Engineering How Do You Calculate Currents in a Circuit Using the Superposition Theorem?

AI Thread Summary
The discussion revolves around calculating currents in a circuit using the Superposition Theorem, focusing on two voltage sources, E1 and E2. The user is attempting to determine the current through resistor R2 with each source individually and then combined. Initial calculations for E1 yielded a current of 6.971 mA, but corrections indicated the use of the wrong current divider formula, leading to a revised current of 10.937 mA. For E2, the user was advised to reconsider the circuit configuration, resulting in a current of -18.319 mA, indicating direction. The final current through R2 when both sources are active is the sum of individual currents, calculated as -7.382 mA.
kevinp
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Just found this forum and found some great help here so far just reading some posts. But I am stuck on a question in my homework. I am asking you to help check my work and make sure I am going in the right direction. I missed my lecture due to illness and have to figure this out on my own. Thank you in advance.

The question asks for the 3 different currents through R2; With E1 as the source only, E2 as the source only, and both E1 and E2.

Here are the variables:
E1= 11V
E2= 8V
R1= 430ohms
R2= 250ohms
R3= 200ohms
R4= 130ohms

Here is the diagram

image001.gif


This is my work with E1 as the only source Voltage

Rt= R1 + ((R2x(R3+R4)) / R2 + R3 + R4)
Rt= 572.2414ohms
Rt (of parallel circuit only is 142.2414ohms)

It= E/Rt = 11/572...
It= 19.223mA

Based on KVL, the current of It = R1 = Ra (the parallel circuit of R2 and (R3 + R4 in series)

Using the current divider, I got

I2= Rt (of parallel) / Rt(prl) + R2 * It
I2 = 6.971mA

So far so good?

With E2 as the only source voltage, I made R1 and R2 the parallel circuit and had the branch of R3, R4 and R2 as my series circuit

Rt = R4 + R3 + (R1xR2 / (R1+R2))
Rt= 488.0882ohms
Rt of parallel = 158.0882ohms

It = E/Rt
It = 16.3905 mA

Current divider rule

I2 = 158.0882 / (158.0082 + 250) ohms * 16.3905mA
I2 = 6.3495

I do not have the answers to this question since its a online test and each section is worth 1% of my final mark. I would like you to help make sure I am doing this question right so I can move onto the other ones.

Where I get stuck is how to find the answer with both voltage sources.

I assume based on the fact there series adding that I would add them together for a total of 13.3205mA but this is where I am unsure.

On top of that, for my future questions, how would I find Rt of the entire circuit with both sources? For the current total, since its a series adding circuit, could I add both to get 3X.XX mA for the It. If that's the case I could find Rt using ohms law, but any help is much appreciated.

Thank you
 
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kevinp said:
It= E/Rt = 11/572...
It= 19.223mA

You got this right.

kevinp said:
Based on KVL, the current of It = R1 = Ra (the parallel circuit of R2 and (R3 + R4 in series)

Using the current divider, I got

I2= Rt (of parallel) / Rt(prl) + R2 * It
I2 = 6.971mA

So far so good?

You've got the current divider wrong. The divider fraction should be:

(R3+R4)/(R2+R3+4)

so that I2 = It * (R3+R4)/(R2+R3+4) = 10.937 mA

kevinp said:
With E2 as the only source voltage, I made R1 and R2 the parallel circuit and had the branch of R3, R4 and R2 as my series circuit

Rt = R4 + R3 + (R1xR2 / (R1+R2))
Rt= 488.0882ohms
Rt of parallel = 158.0882ohms

It = E/Rt
It = 16.3905 mA

Current divider rule

I2 = 158.0882 / (158.0082 + 250) ohms * 16.3905mA
I2 = 6.3495

You need to make the parallel circuit R1 in parallel with (R3+R4); this is then in series with R2. Redo your calculations. You don't need to use a current divider rule here because I2 is It when only E2 is active.

With E2 only active, I get a current I2 = -18.319 mA. The minus sign means the current is directed upward.

Yes, the final answer will be the sum of the two individual currents, which I get as -7.382 mA.
 

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