Circuit Theory 1 · Supermesh Analysis

#19 Supermesh #19 — dependent current source: find V_a

Relates the dependent current source to the mesh variables, then combines source constraints with outer-supermesh KVL to find V_a.

Question

Three-mesh circuit with 4 kΩ and 12 kΩ resistors, a 24 V source, and independent and dependent current sources.
Clockwise mesh currents i_1, i_2 and i_3, the V_a polarity, and the 2V_a mA dependent source are shown.

Use supermesh analysis to find voltage V_a in the circuit containing an independent and a dependent current source.

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. Dependent-source recap

    Three-mesh circuit with 4 kΩ and 12 kΩ resistors, a 24 V source, and independent and dependent current sources.
    Clockwise mesh currents i_1, i_2 and i_3, the V_a polarity, and the 2V_a mA dependent source are shown.

    Dependent-source supermesh

    Same method; one extra dependency relation

    Find: Va

    Narration transcript

    We have now used supermesh analysis when a current source blocks a direct K V L path. The next twist is a dependent current source. The method is the same, but one source value is not a fixed number. It depends on a circuit variable, so we must write one extra relation before solving.

  2. 2. Read the circuit

    Three-mesh circuit with 4 kΩ and 12 kΩ resistors, a 24 V source, and independent and dependent current sources.
    Clockwise mesh currents i_1, i_2 and i_3, the V_a polarity, and the 2V_a mA dependent source are shown.

    4 kΩ | 12 kΩ | 24 V

    Independent source: 12 mA ↑

    Dependent source: 2Va mA ↓

    i1, i2, i3: clockwise

    Narration transcript

    In this circuit, the target is V a. The left resistor is 4 kilo ohms, the right resistor is 12 kilo ohms, and the top voltage source is 24 volts with plus on the V a side. The left current source is 12 milliamps upward. The middle diamond source points downward and has value 2 times V a milliamps. We choose clockwise mesh currents i one, i two, and i three.

  3. 3. Build and solve the equations

    Three-mesh circuit with 4 kΩ and 12 kΩ resistors, a 24 V source, and independent and dependent current sources.
    Clockwise mesh currents i_1, i_2 and i_3, the V_a polarity, and the 2V_a mA dependent source are shown.

    Current-source constraints

    i2−i1=12 mA

    i2−i3=2Va mA

    Va=−4i1VV_{\mathrm{a}}=-4i_{1} V

    8i1+i2−i3=08i_{1}+i_{2}-i_{3}=0

    −4i1−12i3=24-4i_{1}-12i_{3}=24

    i1=−1.5 mA, i2=10.5 mA, i3=−1.5 mA

    Va=−4(−1.5)=6VV_{\mathrm{a}}=-4(-1.5)=6 V

    Narration transcript

    Start with the two current-source constraints. For the 12 milliamp source, the shared-branch relation is i two minus i one equals 12 milliamps. For the dependent source, the relation is i two minus i three equals 2 V a milliamps. Now relate V a to the left mesh current. With the chosen polarity, V a equals negative 4 i one volts, because 4 kilo ohms times i one milliamps gives volts. Substituting this into the dependent-source constraint gives i two minus i three equals negative 8 i one, or 8 i one plus i two minus i three equals zero. The outer supermesh K V L equation skips the two current-source branches and gives negative 4 i one minus 12 i three equals 24. Solving these three equations gives i one equals negative 1.5 milliamps, i two equals 10.5 milliamps, and i three equals negative 1.5 milliamps. Finally, V a equals negative 4 times negative 1.5, so V a is 6 volts.

  4. 4. Method summary

    Three-mesh circuit with 4 kΩ and 12 kΩ resistors, a 24 V source, and independent and dependent current sources.
    Clockwise mesh currents i_1, i_2 and i_3, the V_a polarity, and the 2V_a mA dependent source are shown.

    Dependent source ⇒ extra relation

    Current constraints + outer-supermesh KVL

    Va=6VV_{\mathrm{a}}=6 V

    Narration transcript

    A dependent source does not change the supermesh workflow. It only adds one bookkeeping step: express the dependent source value using the chosen mesh variables. Here, the key link was V a equals negative 4 i one. After that, the current-source constraints and the outer K V L equation formed a normal three-equation system. The final answer is V a equals 6 volts.

Source video: Circuit Theory #19 | Dependent-Source Supermesh Analysis - Find Va (2:39)