Circuit Theory 1 · Supermesh Analysis

#18 Supermesh #18 — Example 3: three mesh currents

Combines the shared current-source constraint with lower-supermesh and top-loop KVL equations.

Question

Three-mesh circuit with 100 V and 50 V voltage sources and an upward 5 A current source in the shared branch.
Clockwise mesh currents i_1, i_2 and i_3 are shown; the middle source gives the constraint i_3−i_1=5 A.

Use the current-source constraint and supermesh analysis to find mesh currents i_1, i_2 and i_3 in the three-mesh 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. Supermesh recap

    Three-mesh circuit with 100 V and 50 V voltage sources and an upward 5 A current source in the shared branch.
    Clockwise mesh currents i_1, i_2 and i_3 are shown; the middle source gives the constraint i_3−i_1=5 A.

    Three mesh currents, one shared current source

    Write the source constraint first

    Then lower-supermesh KVL + top-loop KVL

    Narration transcript

    In the previous lesson, we used current-source constraints in mesh analysis. Now we finish that idea with one compact example that has three mesh currents. The middle 5 ampere current source sits between the lower-left and lower-right meshes, so it creates a supermesh constraint. Then we still need two voltage equations: one around the lower supermesh, and one around the top loop.

  2. 2. Example 3 circuit

    Three-mesh circuit with 100 V and 50 V voltage sources and an upward 5 A current source in the shared branch.
    Clockwise mesh currents i_1, i_2 and i_3 are shown; the middle source gives the constraint i_3−i_1=5 A.

    i1, i2, i3: clockwise

    Middle source: 5 A ↑

    Arrow agrees with i3 and opposes i1

    i3−i1=5Ai_{3}-i_{1}=5 A

    Narration transcript

    The circuit has a 100 volt source on the left, a 50 volt source on the right, and a 5 ampere source in the middle branch. All three mesh currents are chosen clockwise: i one in the lower-left window, i two in the top window, and i three in the lower-right window. Because the middle source arrow points upward, it agrees with i three and opposes i one on that shared branch. Therefore the source constraint is i three minus i one equals 5 amperes.

  3. 3. Build and solve the equations

    Three-mesh circuit with 100 V and 50 V voltage sources and an upward 5 A current source in the shared branch.
    Clockwise mesh currents i_1, i_2 and i_3 are shown; the middle source gives the constraint i_3−i_1=5 A.

    Lower supermesh: bypass the current-source branch

    100−3(i1−i2)−2(i3−i2)−50−4i3−6i1=0100-3(i_{1}-i_{2})-2(i_{3}-i_{2})-50-4i_{3}-6i_{1}=0

    9i1−5i2+6i3=509i_{1}-5i_{2}+6i_{3}=50

    Top loop: 10 Ω, 2 Ω and 3 Ω

    3i1−15i2+2i3=03i_{1}-15i_{2}+2i_{3}=0

    i1=7/4A,i2=5/4A,i3=27/4Ai_{1}=7/4 A, i_{2}=5/4 A, i_{3}=27/4 A

    Check: i3−i1=5 A ✓

    Narration transcript

    First write the lower supermesh equation, skipping the current-source branch. Starting through the 100 volt source as a rise gives 100 minus 3 times the quantity i one minus i two, minus 2 times the quantity i three minus i two, minus 50, minus 4 i three, minus 6 i one equals zero. This simplifies to 9 i one minus 5 i two plus 6 i three equals 50. For the top loop, only the 10 ohm, 2 ohm, and 3 ohm resistors appear. The equation is negative 10 i two, minus 2 times the quantity i two minus i three, minus 3 times the quantity i two minus i one equals zero. So the third equation is 3 i one minus 15 i two plus 2 i three equals zero. Solving these three equations gives i one equals 7 over 4 amperes, i two equals 5 over 4 amperes, and i three equals 27 over 4 amperes. The check is simple: i three minus i one is 27 over 4 minus 7 over 4, which equals 5 amperes.

  4. 4. Method summary

    Three-mesh circuit with 100 V and 50 V voltage sources and an upward 5 A current source in the shared branch.
    Clockwise mesh currents i_1, i_2 and i_3 are shown; the middle source gives the constraint i_3−i_1=5 A.

    Shared current source ⇒ current constraint

    KVL ⇒ supermesh path that does not cross the source

    i1=1.75A,i2=1.25A,i3=6.75Ai_{1}=1.75 A, i_{2}=1.25 A, i_{3}=6.75 A

    Narration transcript

    This example is a good pattern to remember. A current source between two meshes gives a current constraint, not a voltage drop. The K V L equation must go around a path that does not cross that current source. After that, ordinary loop equations can still be used for the remaining meshes. For this circuit, those three equations lead to i one equals 1.75 amperes, i two equals 1.25 amperes, and i three equals 6.75 amperes.

Source video: Circuit Theory #18 | Supermesh Analysis Example 3 - Three Mesh Currents (3:01)