Determine the Norton equivalent of the circuit in Fig. 10.30 as seen from terminals a-b. Use the equivalent to find Io Π2 Ω 10/0° V 8 Ω 4Ω 1Ω -j3Ω τ (+) 22-90° A α b Figure 10.30 Answer: Zy = 3.176 + j0.706 Ω, Ι. 1ο = 985.5/-2.101° mA. 10 Ω -j5Ω = 4.198/-32.68° A.

Find the total/resultant effective voltage, given the following:   e1 = 10 sin ωt   e2 = -15 cos (ωt – π/3)   e3 = 10 cos ωt   e4 = -20 sin (ωt– π/3)

Problem 10.064 - Norton equivalent of a phasor circuit For the circuit shown in the figure below, find the Norton equivalent circuit at terminals a-b, where Is = 9 2 60° A. Please report your answer so the magnitude is positive and all angles are in the range of negative 180 degrees to positive 180 degrees. Is ZN= IN= мее V V 60 92 o a j80 Ω °Ω bo °A 40 92 -j30 92

In an RC parallel circuit, the apparent power is 3400 VA, IR=10A and angle theta=58 degrees. Find XC? Show your work.  a) 12.476 Ω b) 21.443 Ω c) 11.258 Ω d) Ω10.321Ω

The sinusoidal voltage source in the circuit seen in (Figure 1) is operating at a frequency of 44 krad/s. The coefficient of coupling is adjusted until the peak amplitude of 1 is maximum. Figure 20 Ω 100 Ω 5 mH3 75 Ω 20 mH < 300 Ω 1 of 1 40 nF What is the value of k? Express your answer using three significant figures. k= 0.56 Submit ▼ Part B VG| ΑΣΦ | 11 VE X Incorrect; Try Again; 29 attempts remaining i₁ = Previous Answers Request Answer What is the peak amplitude of 21 if vg = 309 cos(4.4 × 10¹t) V? Express your answer to three significant figures and include the appropriate units. vec HÅ Value P Pearson Units ? ? Re

ФЫ В А П Р О ВАПРОЛДЗАПРОЛДЖЭ ЯЧСМИТНЧСМИТЬБНСМИТЬБЮ SOLVE STEP BY STEP IN DIGITAL FORMAT 100 cos (2 x 104t+3°) V 89. Use superposition to determine the voltages vi(1) and v2(t) in the circuit of Figure 10.84. (1) 47 kn WWW mm 112 дн 56 кп :33 μF m 112 H 92 MF (1) 47 kn -т 100 cos (2x 105t-3°) V ЙЦУКЕНГШУКЕНГШЩЗКЕНГШЩзхъ