In a mesh-current method with two meshes sharing an impedance, how is the shared impedance included in the KVL equations?

In a mesh-current method with two meshes sharing an impedance, how is the shared impedance included in the KVL equations?

• A. Exclude the shared impedance from both equations.
• B. Include the shared impedance in only one equation.
• C. Include the shared impedance in both equations with opposite currents terms reflecting direction.
• D. Convert to a nodal analysis instead.

Answer: (C)

When two meshes share an impedance, that element sits on their common boundary and affects both loop equations. The current through the shared impedance is the difference between the two mesh currents, so the voltage drop across it is Z times (I1 − I2) for one mesh and Z times (I2 − I1) for the other. In the KVL equations, this means you include the shared impedance term in both equations, with opposite signs reflecting the opposite directions of the mesh currents. For a shared resistor R, the contributions appear as R(I1 − I2) in the first equation and R(I2 − I1) in the second, which is why the signs are opposite. Excluding the shared impedance or including it in only one equation would ignore or misrepresent the interaction between the two loops, giving incorrect current values. Converting to nodal analysis is a different method and unnecessary for this setup.

When two meshes share an impedance, that element sits on their common boundary and affects both loop equations. The current through the shared impedance is the difference between the two mesh currents, so the voltage drop across it is Z times (I1 − I2) for one mesh and Z times (I2 − I1) for the other. In the KVL equations, this means you include the shared impedance term in both equations, with opposite signs reflecting the opposite directions of the mesh currents. For a shared resistor R, the contributions appear as R(I1 − I2) in the first equation and R(I2 − I1) in the second, which is why the signs are opposite.

Excluding the shared impedance or including it in only one equation would ignore or misrepresent the interaction between the two loops, giving incorrect current values. Converting to nodal analysis is a different method and unnecessary for this setup.