Question

Problem 4 a) Find the Norton equivalent circuit external to points “a” and “b” in figure below. b) Find the magnitude and polarity of the voltage across the 100 N2 resistor. 602 16 VO 100 22 80 Ω 8 A 40 Ω 8 V + b Problem 5 Find the Norton equivalent circuit for the network external to the 4.7 k2 resistor in figure below. 3.3 k92 1 k92 1.2 k92 ş 5 mA î 2.2 k92 ww 3.3 k92 4.7 k 2 12 V "F

Answer 1

IN 16 16 IN A 80 %A from cirluit i, 16 -0.2666 A 60 Apply KeL at node A In = įt8 IN - 0.266678 IN 8.266 A RN 60 80 RN 240 Go 140,80 are in parallel

RN = (60)(40)(80) (60) (40)+ (90)(80) + (80) (60) RN = 192,000 2400+ 3200 + 4800 RN = 14.46112 Norton's Equivalent Circuit + IN = 8.266A too بع 18.461 8 Apply mesh analysis for loop in 8.461117-8.266) + 100 in 850 118.461 in = 160.598 iqa 1.355A Ooltage across 1002 Va = 100 lia) - 100(1-355) Va= 135.50

3.3K IK SMA 126 2.2k 3.3k IN 12 Apply mesh analysis for Loop 2.2klip- 5x103) +3.3ki, tlock (1,-12)+12 = 6.7kii -1.2kiz -11+12 20 6.7ki, -1.2ki2 =-170 Apply mesh analysis for loop iz -12 + 2k lig -:-) +3.37 Ci-IN) = 0

4.5kią - l.2ki, - 3.3K IN = 12 → Apply mesh analysis for loop In 3.3K LIN-12)+ Ik In 30 4.3K IN - 3.3kiz solving % 0,04 ;= 1.058 MA i2= 6.745 mA IN = 5.1764 mA PIN 3.3K Tқ 2.2k 41.2k 3.3k 3.3K 2.2K are in Series Ris 2.2k + 3.3k 5.5 K12 3.3K, 1.2KR art parallel R2 = (3.3k) (1.2%) (5.54 (3.3k) (1.2k) +(1.25) 5.5k) + (5.5k)(3.3K) +

R2 21-78x10 (3.96X16) +(6.6 Xt)+(18.15x604) R 0.7586k 2 ha ik KRN IK, R2 are in Series RN TK +0.7586K RA = 1.758662 Norton's Equivalent : a IN -5.196MA RN 1.75865 4.7k b