Question

Problem One (25 points) 300A. Determine the equivalent resistance as seen by the current source is in the circuit shown below. Show carefully each step of your solution process. B. Find the magnitude of the power delivered by the source in the circuit shown if current source is = 50 mA. OSA 100 150 12 w 75or SOL 5000 750 n 1kn 36000 250 12 3002

Answer 1

Given Circuit is shown below. look 15od MM 7502 & Goor 7 lkna S004 M WW 250-22 3002 Step 1: 2502 below. 6oor and 150 s are in series 750m and are in series. the resulting Circuit is shown Re. = 750+250 = 162 Reza 6oot 1500-7506 1002 750+250 = lk2 z Slka 25002 E Goo+ 150 = 75002 W 3002 lk2 resistors Step 2 : lkoh and are parallel. Re - 7500 Rp = the lkolk =5002 2k and soon are parallel. 750x500 = 750x500 = 3002 750+500 1250 Circuit is loos2 the resulting Shown on the right side. 5000 Z E3ool m 3002

are in series 300 st and loo step 3: 500 and 3002 are in Series. = 500 + 100 = 6oon Rs2 = 3004 300=6002 the resulting Circuit is shown below. E6oor z Good Step 4: Goor and Goor are in parallel. . Reg: 600x600 = 600 - 3002 600 +GOD 2 The resulting circuit is shown @is Z 2002 Rey on the right sid Reg is the equivalent resistance seen by the current source is . . Reg = 3002 B) is, 50 mA Reg: 30052 dissipated on - powes Power the delivered by the source Resistor , Req - 3000. - 3ooch Power delivered a iz Reg = 50x50x100x300 25x10 ² delivered by the Current - 250 mW source is 250 mW powes

2) Given circuit is Voltages, resistances shown below as shown in Label the the currents, figure. M m . Ti s 260.2 3302 Zzou N 20- step! are in parallel. 4. 8002 and 2002 162 loo ww 3son 162 Rp, = 900420 - Too 8082o 80+20 = the resulting circuit is shown on the right side. Rs = 16+4 = 2012 8 i shown Step 2 160 and 402 are in series the resulting circunt is shown on the right side. izitiz uw vir 3300 L 2052 6o step 3 are in series parallel. 302 and 202 30x20 122 con i KO 50 30+20 shown zbor the on 3122 resulting Ciren't is the right side. I i=i+12

Step 4 for and 122 are series. Rs2 = 124822002. in is Circuit the resulting shown on the 1 ig right side. 360- 7 7 520-2. steps gor and zor are parallel. Req= 60X20 60420 0152 Rey Goxee 1200 - 120-40-2 153 60x20 = 80 1200 80 the resulting circuit is shown on the right side & Reg 2 15.02 Seen by Carrent source resistance equivalent Req = 152 If two Resistors R, and Rp are in series, then their equivalent resistance, Rs is given by Rg = R, f R2 If two Resistors R. and R2 are in parallel, then their equivalent resistance, Rp is given by Rp = R1 R2 Rit Rz Current division Rule Let I R and I and R₂ be the Currents Respectively T ? then I = R2 Is & I2 = R1 Is Rit R2 Ryt R2

Given that ig=170 mA From Shep the circuit shown in step 4 60 x 170x10 3 = 127.5 mA 60+20 we know that i liti, as shown in step 3 . the From step 2 , we get i= 20 xi = 20 x 127.5 x10 = 51 mA 30+ 20 SO 30 30+20 . x127 5x10= 76.5 mA 50 3 Label the currents and voltages , nodes as shown in the figure. vi 10-2 v 51 Given that VS va M - 15v= v 32 ve 20. 2400 V = 15V 1 V + O2 + OV₂ = 15 - 4th at Nodes and ② Can be considered as a supes node. KlL at supernode gives. V₂ V3 20 40 10

4 V ₂ - 4 V, + 20v₂ + 2V₂ + ₂ = 0 2 - 44, + 24V₂ + 3 00 - 40,+ 24 V + 3 V₂ = 0 - ② V₂ - V₂ = 5Vs & Vs = V2 2 - V3 = 5 V > zų - 2V₂=512 = OV, + 3V₂ + 2V₂ = 0 - ② - 3V₂ + 2 V₂ = 0 ņi tove to V₂ = 15 - 6 - 4V, + 24 V + 3 U₂ = 0 ov, + 3V, +2 V3 = 0 - © solving , and ® , we get Vi=15 V₂ = 3:08 V V3 = -4.62 v 10 33 TV UY-3:08 - 1192 1+, 154192 = 1646A power delivered by the 151 source = P₂ UxIx P = V. X = 15x 1.646 = 24.69 W

. Given circuit Voltages, nodes is shown below. label Irristance as shown in the currents, the figure. 91-7. In 40 = V Z I vo 3 2 A 2V Given that V=4V and i-7A. a super mesh. KVL on Mesh I and Mesh 2 forms the sup es mesh gives –4 +2 I + I, - Iz + 3 I 2 - 3 I ₂ = 0. - 3 I, + 3 I ₂ - 4 I3 = 4 - 0 I, I 2 = 7 I - I₂ + 0 I 307 - ② x = - 2 I I3 = - 2 Vx = -2x-2 Is = 4 I => 4 T, +0I, - I3 = 0 - solving ①, ② and ③ we get II = -2.5A I2 = -9.5A Izo - lod Vo = 3 (I2 - I3) = 3 [~9.5+10] = 3x015 = 1.5V Vx = -2 I = - 2x - 2.5=5r.