Solve the circuit by mesh analysis and find the current I_x and the voltage across R_2.
mesh analysis problem


Mesh Analysis

There are four meshes in the circuit. So, we need to assign four mesh currents. It is better to have all the mesh currents loop in the same direction (usually clockwise) to prevent errors when writing out the equations.

Update 2019/07/27
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A mesh current is the current passing through elements which are not shared by other loops. This is to say that, for example, the current of R_3 is I_4 , the current of R_6 is I_3 and so on. But how about elements shared between two meshes such as R_2? Current of such elements is the algebraic sum of both meshes. Lets assume that the current of R_2 is defined with direction from right to left, algebraic sum means that its current would be I_4-I_2. If one assume the inverse direction, i.e. from bottom to top, it would be I_2-I_4 because I_2 is passing through R_2 with the same direction of its defined current but I_4 is passing with the reverse direction.

Lets define current directions for all elements and find them in terms of mesh currents:
current directions for mesh analysis



R_1 is shared between mesh #1 and mesh #2, i.e. meshes with currents I_1 and I_2. Therefore I_{R_1} equals to the algebraic sum of I_1 and I_2. To determine signs of mesh currents for I_{R_1}, we need to compare mesh current directions with the I_{R_1} direction. It is clear that I_1 is in the same direction of I_{R_1} and I_2 is in the opposite direction. Thus, I_{R_1}=I_1-I_2.


As we discussed earlier, I_{R_2}=I_2 - I_4.


R_3 is not shared between meshes. It is only in mesh #4 and because I_{R_3} is in the same direction of I_4, I_{R_3}=I_4.


Similar to I_{R_1}: I_{R_4}=I_2-I_3.


Similar to I_{R_3}: I_{R_5}=I_4.


It is similar to I_{R_3} with one exception; the direction of I_{R_6} is opposite to the direction of the mesh current, i.e. I_3. Therefore, I_{R_6}=-I_3.

Voltage Source

Similar to I_{R_6}, since the defined current direction is opposite to the mesh current direction: I_{V_1}=-I_4.

Current Sources

Current sources are known but finding their values in term of mesh currents helps to find mesh current values.


It is not shared between any mesh and in the same direction as the mesh current. Thus I_{S_1}=I_1=2 \text{A}


It is not shared between any mesh and in the reverse direction of the mesh current. Thus I_{S_2}=-I_3=4 \text{A}

Known and unknown mesh currents

Current sources, specially when they are not shared between meshes, are very useful in determining mesh current values. SO far we have found:
I_1=2 \text{A}
I_3=-4 \text{A}
But I_2 and I_4 are still unknown.

Now, lets write the equation for mesh of I_2 (Mesh II). A mesh equation is in fact a KVL equation using mesh currents. We start from a point and calculate algebraic sum of voltage drops around the loop. We try to avoid introducing more unknowns to equations than the mesh currents. For example, instead of I_{R_2}, we use I_2 - I_4. With some practice, you can easily write KVL equations using mesh currents directly. For resistors, the voltage drop equals to the resistance multiplied by mesh currents considering mesh currents in your KVL writing direction with positive sign and for the ones in the opposite direction with negative sign. Lets assume for mesh 2 we start from left-bottom toward top:
(I_2 - I_1)\times R_1+...
I_2 is in the same direction of our KVL and therefore comes with positive sign but I_1 is in the opposite direction and comes with negative sign. Similarly:
(I_2 - I_1)\times R_1+(I_2 - I_4)\times R_2+(I_2 - I_3)\times R_4=0
(I_2 - 2)\times 2+(I_2 - I_4)\times 4+(I_2 +4)\times 3=0
2I_2 - 4+4I_2 - 4I_4+3I_2 +12=0
9I_2- 4I_4=-8 (1)
For Mesh IV, starting from left-bottom toward top and then right:
I_4 \times R_3+...
This is because R_3 is not a shared element.
I_4 \times R_3-4\text{V}+I_4 \times R_5+(I_4-I_2) \times R_2=0
I_4 \times 1-4+I_4 \times 2+(I_4-I_2) \times 4=0
7I_4 -4I_2=4 (2)
Solving equations 1 and 2, we obtain:
I_2=-\frac{40}{47}\text{A}, I_4=\frac{4}{47}\text{A}

Finding I_x

The current of any branch is equal to the algebraic sum of associated mesh currents. I_1 is in the opposite direction of I_x and I_2 is in the same direction as I_x. Therefore:

Finding voltage across R_2

V_{R_2}=I_{R_2}\times R_2=0.9362 \times 4=3.7447\text{V}.

When To Use Mesh Analysis?

It depends on the number of meshes in the circuit comparing to the number of nodes. If there are more meshes than nodes, it is usually better to use mesh analysis.


Solve this circuit using mesh analysis and find I_x and V_{R_2}:
Mesh analysis homework



Published by Yaz

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    1. Mesh is the one which has current source and the loop is the one which have voltage and resistors.

  1. “is defined with direction from right to left, algebraic sum means that its current would be I1−I2 . If one assume the inverse direction, i.e. from bottom to top, it would be I2−I4
    I2−I4 because”

    I think I1 is supposed to be I4-I2

  2. is the homework question answer correct?
    can someone plz confirm coz i’m getting a different value.

  3. Could you please include the steps where you get Ix = -2.851A
    Also what do we do with the equations to come up with the 40/47

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