What is the correct way to analyze a circuit with resistors?

  • #1
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Homework Statement


[/B]
h9_bridge.png


R1 = 42 Ω
R2 = 124 Ω
R3 = 119 Ω
R4 = 105 Ω
Rx = ?
I4 = O A

Homework Equations


R = IV
V = V1 = V2 for parallel Resistors
Iequiv = I1 +I2 for parallel Resistors

Vequiv = V1+V2 for series Resistors
Iequiv = I1= I2 for series Resistors

The Attempt at a Solution



I have to find I1, the current that flows through resistor 1. But I am unsure of how to start. The first thing I tried was getting the combined resistance of R1 and R2 (R12):

(1) 1/R12 = 1/ R1 +1/R2

Since they are parallel I applied the rule that Resistors in parallel have equal voltages:

(2) Vequiv = V1 = V2

but I wasn't sure that would work unless I packed all the resistors together first (and since I don't have a value for Rx yet I am unsure how to proceed in this manner)
So, instead I tried using a ratio. Using V = IR and using the Voltage relationship for parallel resistors I tried:

(3) I1/I2 = R2/R1

I got a value for I1 in terms of I2 and set up an equation using the rule Iequiv = I1 +I2
From this I got

(4) I1/I2 = 62/21 (5) I1 = 62/21(I2)
62/21(I2)+I2 = 12V (voltage of the battery)

solving (4) for I2 I got 252/63
subbing back into the eq (5) I got that I1 = 744/83 but this was wrong.

I think I am misunderstanding how the relationships (Vequiv = V1 = V2 )for parallel and (Iequiv = I1 = I2) for series resistors works. Can someone help clarify and/or explain where I am going wrong in my logic?

** I apologize for formatting. I tried to use LATEX code but I couldn't get it to work properly and with finals next week I simply haven't the time to spend learning it for this question. Please forgive me! :)
 
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Answers and Replies

  • #2
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I have to find I1, the current that flows through resistor 1.
Why?
The first thing I tried was getting the combined resistance of R1 and R2 (R12):

(1) 1/R12 = 1/ R1 +1/R2

Since they are parallel
They are not. There is R4.
You can calculate the voltage drop across R4 to handle that.

62/21(I2)+I2 = 12V (voltage of the battery)
Where does that come from? It is wrong. The left side is a current, the right side is a voltage.

Start with the special role of R4, that will simplify things.
 
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  • #3
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Why?
That is the first Q on the homework assignment. If it was irrelevant I apologize but I thought it might provide context for what I was attempting to do.

They are not. There is R4.
You can calculate the voltage drop across R4 to handle that.
Are you saying that they are in series and not in parallel? If so I am way more confused than I thought I was.

Wouldn't the voltage just be 0 thought across R4? Since V = IR? So I am unsure how this would interact with the rest of the circuit (it seems like it almost wouldn't with the exception of the fact that it likely has a resistance)



Where does that come from? It is wrong. The left side is a current, the right side is a voltage.

Start with the special role of R4, that will simplify things.
I used the fact that resistors in parallel have the relationship that : Vequiv = Va = Vb

So using V = IR for both resistors:

I1R1 = I2R2

I1/I2 = R2/R1 and R2/R1 = 124/42 = 62/21 >> I1 = I2 (62/21)

I understand this might be incorrect because my understanding of how to combine circuits in a manner that allows me to use these special relationships of Vequiv and I equiv to parallel and series circuits is not strong. My instructor uses only a program called smart physics to teach us- we get little to no lecture and all that we have done on this unit is labs with little to no instruction or assistance. So I apologize, but I am nowhere near where I need to be to analyze this properly. Must I first get the resistance of the entire circuit and work backwards from there? That is what smart physics said to do in the "worked example" but that one was straightforward and gave you all resistances ahead of time. Then the first Q they throw at me on my own they omit a resistance; so I am not quite sure how to proceed. If R = I/V and I = 0 and V= 0 for the fourth resistor I have a problem. Again I apologize, I am intelligent and I have a high knowledge of mathematics (through multivariate calc) but with so little instruction I am struggling with physics a lot. If I can get a clear explanation I am able to understand and grasp things easily. Thank your for your help!
 
  • #4
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I was able to solve it by using R3 instead in series with R1:

Requiv = R1 + R3 = 42 Ohm + 119 Ohm = 161 Ohms

I = V/R and I equiv = I1 = I3

12V/161 Ohms = .0745341615 A

However, I am kind of fumbling through it so any insight you can give me regarding why my previous attempt at a solution was incorrect would be very useful. Thanks again!
 
  • #5
gneill
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Wouldn't the voltage just be 0 thought across R4? Since V = IR? So I am unsure how this would interact with the rest of the circuit (it seems like it almost wouldn't with the exception of the fact that it likely has a resistance)
Big hint that's good to remember: A resistor that carries no current can be replaced with a wire or simply removed entirely from the circuit, your choice ;)

This works because with no current flowing the potential difference across the resistor is also zero. That means the two nodes are at the same potential and are effectively the same node. Shorting the two together will not change any of the potentials or currents in the circuit. Also removing the resistor entirely will not change any currents in the circuit.

This trick of combining nodes with the same potential can be quite handy when simplifying circuits with a lot of symmetry, so be on the look out for opportunities to use it.
 
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  • #6
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Big hint that's good to remember: A resistor that carries no current can be replaced with a wire or simply removed entirely from the circuit, your choice ;)

This works because with no current flowing the potential difference across the resistor is also zero. That means the two nodes are at the same potential and are effectively the same node. Shorting the two together will not change any of the potentials or currents in the circuit. Also removing the resistor entirely will not change any currents in the circuit.

This trick of combining nodes with the same potential can be quite handy when simplifying circuits with a lot of symmetry, so be on the look out for opportunities to use it.
Thanks a ton!
 
  • #7
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Another Q if anyone has time:

If I have a circuit like this:

h9_circuit.png


and I need to get Rac (I already have Rab) how do I draw my new diagram to determine whether R1 is parallel or in series with Rab? I guess a better way to ask is: when you have a combined Resistance Rab, how do you determine where Rab resides on the circuit for the purposes of future calculations? Or do I just get to choose?
 
  • #8
BvU
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Series: the current first hast to go through the one, then through the other. Over each of the two (or more) participants the voltage is IR with the same I.

Parallel: the current can split up. It splits up in such a way that the voltage difference over the two (or more) branches is the same. The current through each branch is V/R with for each the same V.
 
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  • #9
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I cannot thank you all enough! That makes it much clearer, @BvU! :w
 

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