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.
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.