Question

The circuit below consists of a thevenin source

Problem #2 - (16%) The circuit below consists of a Thevenin source shown on the left of the vertical dotted line and a load shown on the right of the dotted line. The frequency of the Thevenin source is 600 Hz. The Thevenin voltage is 2 Vpk. R 1:4 70 mH Vin 2 V pk 600 Hz 7502 0.5 UF Source Load a) Compute the load impedance (looking into the ideal transformer from the dotted line). (4%) b) Compute the value of L that will result in maximum power from source to load. (4%) c) Compute the value of R that will result in maximum power from source to load. (4%) d) Compute the power in the 750 82 resistor when the source is delivering maximum power to the load. Hint: if this appears difficult, don't overthink it. (4%)

Answer 1

*12263.89u Lq=Tomh 1:4 2 VPK 600 Hz ' 311 & R2=7500} I 0.5uf = C₂ I Xeq = -} 530.516 Xng =j 27 f L₂ = jQ X TX600 x 70x10 s = ; 263.89 Yen com -o corastice =4530.5163v. Rq|lxce - 750 (-1 530.516) 750 -j 530.516 € 250.116 - j 353.594 r. 29 = XL2 + Rellxc - 1263.89 + 250.116-; 353.594 = 250.116 -8 89.704 m. Zq = X2q + Rallace When secondary impedance za referred to Primary 212 = 22 - (250-116-339.24) ☺? 219 = 15.632 -;5.6065 v. In order to deliver maximum power from source to load, Zg=212 Ry + 8 XL, & R 2 =j42 Equeting imaginary parts XL, = -X42 XL = 5.6065 2TfL = 5.6065 h = 5.6065 v. 2x Ux600 14 = 1.487 mH ]

(By equeting real ports of above equation) I But, Ri = R2 he the R + R2 In the denominator, if R=0 then, the maximum power will transfer from source to load. - so, I R = R = 0.0 K42 Maximum Power PL = to 75or resistor Rst Retî (X2,-X212) 0+ 15.632 + Á (5.6065-5.6065) PL= 0.25588 W ..PL = 255.88 mW The above calculations done when Secondary impedance reffered to Primary. But, the same power (PL) trastered from primary to Secondary so, Power loss in 750N resistor is PL = 0.25588 W or 255.88 mW.