Equivalent Resistance in series-parallel circuit

In summary, the student is attempting to solve an equation for equivalent resistance in series-parallel, but is having difficulty. In addition, the student has attempted to solve the equation, but found it difficult.
  • #1
JustMerc
3
0

Homework Statement


This is a sample problem. I'm trying to figure it out before I do the actual homework. The issue is, the way to figure it out is not included in our notes. I'm absolutely lost (math isn't my strong point) I've also included my attempt at how I think it should be done below, which got my really close. We're rounding to the nearest whole number, but I'm still off by 2. If it's hard to see: R1 = 2900 ohms, R3 = 800 ohms, Req = 3490 ohms. Find R2 in ohms.
Math126_Sample2.jpg


Homework Equations


The only equations given to us for equivalent resistance in series-parallel is GT = G1 + G2 and Rx = 1/GT which is really not helping to figure this out.

The Attempt at a Solution


Here's my attempt. I subtracted R1 from RT (590 or .001694915) and divided by 1. Then subtracted R3 (.00125) from that (.000444915). Divided by 1 and got within 2 of the answer after rounding (2248). I'm so lost on this.
attempt.jpg
 
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  • #2
I got the same thing as you. If you're given that the answer is 2250, it's likely that it simply has to do with the significant figures. Either way, I'd say you're correct.
 
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  • #3
So then I'm actually doing it correctly? The problem with that is, the homework is online. There's only one answer which works. If you're close...it's still wrong and there's no partial credit, because we don't show our work. Regardless to that fact, I appreciate you looking at it and leading me to believe I'm on the right path.
 
  • #4
Agreed. I got 2247.6, which rounds to your answer.
 
  • #5
BTW, a better, more visual way to do this problem is like this:

[tex]\frac{1}{\frac{1}{x}+\frac{1}{800}} = 590[/tex]

Then put the two terms in the denominator over a common denominator, and solve for x. :-)
 
  • #6
berkeman said:
BTW, a better, more visual way to do this problem is like this:

[tex]\frac{1}{\frac{1}{x}+\frac{1}{800}} = 590[/tex]

Then put the two terms in the denominator over a common denominator, and solve for x. :)

Nice, thank you berkeman! Even though I couldn't get the answer (2250) on the sample question, I attempted the homework and thus far it is working. I appreciate the help from you all. It's not fun going into this stuff blindly.
 
  • #7
When you turn in written work, do yourself a favor and don't write stuff like R3 = 800 = 0.00125 because because 800 doesn't equal 0.00125 and R3 isn't 0.00125 ohms. Chances are whoever is grading your work pays attention to mathematical detail and may find what you wrote irritating, and it's best not to irritate the grader.
 
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Related to Equivalent Resistance in series-parallel circuit

What is equivalent resistance in a series-parallel circuit?

Equivalent resistance in a series-parallel circuit is the single resistance value that can replace all the resistors in the circuit and produce the same current flow.

How do you calculate equivalent resistance in a series-parallel circuit?

To calculate equivalent resistance in a series-parallel circuit, you can use the following formula: 1/Req = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn, where Req is the equivalent resistance and R1, R2, R3, etc. are the individual resistances in the circuit.

What is the difference between series and parallel circuits?

In a series circuit, the components are connected in a single loop, so the current flowing through each component is the same. In a parallel circuit, the components are connected in multiple branches, so the current is divided between them. Additionally, in a series circuit, the total resistance is equal to the sum of the individual resistances, while in a parallel circuit, the total resistance is less than the smallest individual resistance.

How do you determine the total current in a series-parallel circuit?

To determine the total current in a series-parallel circuit, you can use Ohm's Law (I = V/R), where I is the current, V is the voltage, and R is the equivalent resistance of the circuit.

What are some practical applications of equivalent resistance in series-parallel circuits?

Equivalent resistance in series-parallel circuits is used in many practical applications, such as in household electrical wiring, electronic devices, and circuit designs for various appliances. It also helps in designing efficient and effective electrical systems that can handle high currents and voltages while minimizing power loss.

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