Analyzing the Circuit: Currents, Potential Differences, and Power Dissipation

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The discussion focuses on analyzing a go-cart lighting circuit with five light bulbs and two batteries, emphasizing the use of Kirchhoff's rules and Ohm's law. Participants debate the number of currents in the circuit, with some suggesting five distinct currents while others argue that some must be the same due to series connections. The potential differences across specific resistors are discussed, concluding that they are equal when measured between the same points. The total power dissipation in the circuit is calculated using the formula p=iv, and the impact of swapping two light bulbs is considered, with the consensus that it would not change the overall analysis if the resistances are equal. The conversation highlights the importance of understanding current flow and potential differences in circuit analysis.
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1. You are part of a team designing a go-cart. A member of the team has given you a picture of the lighting for go-cart. Your mission is to analyze the circuit. The team has not yet decided on what lights or batteries to use yet, so you need to do the analysis using variables (into which numbers will be plugged once they are decided). In the picture below five light bulbs are connected to two batteries. Treat the batteries as sources of potential difference (∆V#) and the light bulbs as resistors (R#).

A) How many different currents are in this circuit? Draw where they are.

B) How does the potential difference across R4 compare to that across R3?

C) Determine the magnitude and direction of each current (in terms of R# and ε#). Suggestion: use Kirchhoff’s rules and Ohm’s law.

D) What is the total power dissipated in the circuit? E) What answers change if light bulbs R2 and R5 are swapped?




Homework Equations


Kirchhoffs Rules


The Attempt at a Solution



B) 5 currents

C) They are equal but have opposite signs
 

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britt said:
B) 5 currents
You mean A). How do you get 5? Are you proposing a different current through each resistor? Can you see that some must be the same?
C) They are equal but have opposite signs
You mean B). Whether they have the same or opposite signs depends which way round you measure them. Since you are not told which way to measure them, the potential differences should be considered unsigned. But out of interest, let's say we measure both from bottom to top in the diagram; what is your answer now?
 
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A) yes I assigned an I to each 5 resistors. Would the series ones be the same? R1 R2 and R5

B) I3=I4 so I34/2
 
britt said:
B) I3=I4 so I34/2
The question asks for a comparison of potential differences.
 
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R3i3=R4i4
 
1/R3 + 1/R4 = 1/R34 -- does this have anything to do with it?
 
britt said:
R3i3=R4i4
Your first answer was "They are equal but have opposite signs" and haruspex questioned this. Now that you've had a chance to review your answer, would you express it any differently?

The answer will be in words.
 
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The potential difference across both are equal?
 
britt said:
The potential difference across both are equal?
Yes.
A) yes I assigned an I to each 5 resistors. Would the series ones be the same? R1 R2 and R5
Yes. What about the current in the wire connecting the batteries?
 
  • #10
britt said:
The potential difference across both are equal?
Correct. Since the two are connected between the same two points, the voltage across each is equal.
 
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  • #11
C) don't all currents go like
_____>
| | |
^ ^ |
| | |
<___V
 
  • #12
help!
 
  • #13
britt said:
C) don't all currents go like
_____>
| | |
^ ^ |
| | |
<___V
Yes, those are the current directions.
What about the current in the wire connecting the batteries?
Scratch that question - I see you answered it already. So yes, the currents through r1, r2, and r5 are all the same. Call it I. If the currents through R3 and R4 are I3 and I4, what two equations can you write down relating them to each other and to I?
 
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  • #14
I34=I3I4R3R4 / R34

I3=I4

Not sure about how to relate them

I = I3 + I4 ?
 
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  • #15
C) ε2-(i4 / r4 + i3 / r3) +ε1-ir=0

d) p= iv = (i+(i3 + i4))(ε12)

e) none would change assuming r2 and r5 have the same resistance
 
  • #16
britt said:
C) ε2-(i4 / r4 + i3 / r3) +ε1-ir=0
wrong

d) p= iv = (i+(i3 + i4))(ε12)
wrong

p=iv is a good start.

e) none would change assuming r2 and r5 have the same resistance
What if R2 and R5 had unequal resistance?

Give some more thought to these.
 
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