Question

a) A coil having resistance of 12 2 and an unknown inductance is connected in series with a 14uF capacitor. The circuit draws maximum current when a voltage of 200 V at 60 Hz is applied. A second capacitor is then connected in parallel to the circuit. i) What should be its value (capacitance), such that the combination acts like a non- inductive resistance, with the same voltage (200V) at 120 Hz? [3.0 marks] ii) Calculate the supply current drawn by the two circuits and draw the phasor diagrams for both cases. [2.0 marks]

Answer 1

Coil Given that R-t-C Seria connected circuit draws mane Current 1447 12.12 D-5025854 when voltage - Joo V 2 60Hz frequency Reconal frequency of this Serice Connection i 60Hz 1 XL = Xc 22fL = 2afc :- 27400) 23 (60) (14x76') L 0.502585 H (2760) (0762) recistance mooong in the equalent part Non inductive there i no inductive impudance term (+j term) at 120 Hz 2: 12+) 378.9401 -j 94.7350 2 12 +j28420501 Не j22.fc2 2 total z 0 0.0001483 -0.003sij +32xfc2 From this equation to make the admitagee having the imaginary trum 27fcz > 0.00351 0400 351 2*** 120 4.6552 MF

In the Quertion it's given that sou total combination is not coneists of inductive receistance C2 > 4.6552 MF give non inductive impudance only, 키 let us coneider E2 = 4.6552 HF only 2,- 12+j284.2050) HE zafa =j 284-2 Zz Zillzz 21 22 (6730.92/98 –284.23) Zitz2 32 200 12 +j2842050) = 0.22966- 30.7024650 -trup zi 0.7030 L-8705822 to neutralize the inductive revistance 1.772 jo. 7024650. 72 v/az j 284.71169 200 tjo.7024650

y - 200 100 라 방 = 6·7030 187.S522 ? 리 이들 5 59ktk.9 =딸 500 V 87.56 229 탄

If you have any doubts please comment

Thank you