Question

Calculate a) Req b) The voltage across each component c) The current through L d) The total power d) The power at R4 and e) The current through each component. F) name explicitly which component is the point of branching based on Kirchhoff's rule g) draw the final Req circuit under the current schematic. Remember what I told you in class in this case. (10 points) 1k2 480pH 3900 4700 7500 V1 12V

Answer 1

According to keL (C&B) e La V = 12v The cincuit can be expressed an 2 42 3904 Yur 12 a 43. MI the current through (1-1) 4702 (1-12275ort each part will be Li<= -1 : faure. " Here, we will consider the steady state Condition Here's an import ant rule: a when an inductor with no-current flowing through it is first switched into a circuit it behaves like an open cineuit. Therefore, equivalent circuit with be . lk 3900 750r n w u m ri Rev =(3904750) 4701 1000 x 470 1000+470 112 V Ž 14602 a Mära Current (im) = 12 A = 8.22 mA 1460 in whom a constant and unchanging current as flowing through an induction the indunton behave like a short cincuit (steady stake condition.). Therefore equivalent curcuit will be KR 390n Hur Roon Roon or w Reg. = 390 * 1000x470 ww 1000+ 470 470n - เอ 12 . Current (in) - 12 A = 16.9 mA - 710

(a) Reg - 14602 (6) Voltage, across 390r = 1390 imx390 23:21 Voltage across inductor and 750r will be VL2 mx 750 = 6'16 V Reg for lar and 470 is 1000x470 ²3200 1000+ 470 voltage across Ikr and 470 r will be 41,410 = 320x i'm= 2.63 v. (c) The current through a is duelove {= 16.9 mA (d) Total power, pa v. ima 12x8,22x10²W = 98.64 mw (d) The power at RRR eiline rodily - i'mx V2 = 8.22 mA x 6.16 V 250.64 mW @ 2.2lim8.22mA 1000 + 470 1410 8:22 - 2.632 5059 na (F) (g) 1750 ? i'm = 8.22 mAh Point B and c in figure The figures are shown above.