Calculate Voltage at Point P - Superposition Help

In summary, you have a circuit with a current source, a voltage source, and two resistors. You need to calculate the voltage drop across the two resistors.
  • #1
192
1
Hi,
I need some help with some circuit analysis.
http://img208.imageshack.us/img208/866/circuitj.png [Broken]

Sorry for the shoddy picture
Problem I need to calculate the voltage between the point P and GND.

So i know that i have to replace the current source with and open circuit
like such:
http://img411.imageshack.us/img411/355/circuit2.png [Broken]

Im not sure how to go about this.
Is it just a matter of using the voltage division principle?
Also I am not really sure about the 12,16 and the 8 ohm resistances, are they all in series?

/James
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
At this stage of your analysis, the 8 ohm resistor is doing nothing; ignore it.

So you have a couple of cascaded voltage dividers. Do you know how to solve that much?
 
  • #3
The Electrician said:
At this stage of your analysis, the 8 ohm resistor is doing nothing; ignore it.

So you have a couple of cascaded voltage dividers. Do you know how to solve that much?

I think so.
The total resistance is ((10+5)*(12+16))/(15+28) = 9,76[tex]\Omega[/tex]
It's a little low, are you sure you don't have to account for the 8 [tex]\Omega[/tex] when calculating the total resistance?

So I am not sure about the total resistance in the circuit but KVL would yield:

I*10 + I*5 = 12 and
I*10 + I*12 + I*16 = 12
 
  • #4
The right end of the 8 ohm resistor isn't connected to anything, so it can't possibly have any effect on the circuit.

Your calculation for the total resistance seen by the 12 volt source is incorrect.

Calculate it like this:

The 12 and 16 ohm resistors can be replaced by a single 28 ohm resistor, which is in parallel with the 5 ohm resistor. Then the 5 ohm resistor can be replaced by a (28*5)/(28+5) = 4.2424 ohm resistor. This is in series with the 10 ohm resistor, so the total resistance seen by the 12 volt source is 14.2424 ohms. The current out of the 12 volt source is therefore 12/14.2424 = .84255 amps. This current causes an 8.4255 volt drop across the 10 ohm resistor, so the voltage at the junction of the 10, 5 and 12 ohm resistors is 3.5745 volts. Then the 12 and 16 ohms resistors act as a voltage divider, so that the voltage at point P is 3.5745*(16)/(12+16) = 2.04255 volts.

Or, you could solve your problem using the loop method.

If you are are going to have 2 KVL equations, then you must have two separate currents. You can't just call all the currents by the single symbol I; you need to use I1 and I2.

You haven't shown which loops you are using, but I think I can guess which they are.

Your two equations should be:

(I1+I2)*10 + (I1)*5 = 12
(I1+I2)*10 + (I2)*(12+16) = 12

which can be rearranged to give:

(10+5)*I1 + (10)*I2 = 12
(10)*I1 + (10+12+16)*I2 = 12

If you solve these, you get I2 (this current passes through the 16 ohm resistor). Its value is .12766 amps. Then the voltage at point P is .12766 * 16 = 2.04255 volts.

This is the first part of your superposition problem. Next you need to replace the 12 volt source with a short and calculate the voltage at point P due to the 5 amp current source, which you add to 2.04255 volts.
 
  • #5
Thank you very much.
Just one last thing regarding the loop method.
So if there were three meshes you'd need to have three equations with three different currents?
 
  • #6
Generally speaking, yes. If there are dependent sources or some other complication, that may change.

Using the nodal method, you will generally need as many equations (and variables) as there are nodes.
 

1. How do I calculate voltage at point P using superposition?

To calculate voltage at point P using superposition, you will first need to determine the voltage contribution from each individual source. This can be done by applying the principle of superposition, which states that the total voltage at P is equal to the sum of the voltage contributions from each source. Simply add up all the voltage contributions from each source to get the total voltage at point P.

2. What is the principle of superposition and how does it apply to calculating voltage at point P?

The principle of superposition states that the total voltage at a point in a circuit is equal to the sum of the voltage contributions from each individual source. This means that when calculating voltage at point P, you can treat each source separately and then add up their individual voltage contributions to get the total voltage at P.

3. Are there any limitations or assumptions when using superposition to calculate voltage at point P?

Superposition is based on the assumption that the sources in the circuit are linear and that they do not interact with each other. This means that the voltage contributions from each source can be added without affecting each other. Additionally, superposition can only be applied to circuits with multiple independent sources.

4. Can superposition be used to calculate voltage at any point in a circuit?

Yes, superposition can be used to calculate voltage at any point in a circuit as long as there are multiple independent sources present. The principle of superposition can be applied to any point in a circuit, not just point P.

5. Is there a specific formula or equation to use when applying superposition to calculate voltage at point P?

There is no specific formula or equation to use when applying superposition to calculate voltage at point P. However, you can use the general formula for voltage (V=IR) and apply it to each individual source to determine its voltage contribution. Then, simply add up all the voltage contributions to get the total voltage at point P.

Suggested for: Calculate Voltage at Point P - Superposition Help

Replies
18
Views
2K
Replies
2
Views
554
Replies
15
Views
1K
Replies
2
Views
851
Replies
1
Views
827
Replies
1
Views
535
Replies
3
Views
713
Back
Top