Two impedances z1-20+j10 and Z2= 10-j30 are connected in parallel and this combination is connected in series with 23=30+jx. Find the value of 'x' which will produce resonance?
Answers
Answer:
Let I = 15 ∠ 0º ; Z1 = 10 + j15 = 18 ∠ 57º Z2 = 6 −j8 = 10 ∠ − 53.1º Total impedance, Z = (Z1Z2)/(Z1 + Z2) = ((10 + j15)(6 - j8))/(16 + j7) Applied voltage is given by V = IZ = 15 ∠ 0º × 10.3 ∠−20.4º = 154.4 ∠− 20.4º I1 = V/Z1 = 154.5 ∠ − 20.4º/18 ∠ 57º = 8.58 ∠ − 77.4º I2 = V/Z2 = 154.5 ∠−20.4º/10 ∠−53.1º = 15.45 ∠32.7º We could also find branch currents as under : I1 = I. Z2/(Z1 + Z2) and I2 = I.Z1/(Z1 + Z2) It is seen from phasor diagram of Fig. that I1 lags behind V by (77.4º − 20.4º) = 57º and I2 leads it by (32.7º + 20.4º) = 53.1º. ∴ P1 = I12R1 = 8.582 × 10 = 736 W; p.f. = cos 57º = 0.544 (lag) P2 = I22 R2 = 15.452 × 6 = 1432 W ; p.f. = cos 53.1º = 0.6 Combined p.f. = cos 20.4º = 0.937 (lead)Read more on Sarthaks.com - https://www.sarthaks.com/493506/two-circuits-the-impedance-of-which-are-given-by-z1-10-15-and-z2-j8-ohm-are-connected-parallel?show=493544#a493544