Supermeshes, method selection & verification

The worked circuit has a 10 V source, three 1 kΩ resistors and an ideal 1 mA current source in the branch shared by the left and right meshes. The current source tells us its branch current exactly, but its voltage VIS is whatever the surrounding network requires.

If we insist on writing separate KVL equations, the left mesh contains the unknown drop VIS:

The right clockwise mesh traverses the current source in the opposite direction:

The new voltage unknown is not wrong, but it defeats the economy of ordinary mesh analysis. The supermesh eliminates it.

Add the two individual KVL equations. The unknown current-source voltage appears once as +VIS and once as −VIS, so it cancels:

This is exactly the KVL equation obtained by tracing the outer perimeter around both meshes while skipping the internal current-source branch:

So a supermesh is not permission to ignore the source branch. It is the sum of two valid mesh equations arranged so the unknown internal source voltage cancels.

With both mesh currents clockwise, I1 crosses the shared branch downward and I2 crosses it upward. The physical downward current is therefore

The source arrow in Figure 5.1 fixes that downward current at 1 mA:

If the source arrow were reversed, the right-hand side would be −1 mA for these same mesh-current references. Always derive the sign from the actual arrows; never memorize “I1−I2” without checking orientation.

For R1=R3=R4=1 kΩ and VS=10 V, the supermesh KVL equation becomes

Using kΩ·mA=V:

Together with the source constraint

we have I1=I2+1 mA. Substitute:

The current-source branch carries I1−I2=1 mA downward exactly as imposed.

R4 belongs only to the right mesh, so its downward current is I2=3 mA. With R4=1 kΩ,

R3 also carries 3 mA from A to B, producing a 3 V drop:

Therefore the ideal current source has 6 V across it from its upper terminal to ground. Its current was known before solving; its voltage is recovered only after the surrounding circuit is solved.

R1 carries 4 mA, so its drop from the 10 V input node is 4 V, independently confirming VA=6 V.

Now forget the mesh solution and analyze the same network by node voltages. At node B:

At node A, count currents leaving A. The current source contributes +1 mA because its arrow points from A toward ground:

Using kΩ·mA=V and collecting terms:

Substituting VA=2VB gives VB=3 V and VA=6 V—the same result obtained from the supermesh. The cross-check uses a different set of unknowns and different Kirchhoff equations, so agreement is strong evidence rather than duplicated algebra.

The three resistor powers are

The current source has 6 V from its upper terminal to ground, and 1 mA enters that upper terminal. Under the passive-sign convention it absorbs

The 10 V source supplies 4 mA out of its positive terminal:

Total absorption is 16+9+9+6=40 mW, exactly matching the voltage source's 40 mW delivery. Like an ideal voltage source, an ideal current source can absorb or deliver power depending on the operating point.

Both methods preserve the ideal source exactly. Neither inserts a fictitious resistance. Both remove an inconvenient internal unknown by summing conservation equations across a larger analysis boundary, then restore the information supplied directly by the source as a constraint.

There is no universal winner. Before writing equations, compare the structure of the circuit and the quantities you need.

A good engineering choice minimizes algebra while keeping the physical constraints visible. For larger automated problems, modified nodal analysis extends the node-voltage framework to voltage sources and other elements; that computational formulation lies beyond this introductory DC-04 course.

1. Topology check. Identify actual nodes, branches and—if using meshes—planarity and elementary meshes.
2. Reference check. Define every unknown voltage/current direction before algebra. Negative answers are allowed.
3. Equation-count check. Match independent unknowns with independent equations and source constraints.
4. Dimensional check. Each KCL term must have current units; each KVL term must have voltage units.
5. Local conservation check. Re-evaluate KCL at solved nodes or KVL around solved loops using recovered physical branch quantities.
6. Source-constraint check. Verify every ideal source arrow/polarity exactly.
7. Global power check. Signed powers of all elements should sum to zero in the lumped model.
8. Independent-route check when practical. Use a simpler equivalent reduction, the alternative nodal/mesh method, or a limiting case.

Mastery problems

1. In the worked supermesh circuit, change the downward current source from 1 mA to 4 mA. Find I1, I2, VA and VB.
2. Keep IS=1 mA but reverse its arrow. Write the current constraint before solving anything.
3. For the original example, recover the current-source voltage from the left individual mesh KVL equation and confirm the 6 V result.
4. A planar network has two unknown node voltages but five meshes. Which method is likely shorter, and what else should you inspect before deciding?
5. A planar network has six unknown node voltages but two meshes and only voltage sources. Which method is likely attractive?
6. A student obtains node voltages that satisfy every written nodal equation, but the total signed power is not zero. List at least three possible causes to investigate.

DC-04 is now conceptually complete. The next course, DC-05, will use these systematic analysis tools to study independent and dependent sources, linearity, source transformations and superposition before reconnecting naturally to the existing Thévenin/Norton course.