V1= - 10 sen(wt + 30°)V y V2 = 20 cos(wt - 45º)V Find= v1+v2
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V1= - 10 sen(wt + 30°)V y
V2 = 20 cos(wt - 45º)V
Find= v1+v2
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- 4. With respect to the following figure, explain why v。 does not follow v₁ until Và – Vỵ and follows only when v₁ ≥ VB - Vy in the positive half cycle? What happens in the negative half-cycle? Explain. Assume a sinusoidal current of losin(ot) from v₁, what is the voltage drop across R when vo follows vi and Vo does not follow v₁. www + R D VI vo VB-Vy VI VoThe circuit in figure 2 is operating in the sinusoidal steady state, I1=0.520 A and V1-50290° V. Use superposition to find phasor response I, L1 R2 j25 500 Ix R1 250 11 C1 V1 -j50 Figure 2In the figure: R1 = 1439.170, R2 = R1 and R3 =2R2, then determine the value of the voltage source E. a 8 mA 10 mA 2 mA RT R1 R2 R3 O 25.33 11.51 O 17.27 15.43 13.82 +
- Q1. For the system shown below T(s)= s5+10s4+45s3+90s2+164s+200 = 0 a. Complete the table Number of poles in RHS Number of poles Number of poles in in LHS jw b. Do you have special case in this question or not? c. Decide the stabilityIn the circuit shown in Figure, V1 = 5 120° Find Z. 20 V1 0.25 0 j1 N -j0.25 N= (1) /0°AVs= 40 V Z1 = 1 0hm Z2 = j2 Ohm Z3 = 2 Ohm Z4 = -j4 Ohm Z5 = 1 Ohm Z6 =j2 Ohm Z7 = 2 Ohm Z8 = -j4 Ohm Vs + 1 Find Vo. * Z1 O (2.41 +j 0.49) V O (1.41-j1.51) V O (36.12-j18.83) V O (1.31 -j41.66) V O (35.78-j17.68) V Z2 Z3 Z4 Z5 www Z6 m Z7 Z8 o+ Vo 10
- Please answer these question complete solution and readable (I'll rate it if it is complete solution) Two sinusoidal currents are given as i1= 10√2 sin wt, i2 = 20√2 sin (wt +60). Find the expression for the sum of these current.Problem Find ₁ and V₁. 12/0° V(+ 202 -J202 + 1:2 Ideal + V₂ 202 -j2 52 {202 Answer I₁ = 3.07/39.81° A. Vo 0 = 3.07439.81 ° V3&page%3D1 American Uni. Girne American Uni.. O portal.gaueng.org My Profile - Zoom Dashboard My Courses This course Find the Norton equivalent of the circuit 60 30 40V ,50 202 A BI E E E E ere to search
- 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?Application: Construct the circuit in figure. R= 2,2KN, L = 1 mH, and C= 10 nF. L Necesssary Formulae: 1 Xc 2.π.f.C X.-2. π.f. L By using the formulea perform the calculations of Xc and XL then fill the table. f (Hz) Xc XL 50000 60000 70000 80000 90000 100000 110000 120000 130000 140000Determine the nodal voltage V1 if l=4L25° A, and R=4 2, for the following figure: -j2 Q 20/-90° V j12 Ref. Oa. 26.56 L 72.507 V Ob. 22.56 L 68.507 V Cc. 28.56 L 74.507 V Od. 24.56 L 70.507 V When the voltage source is compared with the current source which of the fol owing is true Ca. The current source angle leads the voitage source ang e by 135 Ob. The current source angle ags the vo tage source ang e by115 Oc. The current source angle ags the voltage sourde angle by 135 Co. The current source ang e leads the vo tage source angle by115 ww