Finding I1, I2, and power dissipated

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Discussion Overview

The discussion revolves around calculating the currents (I1, I2) and power dissipated in a circuit with multiple resistors. Participants explore various methods for applying Ohm's law and the current divider rule, while addressing potential misunderstandings related to the circuit diagram and voltage across the resistors.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests using the current divider rule but encounters discrepancies in their calculations.
  • Another participant questions the voltage across each resistor and proposes using Ohm's law to find the current.
  • A participant expresses confusion about their calculations, indicating a potential error in their application of voltage and current relationships.
  • Some participants highlight the importance of closely examining the circuit diagram to clarify voltage levels across the resistors.
  • One participant proposes adding voltages (24V and -8V) to determine a total of 32V across the resistors.
  • Another participant confirms the calculation of total current using the equivalent resistance and expresses uncertainty about arriving at the correct answer unintentionally.
  • Power dissipation calculations are discussed, with different methods proposed for determining power across the 4-ohm resistor.
  • Participants note that there are multiple techniques for circuit analysis and power dissipation, emphasizing the flexibility in approaches.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the application of circuit laws and the interpretation of the circuit diagram. There is no clear consensus on the methods used or the correctness of the initial calculations, indicating ongoing uncertainty and exploration of different approaches.

Contextual Notes

Some participants mention issues with the clarity of the provided circuit diagram, which may affect their calculations. There are also references to different methods for calculating power dissipation, suggesting that the discussion is influenced by individual interpretations of circuit analysis techniques.

Josh225
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Homework Statement


See attatched image

Homework Equations


I am assuming to use the current divider rule... But when I use it, I do not get the correct answer

1/Rt = 1/R1 + 1/R2 + 1/R3 .. When I used this I got Rt= 2.18 ohms

I also tried Is= E/Rt... 24/2.18 =11 AI am not sure what to try next..

Thanks!
 

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How many volts are across each resistor?
 
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The Electrician said:
How many volts are across each resistor?
To do that, I would have to apply ohms law: V= I x R , right?

So, i would need to find "I" so I= V/R

12 ohms = 24 v / 12 ohms = 2A
8 ohms = 24 v/ 8 ohms = 3 A
4 ohms = 24v / 4 ohms = 6 A

So now.. V= I x R

12 ohms : 2 A x 12 ohms = 24 Volts
8 ohms : 3 A x 8 ohms = 24 volts
4 ohms: 6 A x 4 ohms = 24 volts

Yeah... That can't be right. My brain must be dead right now or something
 
Josh225 said:
Yeah... That can't be right.
Why? It is wrong but why do you think that can't be right? Is there something wrong with the method or the numbers? See the diagram carefully.
 
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Josh225 said:
To do that, I would have to apply ohms law: V= I x R , right?
Nope, just look more closely at the diagram, as others are hinting.
You've been using 24V, that would be correct if the other side of the resistors were at 0V but that's not what the diagram shows...
 
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Delta V.jpg
 
Delta V-2.jpg
 
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I think I see where to go now. I would have to add the 24 V and -8 V to get 32 Volts. Right?
 
That's it . Good !
 
  • #10
Strange.. So I after I determined that 32 volts went across, I applied the formula Is= E (32)/ Rt (2.18) and got 14.67 A.. Which is the correct answer for both I1 and I2. I don't understand why though since I was calculating for the source current and got the correct answer. I am kind of at that stage where I am seeing which formulas work, and then seeing why afterwards.

I think I get the correct answer because the 12ohm resistor goes with I2 and the 8 ohm and 4 ohm resistor (8+4=12) goes with I1.. So both equal 12 ohms. But that isn't always going to be the case. I didn't really arrive at the answer intentionally either.
 
  • #11
Then for the power dissipated by the 4 ohm resistor, I just do Px=V^2/Rx ... Px= 32^2 / 4 ohms = 256 Watts
 
  • #12
Josh225 said:
I just do Px=V^2/Rx ... Px= 32^2 / 4 ohms = 256 Watts
That's ok .

The diagram supplied with the problem is not very good . Try drawing a new diagram using modern conventions for the layout and symbols .
Two horizontal lines for the power rails with the three resistances set vertically between them . Imagine a ladder on it's side .

Your original method of finding the equivalent resistance and hence the total current is perfectly good .
You could also have found the total current by adding up the individual resistance currents .
 
  • #13
Nidum said:
That's ok .

The diagram supplied with the problem is not very good . Try drawing a new diagram using modern conventions for the layout and symbols .
Two horizontal lines for the power rails with the three resistances set vertically between them . Imagine a ladder on it's side .

Your original method of finding the equivalent resistance and hence the total current is perfectly good .
You could also have found the total current by adding up the individual resistance currents .
Oh nice! Yeah, that involves a little less mess. But how do I know that BOTH I1 and I2 equal the same amount?
 
  • #14
Oh... Current going in = the current going out...
 
  • #15
That's it .
 
  • #16
Thank you!
 
  • #17
Josh225 said:
Then for the power dissipated by the 4 ohm resistor, I just do Px=V^2/Rx ... Px= 32^2 / 4 ohms = 256 Watts
There is many ways in finding power dissipation--you can also do Px=(Iacross-4Ω)2 x R4Ω where
Iacross-4Ω= 36 volts / 4Ω = 8 Amps (since they are in parallel connection, voltage is same across each resistor!)
Then square this and multiply it by the 4Ω..
and still get the correct answer--just pointing out the many different techniques in circuit analysis. :oldsmile:
 
  • #18
Aristotle said:
There is many ways in finding power dissipation--you can also do Px=(Iacross-4Ω)2 x R4Ω where
Iacross-4Ω= 36 volts / 4Ω = 8 Amps (since they are in parallel connection, voltage is same across each resistor!)
Then square this and multiply it by the 4Ω..
and still get the correct answer--just pointing out the many different techniques in circuit analysis. :oldsmile:
Thanks! :) I've added that to my bag of tricks.
 

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