#11 Nodal analysis example 4 — voltage source connected to ground

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

1. 1. What does a voltage source change?

   

   New element: a 6 V source tied to ground

   Its upper terminal is a known +6 V node

   It adds no unknown

   Narration transcript

   In our previous examples, every circuit had only resistors and current sources. That made nodal analysis straightforward — resistor currents are simply voltage over resistance, and current sources plug directly into KCL. Today, a new element appears: a voltage source. The question is, how does a voltage source affect the node method? Let's find out.

2. 2. Read the circuit and known node

   

   Only unknown node: V₁

   $V_{\mathrm{a}}=V_{1}$

   Right resistor lies between V₁ and the known 6 V node

   Narration transcript

   Here is the circuit. A one-milliamp current source on the left pushes current upward into the top node. Three resistors connect that top node to the bottom ground rail: one kilohm on the left, three kilohms in the middle, and two kilohms on the right. But notice the right branch: the two-kilohm resistor is in series with a six-volt source whose positive terminal sits at six volts above ground. This means the junction between the two-kilohm resistor and the six-volt source has a known voltage — it is simply six volts. So we still have only one unknown node at the top, which we call V one. V sub a is measured across the one-kilohm resistor, from V one to ground, so V sub a equals V one. Our task: find V sub a.

3. 3. KCL at node V₁

   

   $V_{1}/1k+V_{1}/3k+(V_{1}-6)/2k-1m=0$

   Multiply by 6k: 6V₁+2V₁+3(V₁−6)−6=0

   $11V_{1}=24$

   Narration transcript

   Let's write KCL at node V one. We sum all currents leaving through resistors and subtract the current entering from the source. The one-kilohm branch carries V one over one k. The three-kilohm branch carries V one over three k. The two-kilohm branch: the far end is at six volts, so the current leaving V one is V one minus six, over two k. The one-milliamp source enters V one, so we write minus one m. Setting the sum to zero: V one over one k, plus V one over three k, plus V one minus six over two k, minus one m, equals zero. To clear the denominators, multiply everything by six k. We get six V one, plus two V one, plus three times V one minus six, minus six, equals zero. Combining: eleven V one minus twenty-four equals zero. So eleven V one equals twenty-four.

4. 4. Find and verify Vₐ

   

   $V_{1}=24/11\approx 2.182 V$

   $V_{\mathrm{a}}=V_{1}\approx 2.182 V$

   2.182+0.727−1.909=1 mA

   Narration transcript

   Dividing both sides by eleven: V one equals twenty-four over eleven, which is approximately two point one eight two volts. Since V sub a is measured from V one to ground, V sub a equals V one, which is also two point one eight two volts. Let's verify. V one over one k is about two point one eight milliamps. V one over three k is about zero point seven three milliamps. V one minus six over two k is about negative one point nine one milliamps. Adding: two point one eight plus zero point seven three minus one point nine one equals one milliamp, which matches the source. Correct.

5. 5. Grounded-source rule

   

   One source terminal at ground ⇒ other node known

   Write KCL only at unknown nodes

   Source between two unknown nodes ⇒ supernode

   Narration transcript

   Here is the key insight from this example. When a voltage source has one terminal connected to the reference node — ground — the other terminal's voltage is immediately known. It does not add an unknown; it actually reduces the number of unknowns. But what happens when a voltage source sits between two non-reference nodes, where neither voltage is known? That creates a special structure called a supernode — and that is exactly what we will tackle in the next video.

Source video: Circuit Theory #11 | Nodal Analysis Example 4 — Voltage Source to Ground (3:44)