The equivalent resistance of the circuit

In summary, to find the equivalent resistance of a circuit, you should follow a step by step method of redrawing the circuit with equivalent resistors and combining them in series or parallel until you reach the final resistance. This method will help you understand the circuit and learn how to solve similar problems in the future.
  • #1
roinujo1
41
1

Homework Statement


Find the equivalent resistance of the circuit shown, for circuit 3.7b, the one in the upper right :
https://scontent-ort2-1.xx.fbcdn.net/v/t34.0-12/14429160_641550399338143_303002326_n.png?oh=ad8d346cfcaff147c9cd7a9d2b509411&oe=57E6C47C

Homework Equations


Rseries=Rtotal =R1+R2+...+RN
Rparallel=(Rtotal)-1=(R1)-1+(R2)-1+...

The Attempt at a Solution


So for my attempt, my assumption was that:
  • The 18 and 12 Ω resistors were in series, so I combined them.
  • I assumed the 20 Ω resistor at the top was in parallel with the 60Ω resistor under it and combined that.
  • Then, the equivalent resistance of the 20 and 60 Ω resistors was in series with the 40 Ω resistor.
Now, I am stuck and confused on where to continue. I was going to assume that the equivalent resistor of 55 Ω(from 60,20, and 40) was in parallel with the diagonal 50 Ω resistor and go from there, but it just seemed wrong. Please, can anyone tell me if I am on the right track or I messed up somewhere?
 
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  • #2
roinujo1 said:

Homework Statement


Find the equivalent resistance of the circuit shown, for circuit 3.7b, the one in the upper right :
https://scontent-ort2-1.xx.fbcdn.net/v/t34.0-12/14429160_641550399338143_303002326_n.png?oh=ad8d346cfcaff147c9cd7a9d2b509411&oe=57E6C47C

Homework Equations


Rseries=Rtotal =R1+R2+...+RN
Rparallel=(Rtotal)-1=(R1)-1+(R2)-1+...

The Attempt at a Solution


So for my attempt, my assumption was that:
  • The 18 and 12 Ω resistors were in series, so I combined them.
  • I assumed the 20 Ω resistor at the top was in parallel with the 60Ω resistor under it and combined that.
  • Then, the equivalent resistance of the 20 and 60 Ω resistors was in series with the 40 Ω resistor.
Now, I am stuck and confused on where to continue. I was going to assume that the equivalent resistor of 55 Ω(from 60,20, and 40) was in parallel with the diagonal 50 Ω resistor and go from there, but it just seemed wrong. Please, can anyone tell me if I am on the right track or I messed up somewhere?

You have the two equations in "Relevant Equations". Now, how can you tell if two resistors are in series or parallel? If they are in series, then the one terminal point of one is directly connected to one terminal of the other. If they are connected in parallel, they have two common terminals.

Starting from the right part of the circuit, the 18 and 12 Ω resistors are in series as you say. Now, it would be a good idea to sketch a diagram with a new resistor to substitute these two. You keep on moving from right to left across the circuit. Do you see immediately which resistors are to be taken into account?
 
  • #3
You cannot take shortcuts. The 20 Ω resister is in parallel with the 60 Ω that is still embedded in the circuit. So you have to determine this imbedded resistance first, it is NOT 60Ω. Continue with your original series of 12 and 18 and calculate them in parallel with the adjoining 60 Ω resister. Calculate that value in parallel with the 30 Ω. Then calculate the resulting value in parallel with the 75 Ω.
.
Just keep whittling down on the circuit.
.
Understanding the overall results should help you be able to get a feel for such circuits be inspection ie once you become familiar with these types of problems, you will be able to make an educated guess before you even start. ie an educated guess would be that the Resistance is probably 15-20 Ωs for you to ballpark your results.
 
  • #4
roinujo1 said:
So for my attempt, my assumption was that:
  • The 18 and 12 Ω resistors were in series, so I combined them.
That's OK. Redraw the circuit after this step and you should see another obvious parallel combination.

I assumed the 20 Ω resistor at the top was in parallel with the 60Ω resistor under it and combined that.

Yes that's also OK but it wouldn't be the most obvious step. See above.

Then, the equivalent resistance of the 20 and 60 Ω resistors was in series with the 40 Ω resistor.

No that's wrong. They aren't in series.

Redraw the circuit after each step, don't keep referring back to the original.
 
  • #5
Thanks for the response. I'm sorry, but I am confused: Can I combine the 60 and 20, or do I leave the 60 till the end.

And is the obvious parallel circuit with the 60 ohm diagonal?
 
  • #6
So, I did what I thought was correct(probably not), but is the awnser 80 ohms equivalent resistance? I apologize again, I don't have an answer book to check.
 
  • #7
I did it real quick and my final answer was 20 ohms.
I could however be wrong though because I'm taking circuits now like you.

My method of solving was to solve that top portion like you did and then ignore it and go back to the right side. If you just keep solving from the right and work your way towards the source you should be fine.
 
  • #8
zr95 said:
I did it real quick and my final answer was 20 ohms.
I could however be wrong though because I'm taking circuits now like you.

My method of solving was to solve that top portion like you did and then ignore it and go back to the right side. If you just keep solving from the right and work your way towards the source you should be fine.
Could you tell me your general method?
 
  • #9
roinujo1 said:
Could you tell me your general method?

Why don't you follow what I and the other persons suggested, i.e going step by step and redrawing the circuit, with an equivalent resistor each time, till you find the final resistance? If you're trying to do it in some quick way, without knowing first the step by step method, how could you learn about solving such exercises/ problems in general?
 
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What is "the equivalent resistance of the circuit"?

The equivalent resistance of a circuit is the single resistance value that can replace all the resistors in a circuit while maintaining the same current flow and voltage drop.

How do you calculate the equivalent resistance of a circuit?

The equivalent resistance of a circuit can be calculated using the formula Req = R1 + R2 + R3 + ... + Rn, where R1 to Rn are the individual resistances in the circuit.

Why is it important to know the equivalent resistance of a circuit?

Knowing the equivalent resistance of a circuit is important because it helps in analyzing and designing circuits. It allows us to determine the amount of current flowing through the circuit and the voltage drop across each component.

What factors affect the equivalent resistance of a circuit?

The equivalent resistance of a circuit is affected by the number and arrangement of resistors in the circuit. In series circuits, the equivalent resistance is the sum of all the resistors, while in parallel circuits, the equivalent resistance is less than the smallest individual resistance.

Can the equivalent resistance of a circuit be negative?

No, the equivalent resistance of a circuit cannot be negative. Resistance is a physical quantity that represents the opposition to the flow of electric current, and it can only have positive values.

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