Simple Nodal Analysis Problem Solution and Explanation

[Magnawolf]

Homework Statement

Homework Equations

KVL and KCL

The Attempt at a Solution

Let V = V1 and the Voltage across the 2 ohm resistor be V2 just for reference.

Step 1: KCL @ Supernode: V1/3 + V2/2 + 21/4 + V3/6 = 0

I think my problem is here somewhere. Does V3 = V2 because they're in parallel? Also, am I wrong to say that 21/4 is a current going out of the supernode?

Step 2: V2-V1 = 9 (pretty sure this is right)

I can't get the correct answer. Any help is appreciated.

 

[gneill]

Homework Statement

Homework Equations

KVL and KCL

The Attempt at a Solution

Let V = V1 and the Voltage across the 2 ohm resistor be V2 just for reference.

Step 1: KCL @ Supernode: V1/3 + V2/2 + 21/4 + V3/6 = 0

I think my problem is here somewhere. Does V3 = V2 because they're in parallel? Also, am I wrong to say that 21/4 is a current going out of the supernode?

Step 2: V2-V1 = 9 (pretty sure this is right)

I can't get the correct answer. Any help is appreciated.

The 2 and 6 Ohm resistors are in parallel so they must share the same potential difference. If V1 is the potential at the top of the 3 Ohm resistor then the potential difference across the 4 Ohm resistor must be V1 - 21.

If you make a supernode of the nodes either side of the 9 V source, then really you need only one voltage variable to write the equation. Suppose you chose v to be that variable (after all it's already specified in the circuit diagram). Then on one side of the source the potential is v and on the other side it's v+9. No need to introduce V2 or V3 etc.

 

[NascentOxygen]

Hi. Your V3 is V2, it's the one and the same node. Keep it as just V2.

The current through the 4Ω is given by (voltage across the 4Ω resistor itself) / 4

So what is the voltage across that 4 ohm resistor? Hint: the expression will involve a subtraction

 

[Magnawolf]

If V1 is the potential at the top of the 3 Ohm resistor then the potential difference across the 4 Ohm resistor must be V1 - 21.

This. This is what I needed. Thanks so much!