Question

oa (50 points) Rş c. L2 - Given the element values R1 = 120 N2, L1 = 10 mH, L2 = 110 mH and w = 4712.39 , find the value of the capacitance C1 that results in a purely resistive impedance at terminals ab. C1 = uF help (numbers) Calculate the value of the input impedance using that value of capacitance. ŽEQ = kl help (numbers)

Answer 1

en Ri= 1202 ; L = lomh ; L2=110mH. W.4712.39 j wli 1200} الباري les 120 rand jul, are in series = 120+; 4712.39x10x103 Za = 120+ j470 1239 This combination is looking like below & jul2 tea Jur = - + jwc & I JwL₂ I tjoc - j 19291 X103 120+; 47. 1239 120-547-1239 +jwc -21.92918103 (128.9214)

120-j47. 1239 tj w C -j 1.92918103 16620.66 Zea is real number Yea also real number If and imaginery juce number ao j4701239 16620.66 +1.9291810 3 j wc = j (2.8352 + 1.929178 10-3 ca 4016436x183 u712-39 c = tool1 ME For this a value zena toa = 16620.66 = 13805 = 001385 kr Zea 120 od a= & 001385 kd ; loolllE Zea=