Question

1. Use the step-by-step method on the circuit below to find the inductor current for all time.

2 ΚΩ t = 0 4 ΚΩ + + 40V Σ12 ΚΩ 20 V 100 mH i,

2. Given the two complex numbers:

a) Express both numbers in rectangular, exponential, and phasor forms;

b) Find the sum, the difference, the product, and the quotient of the numbers.

3. Given the three sinusoids:

a) Find their corresponding phasors using a cosine reference;

b) Does x(t) lead or lag y(t), and by how much?

Answer 1

1 To Calculate Celrrent across inductor 1, we will use fiollowing Reg t L {(t) = [ [0+) - ibo)] e at formula +2.60) 1 w Where 1200+) and illo) is current through inductor L, dime = 0+ and co neobectively. Req: Equivalent Resistance genoss Indudor L. Now At t= 0- : Just before the switch shifts crrcuit look like as in dig 1. 2KR As the inductor in the Sleady state condihon for a Jilo) DC inkeet will act as a short figi At t=0- Circuil, Hence in duchor is shown as short circuled. So Current in the Grewt given by 2.10-) Voldage 20x10 3 Resistance 2X163 + 40V ） 40V = 2 ''. illo-) 20 mA. . Current Across the inducher doesn't change abouktly So, ilot) = {(0-) from egn į (0+) = 20mA 3 CS Scanned with CamScanner

R KR + gov when t>O, Carcuit will look like as shown in dig2 Now thes arcuut goes inlo 4K2 Sleady Dale con dihon at t=00. And at steady 100mH ELL state induchor is short Grouiled hence Crewst es drown in fg 3. fis2 A1 to » As we know that Current 4K (2 І. alisays þar Hoough low Resistance bath so Y I2 12 = 0 460 OK2 and I = (100) So fig 3 · 20 figz Al t=0 illos) 4x103 Ieloo) +dov 3 5x163 4 So 11.60) = 5 MA when Req= R1 + R2 Now cleavy Hoo Registone 4K and 126 ase parallel to induchor L,' and Equivalent Resistence of too parallel resistance given as Rix Re Ri= 4 ke Ra = 12 kg 4x12 4 +12 Req= 3 kr So Req= = 3ke 48 16 3x10 3 2 = 100X10-3 [yiven] (5) Scanned with CamScannet = 100mH

3 Pulling call the values inlo en ? 13x103 x + į.lt) (20-5) e - (Bix! *+) - (311091) IDOXID-3 + 5 15 e + 5 Taking 5 common from Rus - (3x10²t) 2214) = 5(3.e +1] MA Ana @ Rectangular dorm Exponenhal from Phasas form re => Q+j6 jo r Lo h = √a²+62 B = tan-(bla) a where o A 1-@ } - @ a and as 6 r Coso r sino 2) Fig4 V9+4 = @i) 2 = - 3-jd It is in Rectangular fer where a= -3: b= -2 from eq? V = V(32 +62) V13 0= tari) 33.7 Rectangeles lom =) – 3- j2 Exponen del form => j (33.7) із е 113 L(33.7) So Zi Am Scanned with Phasen dorm CamScander

10 Z. = 40 160 C 13 from ean Q = 46360° (2 2 3.46 - b= 4 sin Go & + j (3.46) j60 And 4 e So za Rectangular form => Ex ponen har form Phasas form 4L60° 6 Sum = 21+ Zg = (-3-12)+(2+i (3.46)] - 63 +2) +j/3.46- 2) Sum = -1 +; 1.46 Ans Difference Zi-Zg |-3- j2) - [2 +j[3.46)] = (3-2) -; (2+3.46) Dillerence = -5- ; 5.46|| Ans to find by (3 4 Phasah Product: Peroduel is Produel is easy Let 2,=d, Loi ; Zq hz LO2 then Z ,x2a roxha LOi+ O2 € Product hi and 101-02 ¢ Quotient. 12 I from egn Product VIB 1(33.7)° x 4 160 V13x4 /(33.7)+60 CS Scanned pion CamScan Product = 14.4 1 93.7 And 21/29

5 from 09" Quotient => V13 33.7 4 260 V13 4 | 33.7-60° = Quotient 0.9 LE 26.3) | Am = 1 3 @ 217) 3 Cos 165 +459) ylt 2 Cos (lot-36) 210) 4 sin 10t + 369) Now tongno me tri Identity sino Co1 (90-0) Cos (0-969) Let O = 160€ + 36°) Put this value in ean e sin lwt +300) Cos (wt +30 -30°-90°) sinlwt+ 30%) = Coy 16t-60°)- Puting the value of sin (bot +30) in lecms of cosine in 241) Z(H) 4 Costot-60°) | And since 211) and ylt) altea dy given in Cosine Reference, So alt) 3 69 1Wt +45°) Am YLt) & Cos (ot-30°) (2 750 6 Clearly 211) leading by the y(t) Leading Angle => 45° -L-30) 3 45+ 30 cs = csod valt) leads by an angle of 750 Ant CamScanner