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Can somebody please point me in the right direction here

Im trying to create two equations to solve for IA(left hand loop) and IB(right hand loop)

I need two equations consisting of IA and IB to then solve again using matrices.

What effect does the current source have on the circuit?

schematic

simulate this circuit – Schematic created using CircuitLab

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I decided to label the two loop currents differently, so that I avoid a conflict with your current source's designation:

schematic

simulate this circuit – Schematic created using CircuitLab

Just write out three equations:

$$\begin{align*} 0\:\text{V} - R_1\,I_A-R_2\,I_A-V_{I_1}-R_3\,I_1&=0\:\text{V}\\\\ 0\:\text{V}+R_3\,I_1+V_{I_1}-R_4\,I_B-10\:\text{V}&=0\:\text{V}\\\\ I_1&=I_A-I_B \end{align*}$$

And solve for \$I_A\$, \$I_B\$, and \$V_{I_1}\$.

The main issue is getting the equations written out accurately. The introduction of \$V_{I_1}\$ as a new variable is because we don't know the voltage compliance required of the current source, \$I_1\$. Since it's unknown, using the mesh technique here you need to use KVL in the loop, you need a voltage there. So you create a new variable so that you have one to use.


You can do the solution using Sage (free program):

 var('r1 r2 r3 r4 ia ib vi1 i1')
 az=solve([Eq(-r1*ia-r2*ia-vi1-r3*i1,0),Eq(r3*i1+vi1-r4*ib-10,0),Eq(i1,ia-ib)],[vi1,ia,ib])
 az
{ia: (i1*r4 - 10)/(r1 + r2 + r4),
 ib: -(i1*(r1 + r2) + 10)/(r1 + r2 + r4),
 vi1: -(i1*r3*r4 + i1*r4*(r1 + r2) + (r1 + r2)*(i1*r3 - 10))/(r1 + r2 + r4)}

 az[vi1].subs({i1:.1,r1:7,r2:5,r3:2,r4:10})
4.70909090909091
 az[ia].subs({i1:.1,r1:7,r2:5,r3:2,r4:10})
-0.409090909090909
 az[ib].subs({i1:.1,r1:7,r2:5,r3:2,r4:10})
-0.509090909090909

The negative signs just mean the current directions are opposite the assumptions used to write out the equations.

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  • \$\begingroup\$ Thanks, can that middle branch just be replaced with R3(Ia - Ib) to avoid having to solve for Vi1? \$\endgroup\$ – PMA Jan 21 at 23:50
  • \$\begingroup\$ @PMA \$V_{I_1}\ne 0\:\text{V}\$ in this circuit. So... No. However, \$I_1\$ has \$\infty\:\Omega\$, so \$R_3\$ has no impact on the current and can be replaced by a voltage source equal to \$R_3\,I_1\$ (with the right polarity.) So you could replace \$I_1\$ and \$R_3\$ with a voltage source of unknown value. But you'd still need to keep the 3rd equation I mentioned. Note: \$R_3\,I_1 = R_3\left(I_A-I_B\right)\$. \$\endgroup\$ – jonk Jan 22 at 0:00
  • \$\begingroup\$ @PMA I got a sign wrong in equation 2, so please take note of the correction. \$\endgroup\$ – jonk Jan 22 at 0:23
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Note that the current source has a label of I1 on it which is the same as the current you defined in the left loop. That needs to be fixed (it really doesn't need a label as it is a fixed value). To get your two equations, first note that the 2 loop currents flow into the branch containing the current source. That will give you 1 equation. The second can be derived by using KVL around the outside loop.

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  • \$\begingroup\$ Thanks I've edited now. I need the equations for the left and right hand loops, I've tried simulating the circuit and found I'm looking for IA = 409.09mA, and IB = 509.09mA, I just cant put together the equations to get to those answers. I've also tried removing R3 and found it does not change anything. \$\endgroup\$ – PMA Jan 22 at 0:00

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