Q4 Find Zt, Yt, Is, 11, 12 and 13 for the following circuit: 10 R, 160 E 120 V 201 40 R₁ 50 XC70 15 2
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A: Step 1:Step 2:Step 3: Step 4: Final plot:Computer generated root locus for verification:
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- An AC circuit contains a 24 resistor, a 15.9-mH inductor, and a 13.3F capacitor connected in parallel. The circuit is connected to a 240-V, 400-Hz power supply. Find the following values. XL=XC=IR=AIL=AIC=AP=WVARsL=VARsC=IT=AVA=PF=%=For the following circuit, Vsrc=1.5 V and Rload%3D225 Q. What is Is? Is +. Vsrc VR Rload 30 mA 9 mA 6.67 mA 40 mVFor the circuit shown in figure Req is: * 140 R3 R, R: 22 Rea Ru RC RS R7 33 6 ohm 24.4 ohm 34,4 ohm 44.4 ohm
- Can you also answer this. Please simulate the given figure at multisim.com Will get a like.The figure below shows a simple RC circuit with a 3.30-μF capacitor, a 4.40-M resistor, a 9.00-V emf, and a switch. What are the following exactly 9.00 s after the switch is closed? (a) the charge on the capacitor 13.7204 UC (b) the current in the resistor 0.0110522 x Your response is off by a multiple of ten. HA (c) the rate at which the capacitor is storing energy μW (d) the rate at which the battery is delivering energy μWThe voltage applied to the 622mH inductor is given by v(t)= 15e5t volt and IL (0)=0. Calculate the current. Select one: O a. 14.15[1-es O b. 4.82[1-e5t] O c. 8.57[1-e5t] O d. 20[1-e5t]
- V, = @sin(@t) + 2@ cos(@t). For this 6. A LRC circuit below is used to modify an input voltage applicatio, = 2 ydurāifidrentidi equation gnlist be in terms of %3D charge 9 notes to generate the general solution. No other method will be accepted. Use two decimal point accuracy. You . You must use the complimentary and particular solution method as shown in your are given that -2N@² + 20N = 2@ and -2M@² +20M = 2@¸ which you must use to simplify your particular solution when you equate coefficients. L HH Va VL Vc keo = Im sin(ot) VsInitially relaxed series RLC circuit with R = 150 £2, L=30 H and C= 10 F has de voltage of 150 V applied at time t = 0. Find the equation for current and voltages across different elements. Also draw the waveform under different cases. E R i L -00000 C-9158388cmid%3D3385218&page%3D15 For the circuit shown in the figure, the Norton current (IN) at terminals a-b is: 32 16V 60 2 2V O a. 12 A b. 5.3 A Oc. 5.67 A Od. 4 A. e. -8 A ge
- /Q6 The potential at point P due to Q1 and * :Q2 is = 1x10-12C 0.5 m 0.5 m Q2 = 1x10-12C %3D 50 cm P. 0.01797 V O -0.01271 V O ov O 0.01271 V C 10Fundamentals of Electrical Engineering 2020/2021 Dr. Yaseen H. Tahir Example: Find the equivalent conductance GAB for the circuit in Figure below and then find RAB- 2 mS ww 1mS A GAB - 36mS 3 mS 4mS: Solution: to CGAD= 8.493 MS R-1177.43o2 AB %3DA charging RC circuit has a switched-control DC voltage source of 6V and has a series RC of 20 ohms and 10 microfarads. Determine the following: a. T: Time Constant b. vR at t c. VC at t d. iC at t