Question

1. Find V, in the network in Fig. 1 using superposition. 1 kΩ 2 kΩ R2 R4 2 kΩζv, R1 6 mA R3ξ 2kΩ Is Vs (+) 12V Figure 1.

Find Vo’ due to the current source Is alone. Vo’=______V

Find Vo” due to the voltage source Vs alone. Vo”=______V

Find Vo due to both sources. Vo=______V

Answer 1

2KL R2 2KRZVO Ic Ri 12V * (1 6mA 32k2 ak 2K112K (q Q.1)/ang- Vallage No in the network in figo Shown below ust JK Ry 2k2 V figo IS Voltage No due to current source Is alone is No= Von due to 51К-, figo o ginen ast 1K С. w current source 2Kr. alone. w + EKLEVO It IK to ZK2 (1) 6mA No, Ig Ikr figco & by applying KCL at nade o in fig g) is

(2 ☆ Voiro + Voiro 2K 6 ma 2k - '2Voi GMA XZK 2001 = 120 =12 Noi = 6V . Volbage to due to Current Source. alone is (Vo = Voi : 6 V *I) Voltage to due to vs alone is Voltage source given as Volta Voz =Vo * HV IK2 Ø I 2K-2 Y I Vx²2x2 2K2 Vogl Doc 12V fily applying KCL at Mode & in fugg is given ast Vr-o Vn-o + Vn-12 2 vnt 3vnt vn-36 by BK + 2K 2k Scannedy CS CamScanner 6

V 2 v 2 V 0 3 2 n + 3 VM +3 VM -36 = 0 8Vn3630 = 8. Vn=36 Vn=36 » Vu = 9 8 * Vn=9 . Current from friggs given as! I= (Un-OJA ЗК I= (912-o] I= q me ЗК I= 3 32 MA I=1.5mA By applying KVL in loop om fign A g Voltage Vog'is given as. Voz = 2K-2X I = 2K-2X1.5mA Voq=3 v Voltage Vodne to Voltage source alone * CS Scanning with Camsanne given as V." Vo2 = 3V

* I) Voltage Vo due to both Source that is current source (Is) f Voltage ginen Source Vs 'n en ast K-2 2 w R2+ VS I OMA R2K2 ZK2 www Ru 2K Ri 12V figo *by applying KLL at node D is given as vo 2K + Vo-Vu IK = 6 mA Vo +2V0-2 Vn=12 3 V0-2V1=12 1 > 0 By applying KCL at node @ 'm figo is given as: Vuovo Vn Vu-o 21n-2 No TVnt Vn-12=0 + 1k + Vn-12 2K 2K=0 as y Vn - 2V=12–0

5 * As we see from calculation that both ear of eam are equals that ist 3 Vo-2 Vu 4 Vn-2V 5 Vo= Vr = 5 Vo .' from eano me get 3 Vo-2Vu=12 3 Vo-2 -2 (Ev= 12 = 3 V. =12 + 3V-5Vo = 12 19-5) Vo=36 = 410 = 36 = *|Vo= gul = 6Vn -2 No CS ☆ Scanned with CamScanner