Electrical Engineering Assignment Solution: Circuit Analysis and Power Calculation, Exercises of Electronic Circuits Analysis

Download Electrical Engineering Assignment Solution: Circuit Analysis and Power Calculation and more Exercises Electronic Circuits Analysis in PDF only on Docsity! R4 60 R9 40 R8 20 R1 20 R10 25 R5 20 R4 60 R9 40 R8 40 R1 20 R10 25 R4 60 R9 20 R1 20 R10 25 R10 125 R4 60 R9 40 R7 40 R8 40 R1 20 R3 60 R10 25 R6 40 R2 40 R5 40 R8 40 R6 40 R7 40 R4 60 R5 40 R9 25 R1 20 Solution (ASSIGNMENT NO. 1) Question 1.a:- Q1.b. R4 20 R1 10 R7 40 R6 40 R5 10 R3 20 R2 10 R8 40 docsity.com R6 100 R6 40 R2 20 R4 20 R7 40 R8 40 R3 20 R1 10 R6 40 R4 20 R7 40 R8 40 R3 10 R1 10 R6 40 R4 40 R7 40 R8 40 R6 20 R7 40 R8 40 Q1.c. docsity.com R11 5 V1 60 R10 20 V3=5i + -- + + + Loops = 3 Vi Vi/4 V4 v5 Nodes = 3 2i _ i4 _ _ Branches = 5 v1 = 60 V v2 = 60 V i2 = 60/20 = 3 A i4 = v1/4 = 60/4 = 15 A v3 = 5i2 = 7.5 V By KVL, -60 + v3 + v5 = 0 v5 = 60 – 7.5 = 52.5 V v4 = v5 = 52.5 i5 = v5/5 = 52.5/5 = 10.5 A i3 = i4 + i5 = 15 + 10.5 = 25.5 A i1 = i2 + i3 = 3 + 25.5 = 28.5 A (b) It is now a simple matter to compute the power absorbed by each element: p1 = -v1i1 = -(60)(28.5) = -1.71 kW p2 = v2i2 = (60)(3) = 180 W p3 = v3i3 = (7.5)(25.5) = 191.25 W p4 = v4i4 = (52.5)(15) = 787.5 W p5 = v5i5 = (52.5)(10.5) = 551.25 W and it is a simple matter to check that these values indeed sum to zero as they should. Question 6:- docsity.com 0 0 Vin Vin Rsig=Rsig1 || Rsig2 0 1/gm Rs 0 Rg Rld=Rl || Rd After Simplification Circuit becomes: Vout gm*vi 1 2 + LOOP 1 Vi LOOP 2 _ Apply KVL at LOOP 1 as: -Vin + I1Rsig + I1Rg=0 I1 = Vin / (Rsig + Rg) From diagram we see that, using Ohm’s Law we get: Vi = I1 Rg Vi= Rg Vin / (Rsig + Rg) KVL at Loop 2, we get I1 = gm * Vi Now, Vout = Rld I1 I1 = gm RgVin / (Rsig + Rg) So, Vout = Rld gm RgVin / (Rsig + Rg) Vout = Question 5:- docsity.com R8 2 R7 1 R6 3 R1 2 R2 10 R4 15 R3 40 R5 60 R8 2 R7 1 R6 3 R1 2 R3 8 R5 12 R8 2 R7 1 R6 3 R1 2 R3 8 R5 12 150V 30V 150V 30V 120V For 1 Ohm Resistor Applying voltage division as: V1=120v x1/(2+8+12+3+1+2)= 120 / 28 P1=V2/R=(120 / 28)2 / 1 = 514.28 w For 10 Ohm Resistor Applying current division as: I1=120v P10=I2/R=(120 / 28)2 x 28 = 18.367 w P13=0w (resistors are bi-passed so node voltage and branch currents become zero) docsity.com