- #1
jaus tail
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Homework Statement
Homework Equations
V = IR
The Attempt at a Solution
I used nodal analysis.
Let voltage at central point be Vx
So
1) (10 - Vx )/ 2 is current entering from left branch.
2) Current entering junction from middle branch is 4 Vs
Vs is current in eqn 1 times 2 ohm . This is (10 - Vx ) /2 * 2
Multiply this with 4
and we get current entering from middle branch is 4(10 - Vx)
3) Current leaving junction from rightmost branch is Vx / 8
Equation is 1 + 2 = 3
Which is (10 - Vx)/2 + 4(10 - Vx) = Vx/8
This gives 9(10 -Vx)/2 = Vx/8
This gives Vx = 360/37.
So Vab is half of Vx which is 180/37V.
I did the same using another method.
let current through left most 2 ohm resistance be Ia
Eqn 1 applying KVL in left loop, 10 = 2 Ia - 4 Vs * 2. Negative sign cause we're taking negative direction of current as mention in figure.
Now Vs = 2 Ia
So eqn 1 becomes. 10 = 2 Ia - [4 (2 Ia)] * 2
Solving this we get 10 = -14 Ia.
So Ia = - 10/14.
Vs = 2 Ia = -20/14/
4Vs = - 80/14.
Current in right loop is sum of currents entering junction = Ia + 4Vs = -10/14 - 80/14 = -90/14.
Voltage drop at AB is -90/14 * 4 = -360/14.
This doesn't match answer I got above.
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