Mesh Current Method: Solve for Vx in Network Q11

[TheRedDevil18]

Homework Statement

Obtain the voltage Vx in the network of Fig.Q11, using the mesh current method.
(Ans: Vx=4.35∠-194.5)

Homework Equations

The Attempt at a Solution

I am not getting the correct answer, are my loop equations correct ?

10 - 2*i1 + 2j*i1 -5j(i1+i2) -5(i1+i3) = 0
4.33 + 2.5j - 10*i2 -5j(i1+i2) +2(i3-i2) -2j(i3-i2) = 0
-10*i3 + 2j(i3-i2) -2(i3-i2) -5(i1+i3) = 0

Therefore,

i1(-7-3j) + i2(-5j) -5*i3 = -10......(1)
i1(-5j) + i2(-12-3j) + i3(2-2j) = -4.33-2.5j.....(2)
-5*i1 + i2(2-2j) + i3(-17+2j) = 0.......(3)

 

[berkeman]

10 - 2*i1 + 2j*i1 -5j(i1+i2) -5(i1+i3) = 0
4.33 + 2.5j - 10*i2 -5j(i1+i2) +2(i3-i2) -2j(i3-i2) = 0
-10*i3 + 2j(i3-i2) -2(i3-i2) -5(i1+i3) = 0

Some of the signs in the 2nd equation look wrong to me (but it's been a while since I used KVL -- I usually use KCL).

Can you talk through why you chose the signs that you did for the 2nd equation?

 

[TheRedDevil18]

Some of the signs in the 2nd equation look wrong to me (but it's been a while since I used KVL -- I usually use KCL).

Can you talk through why you chose the signs that you did for the 2nd equation?

Okay, I have labeled the currents in my diagram

4.33+2.5j, That's the voltage of the source in the second loop in complex form (Voltage rise so it's positive)
-10*i2, Voltage drop across the resistor, hence negative sign
-5j(i1+i2), Voltage drop across the inductor, negative sign
+2(i3-i2), Positive sign because the loop is traveling opposite to the current direction as labeled
-2j(i3-i2), Because impedance is negative for a capacitor and loop is traveling opposite in direction to current, so +*- = - sign

 

[NascentOxygen]

One of the difficulties is that we are not told which terminal of each source is (+) and which is (-), i.e., which is the reference terminal.

Perhaps assume the marked current is indicative of the source polarity, though I don't know whether we can generalize that to other problems like this.

I do get the same equations as you, though haven't checked them.

Perhaps reverse the voltage source in the second loop, and see whether you get the book's answer that way?

 

[TheRedDevil18]

One of the difficulties is that we are not told which terminal of each source is (+) and which is (-), i.e., which is the reference terminal.

Perhaps assume the marked current is indicative of the source polarity, though I don't know whether we can generalize that to other problems like this.

I do get the same equations as you, though haven't checked them.

Perhaps reverse the voltage source in the second loop, and see whether you get the book's answer that way?

I tried reversing it but still get the same answer, maybe I am solving it wrong so here are my steps. I used crammers rule

Determinant,
7+3j ...5j...5
5j...12+3j...-2+2j
5....-2+2j...17-2j

Solving that I get, -73+405j

Then for i3,
7+3j...5j...10
5j...12+3j...4.33+2.5j
5...-2+2j...0

Solving that I get, -744.52-100.77j

Therefore i3 = (-744.52-100.77j)/(-73+405j)
= 0.08 + 1.82j

So V = (0.08+1.82j)*10
= 0.8 + 18.24j

Or, 18.26∠87.49