Circuit Theory 1 · Supermesh Analysis

#17 Supermesh #17 — boundary and shared current-source examples

Uses a boundary current source directly, then combines a source constraint with supermesh KVL for a shared current source.

Question

Three-mesh circuit with 6 V and −8 V voltage sources and a 10 A boundary current source.
The boundary source fixes i_3=10 A; V_1 is referenced positive at the top of the shared 1 Ω resistor.

Use current-source constraints and supermesh analysis to find V_1 in the first circuit and mesh currents i_1, i_2 and i_3 in the second circuit.

Written solution and narration transcript(shows the full solution)

Below are all the lines written in the notebook together with the full narration transcript.

  1. 1. Current-source rules

    Three-mesh circuit with 6 V and −8 V voltage sources and a 10 A boundary current source.
    The boundary source fixes i_3=10 A; V_1 is referenced positive at the top of the shared 1 Ω resistor.

    Boundary current source ⇒ directly sets a mesh current

    Source between two meshes ⇒ current constraint

    Missing equation ⇒ supermesh KVL around the shared branch

    Narration transcript

    Now we move from ordinary mesh-current equations to supermesh cases. The trigger is a current source. If a current source is on the outer boundary, it can directly set one mesh current. If a current source lies between two meshes, we avoid writing a voltage across that source. Instead, we write a constraint equation from the source current, and a Kirchhoff's Voltage Law, K V L, equation around the combined supermesh.

  2. 2. Example 1 circuit

    Three-mesh circuit with 6 V and −8 V voltage sources and a 10 A boundary current source.
    The boundary source fixes i_3=10 A; V_1 is referenced positive at the top of the shared 1 Ω resistor.

    i3=10Ai_{3}=10 A

    V1=1(i1−i2)V_{1}=1(i_{1}-i_{2})

    Find: V1

    Narration transcript

    In the first example, the right-side current source is on the outside boundary of mesh three. Because the arrow is downward, and the mesh current i three is also downward through that branch, we immediately get i three equals 10 amperes. The requested quantity is V one across the 1 ohm resistor, with plus at the top and minus at the bottom. So after solving i one and i two, V one will be 1 ohm times i one minus i two.

  3. 3. Solve Example 1

    Three-mesh circuit with 6 V and −8 V voltage sources and a 10 A boundary current source.
    The boundary source fixes i_3=10 A; V_1 is referenced positive at the top of the shared 1 Ω resistor.

    5i1−i2=65i_{1}-i_{2}=6

    i1−10i2=−38i_{1}-10i_{2}=-38

    i1=2A,i2=4Ai_{1}=2 A, i_{2}=4 A

    V1=−2VV_{1}=-2 V

    Narration transcript

    For loop one, the 6 volt source is a rise, then the 4 ohm resistor and the shared 1 ohm resistor are drops. This gives 6 minus 4 i one minus the quantity i one minus i two equals zero, or 5 i one minus i two equals 6. For loop two, the 6 ohm resistor, the shared 3 ohm resistor, and the shared 1 ohm resistor are included, together with the negative 8 volt source. After substituting i three equals 10 amperes, the second equation becomes i one minus 10 i two equals negative 38. Solving the two equations gives i one equals 2 amperes and i two equals 4 amperes. Therefore V one equals 1 times 2 minus 4, which is negative 2 volts.

  4. 4. Example 2 circuit

    Three-mesh circuit with a 2 A boundary source above and a 5 A source shared by the two lower meshes.
    The sources give i_2=−2 A and i_3−i_1=5 A; the two lower meshes form a supermesh.

    i2=−2Ai_{2}=-2 A

    i3−i1=5Ai_{3}-i_{1}=5 A

    Supermesh: loops 1 and 3

    Narration transcript

    In the second example, the top current source is on the boundary of the upper mesh. The source arrow points left, while the clockwise mesh current i two goes to the right through that top branch. So i two equals negative 2 amperes. The 5 ampere source sits between the two lower meshes, so it creates the constraint i three minus i one equals 5 amperes. Now we write one supermesh K V L equation around the outside of loops one and three.

  5. 5. Solve Example 2

    Three-mesh circuit with a 2 A boundary source above and a 5 A source shared by the two lower meshes.
    The sources give i_2=−2 A and i_3−i_1=5 A; the two lower meshes form a supermesh.

    38−4(i1−i2)−(i3−i2)−3i3=038-4(i_{1}-i_{2})-(i_{3}-i_{2})-3i_{3}=0

    i1+i3=7i_{1}+i_{3}=7

    i3−i1=5i_{3}-i_{1}=5

    i1=1A,i2=−2A,i3=6Ai_{1}=1 A, i_{2}=-2 A, i_{3}=6 A

    Narration transcript

    The supermesh travels around the 38 volt source, the 4 ohm resistor, the 1 ohm resistor, and the 3 ohm resistor. The equation is 38 minus 4 times the quantity i one minus i two, minus 1 times the quantity i three minus i two, minus 3 i three equals zero. Using i two equals negative 2, this simplifies to i one plus i three equals 7. Together with the source constraint i three minus i one equals 5, we add the two equations and get 2 i three equals 12. So i three equals 6 amperes, i one equals 1 ampere, and i two equals negative 2 amperes.

  6. 6. Method summary

    Three-mesh circuit with a 2 A boundary source above and a 5 A source shared by the two lower meshes.
    The sources give i_2=−2 A and i_3−i_1=5 A; the two lower meshes form a supermesh.

    Boundary source ⇒ direct mesh current

    Shared source ⇒ constraint + supermesh KVL

    Negative current ⇒ physical direction opposes the reference

    Narration transcript

    These two examples show the two most common current-source patterns in mesh analysis. A boundary current source directly fixes a mesh current. A current source between two meshes creates a supermesh: one constraint equation from the source, and one K V L equation around the combined outside loop. The sign is not decoration. It comes from comparing the source arrow with the chosen mesh-current direction.

Source video: Circuit Theory #17 | Supermesh Analysis Examples - Current Source Constraints (3:47)