Question

Problem 1 (10 points) Considering that the power developed by the power supply of the voltage source Psov < 50 watts. Find the value of the power supply (va) for the 15K A resistor. 101 25K 15K2 45KA w B1oKa 40v. 320K12 35KN ( 100 vs Figure 1

Answer 1

Solution:

Given that

loid . 2KB 15L М. чки + Vo + ib lok 20K 35k loova чоу tо і ф — цо+V +v, D-D 4 26 1БЕ V, чо -VA Vi V, чSь V3 +ve- уіф -ГЕ id : — sla 20 tok. +) уд 20] +) Боокуд _

>>-V1 + 1010 + V3 = 0

V3 - V1 = 1010 + 1

Nodal at V1 + V3 (Super nodal)

Vi 10 + Vi - 40 25 V3 + 500KVA 5 + V3 – 12 45 = 0

= 45V1 + 18V1 – 720 + 90V3 + 45000KVA + 10V3

= 63V1 – 10V2 + 100V3 + 45000KVA

Nodal at V2 :

V2 - 40 15K + V2 20K + V2 - 13 45K = 0

126-480 +9+4%-43=0

= 2512 - 413 480

= 25V2 - 42 10.12 20K + V1 = 480

= 25V2 – 0.00212 + 2V1 = 480 5

from equ 2:

→ 63V1 - 10V2 + 100V3 + 45000K (40 – 12) = 720

= 63V1 – 10V2 + 100V3 + 18 x 10º – 45 x 100V2 = 720

= 63V1 – 10V2 – 45 x 100V2 + 100V3 = 120 – 18 x 10°

= 635-45 x 10+ 100B = -18 x 109 +
from equ 1:

- V3 – V1 = 1010

= V3 - V1 = 10 12 20K

20K V3 20KV1 1012

20K V3 2KV1 V

= 2000V3 2000V1

= 2000V1 2000V3 + 12 = 0 + 4

multiplying equ 3$\times$ 20 :-

1260V1 - 9 x 108V2 + 2000V3 = -3.6 x 1011

2000V1 + V2 2000V3 = 0

3260V1 - 9 x 1082 = -3.6 x 1011 + 6

from equ 5:-

= 25V2 – 0.00212 + 2V1 = 480

N 25V2 + 2V1 = 480

V2 480 – 211 25

from equ 6 :-

3260V1 - 9 x 108 480 - 2V1 25 -3.6 x 1011

3260V1 – 7.2 x 10' V1 = -3.6 x 1011 + 9 x 108

+7.2 x 10' V1 = +3.6 x 1011

Vi 3.6 x 1011 7.2 x 107

Vi 5000

V2 = 480 – 2(5000) 25 -380.8

VA = 40 40 - 12

VA = 40 + 380.8

-420.8

The value of power supply

-4208