Question

4. Find currents 11 and 12 and voltages Vi and V2 for the circuit shown below. R 11 R 12 W | 21 50 V 000 000 L L 11 L2 R 42 1. R3372 av C+1 R350

Answer 1

Rii Iz' L₂ Ry} A {72 sov 5 oc only circuit voltage is applied in the above. so for De voltage. Inductor Replaces with short circuit capacitor Replaces with open circuit - s Because X2 = jwb. For OC W=0 ; So Xe =0.means Coorrent max So short circuit Similarly Y2 = j (0) L 0. 01 CS Scanned cool meang current.10? Open circuito CamScanner

ng in 12 2.2 {am 1 + § 72 I kt. Ę In i ts loop loop so the above circuit has Single So Ij = Iz: - I. Apply KVL in the above loop -50 + 218 + 1.12 + 7 I =0 101 = 50 I = 50 = 5 A. = 3; = 1, · 1: , = 5A I = 56 50 - VAL = 1 CS Scanned with CamScanner 2

- Vogue * = 5 => 50-V, = 10 => -V, = -40 Vogue = 40v Similarly, 40-v2 = 5 2) 40-V4 = 5 --V2 = 5-40 a) Vg =-35 . Vy = 350 - Vn & Vy are nothing but V, and U2. :: V, = 4ov CS Scanifed with 2 = 351 Cam [canner

Multisim Simulation :

Multimeter-XMMA Multimeter-XMM3 5A agn66 5 A XMM4 Set.. Multimeter-XMM2 Multimeter-XMMI 40 V * Set..

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