Engineering Need to find two currents in this circuit

[opticaltempest]

I am trying to find the currents I_D1 and I_D2 labeled in the picture by the arrows. I am stumped on how to obtain the correct values for the currents. I am having trouble using either KCL or KVL. This circuit seems simple but has me stumped.

http://img127.imageshack.us/img127/8600/circuitvz9.jpg

If I write KCL at the node to the right of the I_D1 arrow I get

-I_{D1}+I_{D2}-I_{6V}=0 \implies

I_{D1}=I_{D2}-I_{6V}.

I_{6V}=\frac{6V}{43k\Omega}=140\mu A

I_{D2}=\frac{9V}{22k\Omega}=409\mu A

Is there an easier or different way to do this problem, perhaps using KVL instead of writing a KCL equation at the node I used?

Thanks

 

[Defennder]

Yes you can also do it via mesh analysis. You would have 2 equations here.

 

[CEL]

I am trying to find the currents I_D1 and I_D2 labeled in the picture by the arrows. I am stumped on how to obtain the correct values for the currents. I am having trouble using either KCL or KVL. This circuit seems simple but has me stumped.

http://img127.imageshack.us/img127/8600/circuitvz9.jpg

If I write KCL at the node to the right of the I_D1 arrow I get

-I_{D1}+I_{D2}-I_{6V}=0 \implies

I_{D1}=I_{D2}-I_{6V}.

I_{6V}=\frac{6V}{43k\Omega}=140\mu A

I_{D2}=\frac{9V}{22k\Omega}=409\mu A

Is there an easier or different way to do this problem, perhaps using KVL instead of writing a KCL equation at the node I used?

Thanks

The simplest way to solve the circuit is the one you used.

 

[Andrew123]

Quick solution:

Hey this is a simple mesh anaylisis problem.. I've included a solution. I hope it helps. Cheers