Please, I need help on this question. I feel like I'm on the brink of solving this and need some help with the rest, or If I messed up somewhere, you can help.
CKT: http://i42.tinypic.com/iy067m.jpg
My attempt: http://i40.tinypic.com/2qmm2s5.jpg
Question: Solve the following circuit for the value of the current through the 5K ohm resistor using nodal analysis.Also I can't use source conversion.
Please help, I really need help on this
CKT: http://i42.tinypic.com/iy067m.jpg
My attempt: http://i40.tinypic.com/2qmm2s5.jpg
Question: Solve the following circuit for the value of the current through the 5K ohm resistor using nodal analysis.Also I can't use source conversion.
Please help, I really need help on this
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Start off with a KCL node equation @ node Va: (currents leaving = positive and currents entering = negative)
At node Va, I have a current entering from the left, and currents leaving down and right of the node. so you get:
[ 10 - Va ] / 2k = Va / 1k + [ Va - Vb ] / 4k
expand terms:
10 / (2k) - Va / (2k) = Va / (1k) + Va / (4k) - Vb / (4k)
factoring the terms we get equation 1:
Va * (-7 / 4k) + Vb * (1 / 4k) + 10/2k = 0
KCL @ Vb node: (currents leaving = positive and currents entering = negative)
At this node, I have a current entering from the left, and currents leaving down and right.
[ Vb - 6 ] / [ 1k + 2k ] + Vb / (5K) = [ Va - Vb ] / (4k) ----> remember the 1k and 2k add
expand terms:
Vb / (3k) - 6 / (3k) + Vb / (5k) = Va / (4k) - Vb / (4k)
factoring each Va and Vb term we get:
Va * [ -1 / (4k) ] + Vb * [ 47 / (60k) ] - 6 / (3k) = 0
Now this is a simple linear equation with 2 equations + 2 unknowns. using any method you want, you get:
Va = 3.3757 V
and:
Vb = 3.6305 V
Now the current through the 5k resistor is:
I = Vb / (5k) = 3.6305 / (5k) = 0.7261 mA
At node Va, I have a current entering from the left, and currents leaving down and right of the node. so you get:
[ 10 - Va ] / 2k = Va / 1k + [ Va - Vb ] / 4k
expand terms:
10 / (2k) - Va / (2k) = Va / (1k) + Va / (4k) - Vb / (4k)
factoring the terms we get equation 1:
Va * (-7 / 4k) + Vb * (1 / 4k) + 10/2k = 0
KCL @ Vb node: (currents leaving = positive and currents entering = negative)
At this node, I have a current entering from the left, and currents leaving down and right.
[ Vb - 6 ] / [ 1k + 2k ] + Vb / (5K) = [ Va - Vb ] / (4k) ----> remember the 1k and 2k add
expand terms:
Vb / (3k) - 6 / (3k) + Vb / (5k) = Va / (4k) - Vb / (4k)
factoring each Va and Vb term we get:
Va * [ -1 / (4k) ] + Vb * [ 47 / (60k) ] - 6 / (3k) = 0
Now this is a simple linear equation with 2 equations + 2 unknowns. using any method you want, you get:
Va = 3.3757 V
and:
Vb = 3.6305 V
Now the current through the 5k resistor is:
I = Vb / (5k) = 3.6305 / (5k) = 0.7261 mA