circuit analysismesh
I want to find voltage across 3k resistor using Loop/Mesh analysis.
answer in SPICE = 4.0135
My Solution :
$$ 8=2(I_2)+1.5(I_2-I_3)+3(I_1-I_3) $$
$$ 0=3(I_3-I_1)+1.5(I_3-I_2)+10(I_3) $$
$$ I_1-I_2=0.5 $$
Hence :
$$I_1=1.5980 \\
I_2=0.5304 \\
I_3=4.7973$$
Voltage at 3K $$3(I_3-I_1)=9.5979 $$
Which is wrong answer .
Correct answer is 4.0135V
Where am i wrong ?
Lots of bad advice on this page, so I'm going to answer this even after all this time.
First, you don't have any supermeshes. By definition a supermesh is closed path that does not include a current source on its edges. There are no such paths here.
Furthermore, by inspection I2 = -2A and I3 = -1A. So the current through R5 (going up) is 1A. Also, by KCL, I1 + 2A = 6A. So I1 = 4A.
To find the voltage drop across R5 you need to find two paths form a ref node to both ends of R5 such that these paths pass only through resistors other than R5 (i.e. these paths cannot pass through R5 or any current sources). No two such paths exist. Which means the value of R in this circuit is not fixed/determined by rest of the components (including the current sources).
To convince yourself, assign some arbitrary value to R5 and solve the circuit, say in a simulator to save time. And then try to assign some other value to R5. The circuit is still solvable, you'll just get another voltage drop across R5.
You can even try it yourself here.
You can see it as a current divider consisting of the 2Ω resistor and the other two resistors that add up to 1Ω + 3Ω = 4Ω.
So the proportion of the conductances and thus the proportion of currents of both branches are
1/4 : 1/2 or 1:2,
i.e. 1/3 of the total 6A are going one way and 2/3 the other way, i.e. 2A are goind through the left branch and 4A through the right branch.
If 2A are going through a 1 Ω resistor the voltage across it is 2V.
Best Answer
This is how I solve it.
First I marked the mesh currents
And due to the fact that in mesh one we have a constant current source, we know that:
$$I_1 = 0.5mA$$
Thus the equation for \$I_2\$ mesh will look like this:
$$3(I_2 + I_3 + 0.5) + 10I_2 + 1.5(I_2 + I_3)=0$$
and for \$I_3\$ we have:
$$-8 + 2I_3 + 1.5(I_3+I_2)+3(I_3 + I_2 + 0.5)=0$$
And the solution is: here
$$I_1 = 0.5mA$$ $$I_2 =-0.527027mA$$ $$I_3 = 1.36486mA$$
And the voltage across \$R_3\$ resistor is equal to:
$$V_{R3} = (I_1 + I_2 + I_3)\times 3k\Omega = 4.01351V$$