Engineering Possible Short Title: Possible Mistakes in Textbook Circuit Problems?

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The discussion revolves around a textbook circuit problem where the user is questioning the accuracy of their mesh equations while trying to find the current through an inductive element. They initially presented a supermesh equation that was critiqued for incorrectly treating voltage drops across resistors and inductors. The correct formulation requires consistent directional treatment of voltage drops, adhering to Kirchhoff's Voltage Law, which states that the sum of voltage gains and drops around a loop must equal zero. Clarification was provided that in mesh analysis, one must traverse the loop in a single direction, either clockwise or counterclockwise. The user expressed gratitude for the explanation, indicating a better understanding of the concept.
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Homework Statement


The problem is attached. The question asks: "find the current through the inductive element". I'm starting to wonder how many mistakes are in this textbook. But it might be a mistake I made. Could someone check my Mesh Equations?


Homework Equations





The Attempt at a Solution



SuperMesh 1 & 2: 10V<0 + 1kohm(I1 + 6mA <0) + 4kohm (I2) + j6kohm (I2) = 0

SuperMesh Supplemental: I1 + I2 = 0.1 Vs

Direct Source Supplemental: Vs = (I1 + 6mA <0) * 1kohm

if anyone wants to take a stab at the final answer, I got 7.74mA <171.5, and the book got 1.378mA <-56.31. Way different than what I'm getting
 

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Your supermesh equation is wrongly written.

Assuming that voltage gains are positive values, then in:
10V<0 + 1kohm(I1 + 6mA <0) + 4kohm (I2) + j6kohm (I2) = 0
the part "+ 4kohm (I2) + j6kohm (I2)" is incorrect as clearly there is a voltage drop (according to the direction of I2) across the 4k resistor and 6k inductor.

The correct supermesh equation should be:
10V<0 + 1kohm(I1 + 6mA <0) - 4kohm (I2) - j6kohm (I2) = 0
which should give you the desired answer.
 
So in a supermesh, you treat it as going in one direction the entire way around? I was assuming you treated it as each direction showed

I was taught you add the voltage if there is a voltage drop (voltage drop in relation to the direction your reference mesh current), not subtract as you're showing. And if I'm wrong as you say, wouldn't the other part of the equation (involving the 1k resistor) be incorrect as well? I guess that would really depend on my first question as well
 
Last edited:
Number2Pencil said:
So in a supermesh, you treat it as going in one direction the entire way around?

Yes, because in mesh analysis you are actually applying the Kirchoff Voltage Law which equates all voltage drops (or equivalently voltage gains) around the loop to zero. And yes, you can only traverse the loop in either the clockwise or anticlockwise direction, not both. Supermesh is no different from mesh analysis except that you have now to consider more than one mesh current.
 
thanks, I completely understand
 
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