mesh-current method to find the steady state expression for velt) in the circuit shown. Answer must be in time domain. Below excitation voltage v is given in time domain v(t) 0.75 V,<t 2 Ω ) 5osin(40140°) Problem 3...
Review Part A. Find the relationship between the phasor voltage and phasor current for a resistance The resistor shown here has been transformed into the phasor domain: Ik 4k Suppose that the phasor current is given by IR = 752120 mA. Find the phasor voltage VR. Enter a complex number in polar form, with phase angle in degrees. View Available Hint(s) 3002120 V Submit Previous Answers Correct Part B - Draw the phasor diagram of the resistance from Part A...