# mesh analysis prblm

Discussion in 'Homework Help' started by goldfinger, Jun 28, 2010.

1. ### goldfinger Thread Starter New Member

Jun 26, 2010
3
0
Hi!!
In the prblm below we've to find power dissipated in 2Ω resistor.

If I apply KVL to the outermost loop i'll get : 5 ia= 30
so, ia = 6A, which is wrong soln.(Why )

Similarly,if I apply KVL as below:
30-5 ia -3( ia -2.5)-6 ia = 0, i'll get: ia = 2.67A, which is again wrong(Why)
But, if I apply KVL as below:
30-5 ia -3( ia -2.5)-4(ia -2.5+2)=0, i'll get : ia =3.29A,which is correct.
Once ia is known then voltage acrss 2Ω= 30-6 ia = 10.25 V
Hence, P=(10.25*10.25)/2 = 52.53W Ans.

2. ### Georacer Moderator

Nov 25, 2009
5,174
1,284
This is not quite correct. A current source doesn't produce zero voltage on its edges. It produces as much voltage as it needs to let a current equal to its specified value run through it, and generally this voltage is unknown. This is the reason KVL doesn't yield the correct result. In your third mesh, where you didn't incude a current source, you didn't do that mistake and got the correct answer.

3. ### goldfinger Thread Starter New Member

Jun 26, 2010
3
0
Thanx a ton !!!
I've understood the reason..i was losing sleep over these type of prblms.
Thanx again.