please see attached pdfs

 

please see attached pdfs

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:
\( i_\beta = 10^{-3} A \)
\( v_c = 6 V\)

 

thank you very much but can yu tell what is wrong with the equation i come with?

 

thank you very much but can yu tell what is wrong with the equation i come with?

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?

 

thank you mjakov the 15 value is the 25 by mistake and you are right thank you