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I want to find voltage across 3k resistor using Loop/Mesh analysis.

answer in SPICE = 4.0135

enter image description here

My Solution :

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$$ 8=2(I_2)+1.5(I_2-I_3)+3(I_1-I_3) $$ $$ 0=3(I_3-I_1)+1.5(I_3-I_2)+10(I_3) $$ $$ I_1-I_2=0.5 $$ 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 ?

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  • \$\begingroup\$ There are three loops. \$\endgroup\$
    – aMike
    Commented Feb 2, 2021 at 18:39
  • \$\begingroup\$ @aMike But a loop is not necessarily a mesh. Not sure what your point is here. \$\endgroup\$
    – Elliot Alderson
    Commented Feb 2, 2021 at 19:33

2 Answers 2

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This is how I solve it.

First I marked the mesh currents

enter image description here

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:

$$V_{R3} = (I_1 + I_2 + I_3)\times 3k\Omega = 4.01351V$$

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    \$\begingroup\$ Excellent approach. It keeps it down to two equations. One could include the \$I_1\$ loop equation by creating a voltage variable to recognize the drop across the current source and solve three: \$I_2\$, \$I_3\$, and the current source voltage value as unknowns, substituting in \$I_1\$'s value, later. (But when attempting that it is important that no two loops enclose the same current source as its loop current.) \$\endgroup\$
    – jonk
    Commented Feb 2, 2021 at 19:20
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    \$\begingroup\$ Thenk for comment @jonk.And very good point about current source and that only one mesh current loop is allowed for the same current source. \$\endgroup\$
    – G36
    Commented Feb 2, 2021 at 19:29
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    \$\begingroup\$ @ElliotAlderson I think this is one of those cases where the OP has carefully shown their work and where the harm in showing a completed result is nil. In fact, it is actually better than otherwise as it allows the OP to study a now-worked example and learn for themselves several new things they missed. I can't see any harm to the OP here. Not at all. I think in this case you are administering "by rote" (automatic pilot, so to speak) rather than examining each situation and weighing its merits. \$\endgroup\$
    – jonk
    Commented Feb 2, 2021 at 19:48
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    \$\begingroup\$ @jonk Sorry, I disagree. We have no knowledge of whether the OP will carefully study the answer to learn from their mistakes or they will copy down the provided equations and hand in the exam with no thought whatsoever. I think a much better approach is to say "Your second equation is incorrect, you forgot about..." and give the OP a chance to work through the rest of the problem. The potential downside of giving an answer is that it is unfair to students who don't cheat, so I try to avoid that. \$\endgroup\$
    – Elliot Alderson
    Commented Feb 2, 2021 at 19:59
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    \$\begingroup\$ @ElliotAlderson Neither of us can read the OPs mind and different people will come to different conclusions about that. So the point is moot. But I see a serious person there. If you don't, then your conclusion about the OP is itself perhaps harmful and at the very least, judgmental. But we will have to leave it there. We just disagree. (My opinion is getting stronger because just last fall I taught about 50 students EET221 and I've been getting more of a feeling for where they are at.) \$\endgroup\$
    – jonk
    Commented Feb 2, 2021 at 20:01
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The above problem can also be solved directly by only two equations using node analysis. Shown below is the the setup for Mathematica and result. V2 is the voltage across the 3k resistor.

Solve[{-0.0005 + v2/3000 + (v2 - v3)/1500 == 0, (v3 - 8)/2000 + (v3 - v2)/1500 + v3/10000 == 0},

{v2 -> 4.01351, v3 -> 5.27027}}

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