# Node-voltage method problem, to find voltage.

• Zarathuztra
In summary, the conversation discusses using the node-voltage method to find the voltage output and power delivered by a current source in a circuit. It provides instructions for using different reference nodes and discusses the use of equations and principles such as Kirchoff's Laws and Ohm's Law. The conversation also includes a problem-solving process and an explanation for a mistake made in the solution.
Zarathuztra

## Homework Statement

Use the node-voltage method to find vo and the power delivered by the current source in the circuit in figure(Figure 1) if the v = 30V and the i = 1.6A.
#1: Use the node a as the reference node to find vo
#2: Use the node b as the reference node to find vo
#3: Find the power delivered by the 1.6A current source

I am stuck on the first one at the moment.

http://imageshack.us/a/img152/2156/eeehomework04p4.png Left is the Given with my adjustment notes, right is modified result.

## Homework Equations

Kirchoff's Volt Law
Kirchoff's Current Law
Ohm's Law
Node-voltage method
supernode & mesh analysis

## The Attempt at a Solution

In the pic above you can see I combined the series resistances, 20+55=75. Then I combined the parallel resistors, 150 and the new 75 to get (150*75)/(150+75)=50. This problem instructs me to use node a as the reference, so I put the ground/ref. there for #1.

This is when I setup my node-voltage method equation.
node b: 1.6 + (1/50)Vb + (1/50)(Vb - 30) = 0

This was the only equation I was able to make, and it gives me 0 for Vb which is wrong... I'm probably handling the voltage source incorrectly in the problem.. Is the modified picture correct? If so, did I make an error with my equation for Vb, or should I have a 2nd equation from node C?

Last edited by a moderator:
Your node b equation looks okay. Perhaps an algebra problem while solving for Vb?

yup, that was the problem, after correcting it, I got -25 for Vb

Zarathuztra said:
yup, that was the problem, after correcting it, I got -25 for Vb

Yup. That looks fine

for the 2nd part, with node b as reference, will I get 2 equations this time or just 1 again? I'm getting:
node a: -1.6 + (1/50)Va + (1/50)(Va-30) = 0

Zarathuztra said:
for the 2nd part, with node b as reference, will I get 2 equations this time or just 1 again? I'm getting:
node a: -1.6 + (1/50)Va + (1/50)(Va-30) = 0

Two things. First, note the polarity of Va (how it's "measured" as per the schematic). The Va you've written into your equation has the opposite polarity. Second, note that the voltage source is increasing the potential between node a and the reference node b; you've written it as decreasing the potential difference.

EDIT: Never mind! I was thinking that you were solving for Vo, but you've in fact written a correct equation for the node potential Va with respect to node b except for the slip with the voltage source. You can translate Va to Vo by noting their relative polarities on the schematic.

Last edited:
so to correct my eq for that...

-1.6 - (1/50)Va + (1/50)(Va + 30) = 0

That should adjust to the polarity, and the Voltage gain from the source.

Zarathuztra said:
so to correct my eq for that...

-1.6 - (1/50)Va + (1/50)(Va + 30) = 0

That should adjust to the polarity, and the Voltage gain from the source.

D'oh! See my edit above.

For the third portion, I thought I should be summing up the power across the resistors, I know Vo is -25V so that's the one, and the V over the other resistor should be Vb-30, which is -55V

So, 25^2/50 + 55^2/50 = 73W ... This isn't correct, am I calculating it wrong or am I missing an additional source of power?

Zarathuztra said:
For the third portion, I thought I should be summing up the power across the resistors, I know Vo is -25V so that's the one, and the V over the other resistor should be Vb-30, which is -55V

So, 25^2/50 + 55^2/50 = 73W ... This isn't correct, am I calculating it wrong or am I missing an additional source of power?

The voltage source is another possible source (or sink) for power.

If you have the potential across the current source, and if you know its current, then you can determine the power it delvers or sinks directly. P = I*V. It's delivering power if the current emerges from its more positive terminal.

## 1. How do I apply the node-voltage method to a circuit?

To apply the node-voltage method, you must first identify all the nodes in the circuit and assign a variable for the voltage at each node. Then, write Kirchhoff's Current Law equations for each node and solve the resulting system of equations to find the node voltages.

## 2. Can I use the node-voltage method for any circuit?

Yes, the node-voltage method can be used to solve any type of circuit, including circuits with voltage sources, current sources, and complex components such as capacitors and inductors.

## 3. Do I need to use the node-voltage method if I already know the voltage at some nodes?

No, if the voltage at some nodes in the circuit is known, you can use those values to simplify the circuit and then use other methods, such as the branch-current method, to solve for the remaining node voltages.

## 4. How do I handle dependent voltage sources in the node-voltage method?

Dependent voltage sources can be handled by writing a KCL equation at the node where the source is connected, using the dependent source's controlling variable as one of the currents. Then, the resulting equation can be used in the system of equations to solve for the node voltages.

## 5. Can I use the node-voltage method for circuits with multiple voltage sources?

Yes, the node-voltage method can be used for circuits with multiple voltage sources. Each voltage source will add an additional equation to the system, and the resulting system can be solved using standard methods such as substitution or elimination.

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