Using KCL in Mesh Analysis

Question

I previously asked a question about how KVL is used in node analysis. I'm betting that KCL is also used in mesh analysis.

Could someone explain if and how KCL is used in mesh analysis?

Answer 1

In fact, circuit analysis by the method of the meshes, is not based on the KCL, if not in the KVL.. Suppose the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

To perform mesh analysis, we propose the three currents displayed. Keep in mind that these three currents are actually proposed, which does not mean they are the actual currents.

Given these currents, write the KVL to each mesh

$$ V_1 = I_1\cdot R_1 + (I_1 - I_2)\cdot R_2 \\ 0 = (I_1 - I_2)\cdot R_2 + I_2\cdot R_3 + (I_2 - I_3)\cdot R_4 \\ (I_2 - I_3)\cdot R_4 + I_3\cdot R_5 - V_2 = 0 $$

Applying a little algebra, we can rewrite the system of equations as:

$$ I_1\cdot R_{11} - I_2 \cdot R_{12} = V_1\\ -I_1\cdot R_{21} + I_2\cdot R_{22} - I_3\cdot R_{23} = 0\\ -I_2\cdot R_{32} + I_3\cdot R{33} = V_2 $$

Where

$$ R_{11} = R_1 + R_2\qquad R_{22} = R_2 + R_3 + R_4\qquad R_{33}=R_4 + R_5\\ R_{12} = R_{21} = -R_2 \qquad R_{23} = R_{32} = -R_4\qquad R_{13}=R_{31}=0\\ $$

And this system accepts a matrix representation:

$$ \left(\begin{matrix} R_{11} & R_{12} & R_{13}\\ R_{21} & R_{22} & R_{23}\\ R_{31} & R_{32} & R_{33} \end{matrix}\right)\cdot \left(\begin{matrix} I_1\\ I_2\\ I_3 \end{matrix}\right)= \left(\begin{matrix} V_1\\ 0\\ V_2 \end{matrix}\right) $$

Solving this system we find the three current proposed.

But we must not forget that these currents are proposed; for example, if \$I_1\$ give a negative result means that the actual current in the circuit is counterclockwise. Moreover, to find the actual current through \$R_2\$ should subtract the currents \$I_1\$ and \$I_2\$.
You see, an analysis by the method of the mesh, implies apply KVL.

Answer 2

Could someone explain if and how KCL is used in mesh analysis?

Yes, it is implicitly used in mesh analysis. Consider the following simple example circuit.

^{simulate this circuit – Schematic created using CircuitLab}

Formally writing the KVL equations around mesh 1 and 2 yields

$$V_1 = V_{R1} + V_{R3}$$

$$V_{R3} = V_{R2} + V_2$$

But, to solve for the mesh currents \$I_1\$ and \$I_2\$, we must write the voltage variables in terms of the mesh currents.

By Ohm's law, we have

$$V_{R1} = I_1 R_1 $$

$$V_{R2} = I_2 R_2 $$

$$V_{R3} = I_{R3} R_3 $$

However, for \$I_{R3}\$ we must (at least implicitly) apply KCL at the top center node:

$$I_1 = I_{R3} + I_2 \Rightarrow I_{R3} = I_1 - I_2$$

thus

$$V_{R3} = (I_1 - I_2)R_3 $$

Yes, this an almost trivial example but it nonetheless shows that KCL is applied to find the current through circuit elements that share mesh currents.