#10 Nodal analysis example 3 — current and power with three nodes

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

1. 1. Read the circuit and targets

   

   Unknowns: V₁, V₂, V₃

   Target: i=(V₃−V₂)/4

   Power in the red 6 Ω resistor

   Narration transcript

   Here is our most complex example so far — a circuit with three unknown node voltages. Node V one connects to node V three through a six-ohm resistor — highlighted in red — and to node V two through a twelve-ohm resistor at the top. Node V two also connects to V three through a four-ohm resistor, and to ground through a six-ohm resistor on the right. Node V three connects to ground through an eighteen-ohm resistor. A five-ampere current source enters V one from below. Our task: find the current i through the four-ohm resistor, and the power dissipated by the red six-ohm resistor.

2. 2. KCL at node V₁

   

   $(V_{1}-V_{3})/6+(V_{1}-V_{2})/12-5=0$

   Multiply by 12 and simplify

   $3V_{1}-V_{2}-2V_{3}=60$

   Narration transcript

   Let's write KCL at node V one. Three terms. The six-ohm resistor to V three gives V one minus V three, over six. The twelve-ohm resistor to V two gives V one minus V two, over twelve. The five-ampere source enters V one, so we write minus five. Putting it together: V one minus V three over six, plus V one minus V two over twelve, minus five, equals zero. Multiply through by twelve to clear denominators. We get two times V one minus V three, plus V one minus V two, minus sixty, equals zero. Combining: three V one minus V two minus two V three equals sixty. That is our first equation.

3. 3. KCL at node V₂

   

   $(V_{2}-V_{1})/12+(V_{2}-V_{3})/4+V_{2}/6=0$

   Multiply by 12 and simplify

   $-V_{1}+6V_{2}-3V_{3}=0$

   Narration transcript

   KCL at node V two. Three branches, no current source at this node. The twelve-ohm resistor gives V two minus V one, over twelve. The four-ohm resistor to V three gives V two minus V three, over four. The six-ohm resistor to ground gives V two over six. Equation: V two minus V one over twelve, plus V two minus V three over four, plus V two over six, equals zero. Multiply by twelve. We get V two minus V one, plus three times V two minus V three, plus two V two, equals zero. Combining: minus V one plus six V two minus three V three equals zero. Equation two.

4. 4. KCL at node V₃

   

   $(V_{3}-V_{1})/6+(V_{3}-V_{2})/4+V_{3}/18=0$

   Multiply by 36 and simplify

   $-6V_{1}-9V_{2}+17V_{3}=0$

   Narration transcript

   KCL at node V three. Again three branches, no source. The six-ohm resistor gives V three minus V one, over six. The four-ohm resistor gives V three minus V two, over four. The eighteen-ohm resistor to ground gives V three over eighteen. Multiply by thirty-six to clear all denominators. We get six times V three minus V one, plus nine times V three minus V two, plus two V three, equals zero. Combining: minus six V one minus nine V two plus seventeen V three equals zero. Equation three.

5. 5. Solve and verify the system

   

   $45V_{2}-35V_{3}=0 \Rightarrow V_{3}=(9/7)V_{2}$

   $V_{1}=45 V, V_{2}=21 V, V_{3}=27 V$

   Check at V₁: 3 A+2 A=5 A

   Narration transcript

   We now have three equations with three unknowns. Equation one: three V one minus V two minus two V three equals sixty. Equation two: minus V one plus six V two minus three V three equals zero. Equation three: minus six V one minus nine V two plus seventeen V three equals zero. To solve this system we use elimination. Multiply equation two by six and subtract equation three. V one cancels and we get: forty-five V two minus thirty-five V three equals zero. So V three equals nine-sevenths of V two. Substitute back: V two equals twenty-one volts, V three equals twenty-seven volts, and V one equals forty-five volts. Quick verify at V one: forty-five minus twenty-seven over six is three amperes. Forty-five minus twenty-one over twelve is two amperes. Three plus two equals five, matching the five-ampere source. Correct.

6. 6. Branch current and resistor power

   

   $i=(V_{3}-V_{2})/4=(27-21)/4$

   $i=1.5 A$

   $P=(V_{1}-V_{3})^{2}/6=18^{2}/6$

   $P=54 W$

   Narration transcript

   Now for the two quantities the question asks. The current i flows through the four-ohm resistor from V three toward V two. By Ohm's law: i equals V three minus V two, divided by four. That is twenty-seven minus twenty-one over four, which equals one point five amperes. For the power dissipated by the red six-ohm resistor between V one and V three: the voltage across it is V one minus V three, which is forty-five minus twenty-seven, equals eighteen volts. Power equals V squared over R: eighteen squared over six equals fifty-four watts.

7. 7. Method summary

   

   1. Choose nodes 2. Write KCL

   3. Solve the system 4. Verify

   Branch current: ΔV/R

   Resistor power: V²/R

   Narration transcript

   The main takeaway: the four-step nodal method scales directly from two nodes to three or more. The only difference is bigger algebra — three equations instead of two. Once you have the node voltages, Ohm's law gives any branch current, and P equals V squared over R gives any resistor power. In the next video, we'll see one more nodal analysis example.

Source video: Circuit Theory #10 | Nodal Analysis Example 3 — Three Nodes, Finding i and P (6:15)