# Find the power supplied by the 1V source in the circuit using nodal analysis

Nodal analysis

## Homework Equations:

I in = I out
My attempt:

At node 1 > 4 = 6 + I1
Node 2 > I1 + I2 = I 2
Node 3 > 6 = I 3 + I 4

I1 = (V1 - 4)/21

In node 1,
Using the equation for I1, V1 = -38V.

I2 = (V1 - 4 - 1)/2
So, I2 = -21.5A

Power supplied by the 1V battery is
P=VI
P=-21.5 x 1
P=-21.5W

I am not sure of my method, I just started with Circuit analysis class.
Thanks.

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lewando
Homework Helper
Gold Member
Edited (and repaired) for clarity:
Node 1 > 4 = 6 + I1
Node 2 > I1 + I3 = I2
Node 3 > 6 = I3 + I4
There is another node to consider.

This is a problem:
I2 = (V1 - 4 - 1)/2
It implies that V2 = V1 - 4 (which you indicated in red in one of your attachments) which is not correct.

Is there a way to approach a question like this, I really have problems with voltage source in between nodes.

lewando
Homework Helper
Gold Member
For a voltage across two nodes you have to consider all elements and their voltage change contributions. In your case, you have not included the voltages across the 14 and 7 ohm resistors, as a function of I1.

lewando
Homework Helper
Gold Member
One other important thing-- when you indicate a node voltage: V1, V2, or V3, you need to establish a reference point for that particular voltage. Commonly, a 0 V reference point or "ground" reference is identified for the circuit. You don't show one on your circuit.

gneill
Mentor
Is there a way to approach a question like this, I really have problems with voltage source in between nodes.
When there is a voltage source along with other components (such as resistors) in a branch between two nodes, as is the case in the following excerpt from your circuit,

then you can consider that the source raises or lowers the effective potential of one of the nodes depending upon the source orientation. So in the above scenario if we want to write an expression for $i_1$ for node $V_1$ we can write:

$i_1 = \frac{V_1 - (V_2 + 4)}{14 + 7}$

or

$i_1 = \frac{(V_1 - 4) - V_2}{14 + 7}$

When there is a voltage source along with other components (such as resistors) in a branch between two nodes, as is the case in the following excerpt from your circuit,
View attachment 242541
then you can consider that the source raises or lowers the effective potential of one of the nodes depending upon the source orientation. So in the above scenario if we want to write an expression for $i_1$ for node $V_1$ we can write:

$i_1 = \frac{V_1 - (V_2 + 4)}{14 + 7}$

or

$i_1 = \frac{(V_1 - 4) - V_2}{14 + 7}$
Thank you, I get it now!

gneill
Mentor
Thank you, I get it now!
Fantastic! Cheers.