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- a. The following 40Ω capacitive reactance and 60Ω inductive reactance are in series with a 30Ω resistor in a circuit. The applied voltage is 80mV. Find the total impedance and the power factor of the circuit. Draw the circuit and find the real power that is going through the circuit. c. A signal generator generated a signal according to the function below. Calculate the strength of the signal: g(t) = cos2t – sint 0 ≤ t ≤ πUse the T-equivalent circuit for the magnetically coupled coils shown in Fig below to find the phasor currents 4 and 12. The source frequency is 100 rad/s. 100 D a 400 D 500 Ω 0.5H 2H 800 Ω 5 H v2 i2 8H 90LOV SµF VIThe equation of v2 in the figure shown is * F 3sin1667nt volts O 1.5sin5236t volts None of the given option 3sin(5000π/3)t volts O 1.5sin1667t volts P1 V2 H₂ = 200μs/div and V₂ = 0.5V/div
- 9 62 802 5 mH 一 80V 156 1oov o=t エ has been n positien it moves to position The Switeh n the civauit in for a log time. At +=0 b. Find reltl for t70.1. For the circuit shown in the figure, assume the inputs are Add/Subt = 1, A = 1011, and B = 1001. What is the output. %3D A3 B3 A2 B2 A, BỊ Ao Bo Add/Subt. O А В С A B Cn А В С A B Ci Cout Σ |Cout Cout |Cout Σ Σ Σ ΣCalculate the difference A B if A=10236.87 and B=10453.13°. Show your solution graphically.
- A student set up the circuit shown for her electronics class. Assuming Ɛ = 9.00 V and R = 4.60 N, find the following quantities. 12.0 V 4.00 2 2.00 Q R (a) the current in the 2.000 resistor (Enter the magnitude in mA.) 590 Redraw the circuit, labeling currents, and apply Kirchhoff's rules to determine equations in terms of the given values. Solve your equations for the current in the 2.00-0 resistor. mA (b) the potential difference (in V) between points a and b Vfor the circuit shown below If v_in=5 sin2m10t then corner frequency will be ......Hz 1 UF 2 k Ohm 1 k Ohim 050 060 070 080 vinMinistry of Manpower Directorate General of Technological Education Salalah College of Technology Zlectrical Inginearina Problem - 8 Refer to the circuit below and -25V C2 N. compute the following: i) Total capacitance ii) Voltage across C, iii)Charge across C, iv)Voltage across C5 5 uF C1 5 uF 5 uF C4 5 uF C6 5 uF -16-65V. C5 5 uF (Take V, as 25V) 8-35V VT 25V
- PFOY Circuitin the Fig-1 Find lett) for tyo then Find ie(t=2T) (T=tintecoustaut 602 60 20 V 60 2Ź J Velő)=0 4mf Fight.2. Given: v(t) = 325.269 sin( 376.991t + 36.87 ) V Find : e. V when t = 5us f. V when o = 30° g. The first angle after = 0 when V = -100V h. The first time after t 0 when V = -100V Answer: e. V = 195.652 V f. V= 299.126 V g. 0 = 161.035 h. t= 7.455 msecj32 -j6 2 4Ω ww 62 Vo I V /0° V a三 E100 j8 Q I2 E j4 2 For the circuit shown, if V-276, the value of V. in V is O A. -8.998+j3.486 O B. 17.024-j27.294 O C.-53.986+j20.921 O D. 68.095-j109.174 O E. None