# Mesh current analysis voltage and current source

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?

simulate this circuit – Schematic created using CircuitLab

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

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.

• Thanks, can that middle branch just be replaced with R3(Ia - Ib) to avoid having to solve for Vi1? – PMA Jan 21 at 23:50
• @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: $R_3\,I_1 = R_3\left(I_A-I_B\right)$. – jonk Jan 22 at 0:00
• @PMA I got a sign wrong in equation 2, so please take note of the correction. – jonk Jan 22 at 0:23

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.

• 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. – PMA Jan 22 at 0:00