Calculating Current Using Superposition Theorem

In summary, you would calculate the current in the right-most branch in the following circuit as follows: Attempt at solution. I basically followed the theorem, Both voltage sources in the circuit have the same polarity so the 2 resulting currents calculated will have to be added together. The problem I come to most often with these questions is whether resistors are in parallel or series combinations. This is how far I got, the answer is said to be 845microA. After firstly making VS2 a short, I calculated the total resistance. Rt = R1+R2//R3+R4//R5 = 3200//2000//2200 which I calculated to being 790ohms.( // denotes parallel relationship
  • #1
shifty101uk
17
0
1. Using the superposition theorem, calculate the the current in the right-most branch in the following circuit.
http://img329.imageshack.us/img329/1190/superpositionmp1.png



Attempt at solution. I basically followed the theorem, Both voltage sources in the circuit have the same polarity so the 2 resulting currents calculated will have to be added together. The problem I come to most often with these questions is whether resistors are in parallel or series combinations. This is how far I got, the answer is said to be 845microA

After firstly making VS2 a short, I calculated the total resistance

Rt = R1+R2//R3+R4//R5 = 3200//2000//2200 which I calculated to being 790ohms.( // denotes parallel relationship)

It(VS1) = VS1/Rt = 2/790 = 2.53mA

I then applied the Current divider formula to work my way to finding Ir5.

Ir5(VS1) = 2000/4200 x 1.557mA = 741microA

Now when VS1 is a short, I got the total resistance from VS2 to be

Rt = R4//R5//R3+R2//R1 = 1000//2200//3200//1000 = 361.4ohms

It(VS2) = 3/361.2 = 8.3mA

Ir5(VS2) = 1000/3200 x 8.3mA = 2.59mA

Ir5 = Ir5(VS1)+Ir5(VS2) = 2.59mA + 714microA = 3.304mA

I think I have gone wrong mainly in my execution of the Current divider method, I also have doubts over my calculations of Rt in both circumstances. If anyone could provide me with any help I would be very grateful.
 
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  • #2
Not only your current divider is wrong, but your calculation of Rt for Vs2 is also incorrect. R4 is in series with the rest of the circuit and not in parallel.
 
  • #3
Ok thanks, I will give it another shot with that info.
 
  • #4
Ok well I had another go, here's what I got

Make VS2 a short

Rt = R1+R2//R3+R4//R5 = 3200//2000//2200 = 790ohms (I assume this is correct).

It(vs1) = 2/790 = 2.531mA

Therefore IR5 = 790/2200 x 2.531mA = 908microA(is this correct? as I5 = Rt/R5 * It(vs1)).

Make VS1 a short.

Rt = R4+R5//R3+R2//R1 = 615ohms

It(VS2) = Vs2/Rt = 3/615 = 4.87mA
(is the value for Rt while VS1 is made a short correct? Could someone direct me to any tutorials on series parallel relationships? I am going to have a look through the chapter again tomorrow but the more examples I can find the better really.)

IR5 = Rt/R5 x It(vs2) = 615/2200 x 4.87mA = 1.361mA

Ir5(total) = Ir5(VS1)+Ir5(VS2) as they are the same polarity and will be flowing in the same direction.

1.361mA+908microA = 2.269mA(around 5 times larger than the answer given :S)

is this any closer to the correct answer? If someone could tell me where and how I went wrong then Id be very grateful.
 
  • #5
shifty101uk said:
Ok well I had another go, here's what I got

Make VS2 a short

Rt = R1+R2//R3+R4//R5 = 3200//2000//2200 = 790ohms (I assume this is correct).

It(vs1) = 2/790 = 2.531mA

Therefore IR5 = 790/2200 x 2.531mA = 908microA(is this correct? as I5 = Rt/R5 * It(vs1)).

Make VS1 a short.

Rt = R4+R5//R3+R2//R1 = 615ohms

It(VS2) = Vs2/Rt = 3/615 = 4.87mA
(is the value for Rt while VS1 is made a short correct? Could someone direct me to any tutorials on series parallel relationships? I am going to have a look through the chapter again tomorrow but the more examples I can find the better really.)

IR5 = Rt/R5 x It(vs2) = 615/2200 x 4.87mA = 1.361mA

Ir5(total) = Ir5(VS1)+Ir5(VS2) as they are the same polarity and will be flowing in the same direction.

1.361mA+908microA = 2.269mA(around 5 times larger than the answer given :S)

is this any closer to the correct answer? If someone could tell me where and how I went wrong then Id be very grateful.

You forgot some parentheses in your calculatios of Rt.

With Vs2 shorted you have:
Rt = R1+R2||(R3+R4||R5)

and with Vs1 shorted
Rt = R4+R5||(R3+R1||R2)
 
  • #6
Ok well I am starting to get the hang of all this now, thanks a lot for your help. Ill go back and review series parallel relationships as they seem to be my biggest weak spot at the moment.
 

1. What is the Superposition Theorem Problem?

The Superposition Theorem Problem is a method used in electrical circuit analysis to determine the voltage or current at a specific point in a complex circuit. It states that the total voltage or current in a linear circuit is equal to the algebraic sum of individual voltages or currents caused by each source acting separately.

2. How is the Superposition Theorem Problem used?

The Superposition Theorem Problem is used to simplify complex circuits into smaller, more manageable parts. By considering each voltage or current source separately and taking their effects on the circuit into account, the overall behavior of the circuit can be determined. This method is especially useful in circuits with multiple sources and can save time and effort in calculating the final values.

3. What are the limitations of the Superposition Theorem Problem?

The Superposition Theorem Problem can only be applied to linear circuits, meaning that the voltage-current relationship of the components in the circuit must be constant. It also assumes that all sources are independent and can be turned off individually. Additionally, the theorem does not account for any nonlinearities in the circuit, such as diodes or transistors.

4. How do you solve a Superposition Theorem Problem?

To solve a Superposition Theorem Problem, you must follow these steps:

  1. Turn off all but one of the voltage or current sources in the circuit.
  2. Calculate the voltage or current at the desired point in the circuit using traditional circuit analysis methods.
  3. Repeat step 1 and 2 for each source in the circuit.
  4. Add or subtract the individual values obtained in step 2, depending on their direction and polarity, to get the final value at the desired point in the circuit.

5. What are some real-world applications of the Superposition Theorem Problem?

The Superposition Theorem Problem is commonly used in the design and analysis of electronic circuits, such as in the development of power supplies, amplifiers, and communication systems. It is also used in the optimization of electric power systems, such as in determining the voltage and current flow in a power grid. Additionally, the theorem can be applied in the study of biological systems, such as the electrical activity in the heart or brain.

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