Question

1. Find 1:4 56 3 162 "g(t) = m con 2004 Volts VER = 80 Volts. RMS 2. (25) Find I, I, 321 D zex tu M 212 2 212

Answer 1

Given circuit is 12 1:4 1612 o Volt) Vg (t) in Cos zoot volt Vibe = 80 volt, RMS This circuit comes under Transformer as an Impedance Matching Device Proof: (or) Derivation: I2 Il 1:n 11 + + | ZL V₂ Zin VA w 7N2 zin فيما N,:N2 =1: N2 V2 N2 En NA = 1:n NA I 1 Iz * B I2 = Vl. It » Pout Pin Transformar ie passive, reciprocal ckr sils HI 11 » EL. Iz zin. It Iz ZL - [E] - n2 zin >> Zin ZL n Rin RL 2 ท”

coming back our problem, Iz 12 n=4 3116 162 Ng =Vm Lo Qr w=2000& Zin Zin EL Zin 11 From the previbus derivation, 거라 Zin' 16 = 12 (432 below The abore circuit can be redrawn as 12 12 Total Zin = (1+1)2 Vg = Vm Lil Zin =22 ws 2007/s I1 = VnLoo Vm A 2 uitleri an Is = I 2 = [m] . 1 Iz Vm A E 8 Vibrams -80V 11 V2 = I(16) → 80 = - Me [16] - 400 1 [Anoever] Vm= Vg = = 40 Cos 200t (V)

(2) It Node A + 2.2 & D1 20V w 212 222 &2 To check whether a particulas diode operating region, anode potentials . look at the cathode and 20V + 4 Dy te current flows » Active Region » Short circuit. Less than 2ov 0< voltage c20V # D2 + Cathode OV Anode potential is less than * No current flow Open circuit Reverse Bias » Circuit can be redrawn as Il A A node R1322 + 200 B mre ta 212 RS 22 22 Ry This 21. resistor can be repositioned as Is IL 62 ww + 20V R232.2 22 22 RA + 20V 2.2 62 692 R3}22 RAB B → B A in the RAC = Rq tR3+ A. R3 22 R2 E RAC RBC 322 3R3 11 4 +262 RAB = 2 +2 + 22 lly RBc = R. = 6.2 RAC

TA. + 20V 인 6 + SSD = 7.SD. TA. + 20V 67 7. Gtts 불어 By Ohm's Low, VeIR 20 A 2 DE 2043 0 포 11 IqF 10 R > current If = I = GA