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How to find power dissipation of a resistor in this circuit?

  1. Aug 29, 2016 #1
    Here is an image with information: https://i.snag.gy/UMCRig.jpg

    I have calculated the total power dissipation as 11W. What I don't see is why R1 has a higher dissipation rate than R2/R3, and my main question is how do I calculate the power dissipation of R1?

    I get that R2/R3 are in series so have a lower resistance apparently, but each one has more ohms so should surely result in more power dissipation as the current passes them.

    Is the power dissipation referring to joules lost as heat per second, or useful electrical power in joules per second?

    Also, does this circuit refer to conventional energy flow from p to n, or real life flow from n to p? Not too clear, and there's no more information regarding the question.

    Appreciate all help,
    thank you.
     
  2. jcsd
  3. Aug 29, 2016 #2
    R2 and R3 are connected in parallel also notice that R2 and R3 current summed-up in R1 resistor current. So IR1 = IR2+IR3.
    And P = I*V = IR1*VR1
     
  4. Aug 29, 2016 #3

    berkeman

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    They are in parallel. I think it's just a simple typo on your part.

    EDIT -- Beaten to the punch by Jony! :smile:
     
  5. Aug 29, 2016 #4
    Yes I meant in parallel. Anyone able to answer those question? I'd be grateful, especially for the first one.
     
  6. Aug 29, 2016 #5

    davenn

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    so firstly work out the combined resistance RT of R2 and R3 in parallel, do you know how to do resistors in parallel ?
    http://www.sengpielaudio.com/calculator-paralresist.htm and note the formula used

    once that is done, you can then easily work out the current in the circuit
    Hint, the current in a (now) series circuit is the same everywhere

    once you know the current in the circuit, you can work out the power in Watts dissipated by RT and or R1

    Power is usually measured in Watts for this sort of problem
    Joney130 in post #2 gave you the formula for working out the power

    irrelevant, wont make any difference to the calculations .... that's why there is no info given


    Dave
     
  7. Aug 29, 2016 #6
    Much appreciated, this helped me out. My main reason for struggle was the fact I wasn't aware that the current of a series circuit was constant. With that in mind, it made answering the questions easier. My issue is, I had always thought against this. Surely adding a resistor would decrease the current, and believing that made answering the question of power dissipated by each resistor impossible. Why is it that that the current remains constant in a series circuit? That defies everything I had previously learned, and makes resistors seems utterly useless. There must clearly be a concept I am missing here.
     
  8. Aug 29, 2016 #7

    davenn

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    yes it does but once that second resistor is added then the NEW current in the circuit will be the same where ever in the circuit it is measured
     
  9. Aug 29, 2016 #8

    berkeman

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    The statement applies to all of the components connected in series in a branch, not to a branch where you are changing components and comparing before and after the change.

    In a series circuit, the flow of electrons is the same at all points in that branch. If you have 1A through one resistor in a string, you will have that same 1A through all of the resistors connected in that string. But as Dave says, if you then change the value of a resistor in that string and have the same voltage source driving the string, the 1A will change to some other value, which will be constant throughout the new string.

    Makes sense?
     
  10. Aug 29, 2016 #9

    davenn

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    not at all, as I showed in my last post, adding extra resistance lowers the overall current in the circuit

    Resistors are essential for current limiting in circuits such as ones with LED's (Light Emitting Diodes)
    LED's are a current driven device, That is, they require xx amount of current, any more will zap them

    Take a standard 5mm red LED, typically 3V drop across it and a max of 25mA put across a 12V PSU.
    A calculation is needed to work out the series resistor needed so that the LED sees a max of 25mA

    Using the values we have, R = (12 - 3.0) / 0.025 = 360 Ohms. In this case we might pick a 330 ohm resistor, it would be close enough



    Dave
     
  11. Sep 8, 2016 #10
    So from this I understand that after current passes through a resistor and its value decreases, that value remains constant in a series circuit.

    What I don't understand, is why adding more resistors after the first one has no further effect on the current. Why don't additional resistors change the current's value?

    Also, why is it that the current's value doesn't remain constant after the initial resistor in a parallel circuit? Is the curren't equally divided between each node?

    I appreciate the responses.
     
  12. Sep 8, 2016 #11

    berkeman

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    The current does not decrease in passing through a resistor. The voltage drops some, so that the full source voltage divides among the series resistors. The series current in a branch is constant, given some configuration of resistors and some set source voltage.
    Adding resistance to a branch decreases the current as compared to the current before the extra resistance was added. This follows from V=IR, where V is being held constant between the two configurations, and the R is increasing between the two configurations.
    Currents divide in parallel circuits, current does not divide in a series circuit; it is the same throughout the series branch.
     
  13. Sep 8, 2016 #12

    sophiecentaur

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    You are in a muddle here and I'm surprised that no one seems to have picked up on it fully. (Too polite, I guess :wink:)

    Resistors would certainly be "utterly useless" if things worked the way you have "learned". I think you need to start at the beginning again and make no assumptions based on what you have understood so far.
    To understand the relationship between Current and Volts in a resistive circuit there are two rules that apply: Kirchhoff's First and Second Laws. These are the "concept" that you need to learn and apply.

    Google Kirchhoff's Laws and trawl around till you find a site that you are happy with. A .edu or .org URL will usually be reliable. Steer clear of most forums if you want to be sure to get your basics right. PF is pretty reliable because BS is usually spotted and dealt with. :smile: But the following is more or less what you will learn.

    The first law says that charge cannot 'build up' in any point of a resistive circuit so the total current flowing in and out of a junction ('node') must be zero. The second law says that the total energy put into a charge (from a battery, for instance) that flows round a loop within that circuit must be dissipated by the resistors in that loop, according to their Resistance. That implies nothing about a 'constant current' around a circuit.

    If any resistor in a circuit is changed, you can expect it to affect the Volts and Currents throughout that circuit. So no resistor is "useless". Circuits are full of resistors that will either produce a desired current from a source of Voltage or a desired Voltage by suitably sharing a supplied voltage between two resistors.

    Working out simple circuits is easy if you approach things in the right order. You can often make things easier by spotting where two resistors can be reduced to just one resistor for analysis- where there are two (or more) in parallel between the same two nodes, you can always reduce them to one equivalent resistor and where there is a chain of two or more resistors joining two nodes (with no other possible paths) you can reduce them to a single equivalent resistor.

    So in your OP diagram, the parallel pair can be reduced to a single resistor and that leads you to two resistors in series. That will allow you to work out the current flowing from the battery (I = V/R) Ignore any resistor values until you have worked out the principle. K1 tells you that the total current through R2 and R3 is the same as the current through R1. You can then calculate the Volts across R1. K2 tells you that the Volts across R2 and R3 is the left over Volts from the battery volts.

    That info is enough to work out the Power dissipated by R1, R2 and R3.
    I can't find any figures in your post but you only need to slot in the values that you have and Bob's Your Uncle.

    PS You will very soon find that you can look at a circuit and do a lot of this stuff in your head, writing down intermediate answers directly.
     
  14. Sep 9, 2016 #13
    @Metals, there is a flaw in your understanding causing confusion. If you spend some time studying nodal analysis it should become clear. Go to the math first as already pointed out. Once you understand the math things will fall in place conceptually. It's pretty much always the case in understanding electricity and electronics, you have to get a feel for relationships on paper then it becomes easier to understand at the conceptual level.
     
  15. Sep 13, 2016 #14
    This says that resistors change the current, as I believed.

    This says that resistors change current, as I believed.

    This point then contradicts what the two previous ones stated, saying that it's the voltage which is affected by a resistor. I believed that voltage remained constant, and the changes would occur in current/resistance only.

    Then it is being stated that current changes due to resistance, as I had always believed.


    @sophiecentaur What I had learned was what is true, what I said is that the new information I received from a previous point contradicted my prior (and current) knowledge. I had looked into Kirchoff's first and second laws before, they were simple, didn't look into his third law though. I appreciate your effort to help, though my level of knowledge in this field isn't that low, and it's the previous posts which have confused me causing me to take a few steps back. As seen in this current post, there is a slight contradiction. I am confident that those replying to me are most likely professionals, and fully respect their knowledge in this field, though there has clearly been increasing confusion as this thread has developed. Lets say, my confidence on electrical engineering has been disordered like entropy.
     
  16. Sep 13, 2016 #15

    berkeman

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    Nothing I said is contradictory. I think you are still just misunderstanding V=IR a bit. Just keep on learning more and more about circuit analysis, and I think you will get the hang of it. :smile:
     
  17. Sep 13, 2016 #16

    Grinkle

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    It might help to think of water in a pipe.

    1. Water flows from one end of the pipe to the other. You should conclude easily that anything entering the pipe must exit the pipe. There is no way the pipe can store water as long as the water is actually flowing. The water pressure drop from the entrance of the pipe to the exit is analogous to voltage, the water flow to current.

    2. Now take pliers and put a slight crimp in the pipe somewhere. The total water flow through the pipe decreases, but still, the crimp does not store any water, it can't.

    3. Now take the pliers and add a second crimp somewhere else on the pipe. The flow decreases more, but still, neither of the two crimps can store any water.

    The crimps are the resistors in this analogy.

    I hope that is some help to developing an intuition for why flow is the same at all parts of a resistor loop.
     
  18. Sep 13, 2016 #17

    sophiecentaur

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    The value of a resistor will affect the current flowing in a circuit BUT the current IN will always be the was as the current OUT. You have to analyse the whole circuit to find the effect of each (significant) resistor. So a Change in resistor value will involve a Change in current.
    But you have to let the circuit settle down after it's switched on. The speed of light is the limit to how quickly each component makes its presence felt and involves multiple reflections inside the circuit of the 'step change' when the switch is connected. This is why we choose to deal with the (eventual) steady state situation in our treatment of most circuits.
     
  19. Sep 13, 2016 #18

    davenn

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    You still seem to be missing the point :wink: NOT after the resistor, the current in the WHOLE circuit decreases, BEFORE and AFTER the resistor


    that's still not right
    it does change the current !! again, in the WHOLE circuit.

    a circuit with 10 V and 30 Ohms resistance with have a current of 0.33 Amps flowing through the whole circuit
    it doesn't matter if that is a single 30 ohm resistor or 3 x 10 Ohm resistors, the current throughout the whole circuit will be the same ... 0.33 Ohms

    If, for example you add another 30 Ohm resistor, you now have 60 Ohms total and the current will be half of the original 0.33 A
    10V / 60 Ohms = 0.166 A

    a little pic I put together for you to ponder on .....

    series resistors.GIF

    The measured current at any of those points in either circuit will be the same 0.33 A


    Dave
     
  20. Sep 13, 2016 #19

    davenn

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    now to take it a step further

    a single 30 Ohm resistor we know that the current in the circuit is 0.33 A

    I = V/R
    I = 10 / 30
    I = 0.33 A

    we ADD a second 30 Ohm resistor
    I = V/R
    I = 10 / 60
    I = 0.166 A

    2 x series resistors .GIF

    so you can see, adding a resistor does decrease the current
    in the first circuit, the measured current at A or B will be the same = 0.33 A
    in the second circuit, the measured current at A, B or C will be the same = 0.166 A


    Dave
     
  21. Sep 13, 2016 #20

    sophiecentaur

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    @Metals I think you have your own private view of what Electric Current actually is. Hence you are confused by the apparently contradictory statements that you have read. Rather than trying to fit your personal model to what you are reading, just follow the well known 'rules' that apply to Current.
    Resist the temptation to make Physics fit your view and approach things believing that the Physics you read is Right. Then bend your view to fit in with what it says. Avoid the "yes but" reaction and just accept that it all works consistently. Believe that you are going wrong somewhere along the line and follow the line of accepted Physics.
     
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