# node analysis

Discussion in 'Homework Help' started by yiannistamv, Sep 24, 2014.

1. ### yiannistamv Thread Starter Member

Jul 23, 2014
35
0

File size:
313 KB
Views:
43
• ###### 2.pdf
File size:
323.8 KB
Views:
32
Last edited: Sep 24, 2014
2. ### mjakov New Member

Feb 13, 2014
20
5
Hi!

The branch current method could be used for Figure P2.28. This method is similar to nodal analysis. Let the node between the 0.8V voltage source and the dependent current source be called C and let the lowermost node in the picture be the reference node.

Kirchoff's Current Low at C:
$i_\beta + 29 i_\beta = 30 i_{\beta}$
Branch Current Equations
From node C to ref. node:
$\frac{v_c}{200} = 30 i_\beta$
Left branch:
$\frac{15.2 - (v_c - 0.8)}{10^4} = i_\beta$
It follows:
$i_\beta = 10^{-3} A$
$v_c = 6 V$

yiannistamv likes this.
3. ### yiannistamv Thread Starter Member

Jul 23, 2014
35
0
thank you very much but can yu tell what is wrong with the equation i come with?

4. ### mjakov New Member

Feb 13, 2014
20
5
Hi!
Is your equation in Fig. 2.28 obtained by the KVL of the outer circuit loop? If so then, have you considered the voltage drop at v_y and the 25 V source. Where have you got the value 15 in your equation from?

yiannistamv likes this.
5. ### yiannistamv Thread Starter Member

Jul 23, 2014
35
0
thank you mjakov the 15 value is the 25 by mistake and you are right thank you