One place to start is to draw the circuit and label the three currents that flow in the three branches.

For example, draw an arrow beside the 5.1 ohm resistor and label it I1. Then draw an arrow beside the 10.3 ohm resistor and label it I2. Then draw an arrow beside the 4A current source and label it I3.

I say three branches because I am ignoring the current source on the right and R1.

Then post your drawing here.

hgmjr

 

where did u get 4A from...isnt it 2A??

so would it be

5+2(I1+I2+I3)...like this???

 

im still on confuse

 

The 2A source and the resistor R1 can be set aside since you have been given the fact that voltage at Va is given.

I was referring to the the 4A current source on the right side of the circuit.

hgmjr

 

so watz wrong with my equation?????????

 

The equation you have formed is not quite correct.

By sketching the circuit and annotating it with the current arrows as I described, I will be able to walk you through the rest of the steps used in the derivation of Vb.

hgmjr

 

well i followed wat u said....is it wrong???

 

Post you sketch with the arrows here so that I can see what you have done. Then I can help you with the next step.

hgmjr

 

i dont have a scanner...to put it online

 

So you can't post the sketch?

hgmjr

 

noo....what is the correct equation???

 

can anyone help me??

 

HINT:

\({\LARGE i}_{\small 1}\ -\ {\LARGE i}_{\small 2}\ +\ {\LARGE i}_{\small 3} = 0\)

This equation is taken from Kirchoff's Current Law for the sum of the currents flowing into and out of the node Vb.

Where:

\({\LARGE i}_{\small 1}\) equals the current in the 5.1 ohm resistor.
\({\LARGE i}_{\small 2}\) equals the current in the 10.3 ohm resistor.
\({\LARGE i}_{\small 3}\) equals the current in the 4 Amp current source.

This equation along with the diagram I suggested you draw should give you a starting point. All you need to do is write the equations for the three currents in terms of the values given and you will be well on your way to a solution.

hgmjr