#16 Mesh currents #16 — equivalent resistance and a dependent source

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

1. 1. Roadmap for two examples

   

   Two new mesh-current examples

   Example 2: R_eq from a test source

   Example 3: dependent source and i_x

   Narration transcript

   In the previous lesson, we introduced mesh currents with one complete worked example. Now we continue with two more examples. The first one uses a 12 volt test source to find an equivalent resistance. The second one includes a dependent voltage source, so the controlling current must stay visible until the end. The workflow is the same in both cases: choose clockwise mesh currents, write K V L, solve the linear system, then interpret the requested current or resistance.

2. 2. Example 2 circuit

   

   V_test=12 V; all five resistors are 10 Ω

   $i=i_{1}$

   $R_{\mathrm{e}}q=12/i_{1}$

   Narration transcript

   Example two asks for the equivalent resistance seen from terminals a and b. We apply a 12 volt test source at the terminals and define the input current i. All five resistors are 10 ohms. There are three meshes, so we choose i one, i two, and i three clockwise. Because the current through the source branch is i one, the equivalent resistance will be 12 divided by i one.

3. 3. Solve Example 2

   

   $20i_{1}-10i_{2}-10i_{3}=12$

   $-10i_{1}+30i_{2}-10i_{3}=0$

   $-10i_{1}-10i_{2}+30i_{3}=0$

   $i_{1}=1.2 A, i_{2}=i_{3}=0.6 A$

   $R_{\mathrm{e}}q=10 \Omega$

   Narration transcript

   The three mesh equations are written in positive diagonal form. Loop one gives 20 i one minus 10 i two minus 10 i three equals 12. Loop two gives negative 10 i one plus 30 i two minus 10 i three equals zero. Loop three gives negative 10 i one minus 10 i two plus 30 i three equals zero. Solving this system gives i one equals 1.2 amperes, while i two and i three are each 0.6 amperes. So the equivalent resistance is 12 volts divided by 1.2 amperes, which is 10 ohms.

4. 4. Example 3 circuit

   

   Independent source: 50 V

   Dependent source: 15i_x

   $i_{\mathrm{x}}=i_{1}-i_{3}$

   Find: i_x

   Narration transcript

   Example three adds one new ingredient: a dependent voltage source. The left source is 50 volts. The right dependent source has value 15 times i x. The controlling current i x is drawn downward through the 20 ohm resistor. With the mesh currents shown, that branch current is i one minus i three. This relation is the key point; if we lose it, the dependent source equation becomes unclear.

5. 5. Equations for Example 3

   

   $25i_{1}-5i_{2}-20i_{3}=50$

   $-5i_{1}+10i_{2}-4i_{3}=0$

   $-5i_{1}-4i_{2}+9i_{3}=0$

   Narration transcript

   Now write the equations. Loop one contains the 50 volt source, the 5 ohm resistor, and the 20 ohm shared resistor. It gives 25 i one minus 5 i two minus 20 i three equals 50. Loop two is the upper mesh with the 1 ohm, 5 ohm, and 4 ohm resistors. It gives negative 5 i one plus 10 i two minus 4 i three equals zero. Loop three includes the dependent source. After substituting i x equals i one minus i three, the simplified equation is negative 5 i one minus 4 i two plus 9 i three equals zero.

6. 6. Solve Example 3

   

   $i_{1}=29.6 A$

   $i_{2}=26 A, i_{3}=28 A$

   $i_{\mathrm{x}}=i_{1}-i_{3}=1.6 A$

   Physical direction: downward

   Narration transcript

   Solving the three equations gives i one equals 29.6 amperes, i two equals 26 amperes, and i three equals 28 amperes. The question asks for i x, not directly for the mesh currents. So we return to the controlling current relation: i x equals i one minus i three. That is 29.6 minus 28, so i x equals 1.6 amperes. The result is positive, which means the actual current direction is the same as the downward arrow in the circuit.

7. 7. Method summary

   

   Test source ⇒ R_eq=V_test/I_test

   Shared resistor ⇒ difference of neighboring mesh currents

   Dependent source ⇒ express its control variable using mesh currents

   Narration transcript

   The important lesson is not just the final numbers. In example two, a test source turned equivalent resistance into a mesh-current problem. In example three, the dependent source forced us to keep the controlling current relation i x equals i one minus i three. For mesh analysis, shared resistors always create difference terms. Dependent sources are handled the same way, as long as the controlling variable is written in terms of the mesh currents.

Source video: Circuit Theory #16 | Mesh-Current Examples — Equivalent Resistance + Dependent Source (4:15)