Question

Part G The phasor method for analyzing oscillations is useful in many areas of physics. A phasor is a rotating vector with constant angular speed w, which can also be defined as a complex number that represents the amplitude and phase of a sinusoid. The standard convention is to use the real part of the phasor to represent the sinusoid. For the circuit in (Figure 1), vg = 2.9 cos( 4 t + 1.08 )V, R= 92, C = 5 x 10-2 F, L= 8.7 H , find the reactance X seen from the source. Keep 3 significan figures. A complex number can be represented in four ways: (1) rectangular form z= x +iy, (2) polar form z=r20, (3) expoenetial form z=rexp(io), (4) geometrical form. IVO AQ R O 2 ? Submit Request Answer Figure < 1 of 1 > W Part 1 R R The phase difference between Vs and is in Part G, A0 = 0, – 0; is IVO AQ o 2 ? A0 = rads

Answer 1

C solution: Giver data Vs=2.9 CosR4E+1.08), p=92, c=5010²F, L=8-7H. source voltage in phasor form is 8=2911.08 v. w = 4 rad/sec. Capacitive reactence of 5000 F is to Xc=- ) = -j5r 4x5x16² stva stѕ сhаr os - те у ј» past : x= 34 x 8.7= 334.82 non cirwit is shown i grequency domain as below. → Total impedance sees mar ge IR grom source side is Z = 9+ 334.8)19-35) = 94 334.8 XR-j5) fig. circuit in frequency 334.849-35 dorrain. Ag+ 34-8)98 x 9.488€) -29,05° - 31.131 73.190 = 94 10.60521-12-2y =94 10.3641-j 2.2484 2.9. $334827-352 z = 19.3641 -j 2.2484 The reactance seen from source ? side is x = 2.248 ur capacitive reactance

past A hi source correst Is = 2-921.08 19.3641 -32.2484 Is = 2.921080 19.49425-6.67506-148917. Hos Op = 7.7o3 voltage source angle ou = 1.88 phase differesa 40 = 0o-Op = 1.08-7.703 40=-6.623 = 6.623 leading - 6.6237 180 40 = s.116 rad