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Finding Vmax

Discussion in 'General Electronics Discussion' started by bfickes, Jul 22, 2009.

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  1. bfickes

    bfickes

    3
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    Jul 22, 2009
    I am a electronics student (self-teaching) with Grob's Basic Electronics Textbook. In the problems manual I've come across a word problem that I cannot solve. The problem deals with Vmax. Here is the problem: A 1-k Ohm 1/4-W resistor and a 1.5-k Ohm 1/8-W resistor are in series. What is the maximum amount of voltage that can be applied to this circuit without exceeding the wattage rating of either resistor? The formula for for Vmax is the square root of the P-rating x resistance. Solving for the 1-k Ohm 1/4 W resistor I get 15.8 V and for the 1.5-k Ohm 1/8-W resistor I get 13.7 V. When added together they total 29.5 V. The answer in back of the book is 22.82 V. Does anybody know how to solve for this problem? :confused:
     
    Last edited: Jul 22, 2009
  2. jasonben

    jasonben

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    Jul 15, 2009
    I can’t answer the question. I’m looking at the equation for power, P = (I^2)*R = IV. According to it, is seems to me – if I am right, that because I does not change in this series circuit, the power dissipated is going to be directly proportional to the resistance for each resistor. Current into a part of a circuit equals current out. That would seem to me to mean that power is going to be dissipated twice in the circuit. Once, across the first resistor. A second time across the second. It is dissipated with the voltage drops. All the voltage drops in the circuit add up to the total voltage of the circuit. Is this right? If it were, the value for P would be different for each resistor. That is, the power dissipated would be different for each resistor. Is the power dissipated across each resistor effected by the rating for each resistor? I’m guessing that there is a certain amount of power dissipated across each resistor, and the power rating of each resistor might usually be greater than this according to a safety factor equation – if the circuit was not supposed to produce too much heat or decrease the capability any resistors that the circuit was comprised of. However, I wasn't thinking that this power rating would change significantly the power dissipated across the resistor in most situations. So, maybe P could mean two different things: the power rating of the resistor and the power dissipated by the resistor - depending on how it was used each time an equation was used.
     
    Last edited: Jul 24, 2009
  3. jasonben

    jasonben

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    Jul 15, 2009
    I went ahead and worked on this problem. I got the same answer as the book – though I can’t guarantee that I got it the right way. Hopefully bfickes has solved the problem by now. I could still use help if I got it wrong. I also went over bfickes work.
    The equation
    V = (P*R)^(1/2)
    (* means times, / means divided by, ^ means to the power of, x^(1/2) means the square root of x) seemed to have been used. I tried to figure out how this equation was derived. Taking the square of both sides results in
    V^2 = P*R.
    Multiplying by P gives
    P*V^2 = (P^2)*R.
    Dividing by V^2 gives
    P = (P^2/V^2)*R (equation 1).

    I had gotten
    P^2/V^2
    In terms of I by squaring both sides of
    I = PV
    to give
    I^2 = (P/V)^2 = P^2/V^2 (equation 2).

    I had originally worked out the equation from
    P = (I^2)*R (equation 3).

    Equation 3 is one of the equations that I think is usually derived in some textbooks using
    P = IV
    and
    V = IR
    And substituting IR for V in P = IV to give
    P = IV = I*(IR) = (I^2)*R.

    Substituting I^2 for P^2/V^2 (according to equation 2) in P = (P^2/V^2)*R (according to equation 1) does give equation 3 – which is
    P = (I^2)*R.
    However, can this substitution be made? It might work when there is only one resistor present. I don’t know. In this case, I^2 is defined as P^2/V^2. But the P in P^2 and the V in V^2 in P^2/V^2 in this situation – when each resistor is solved for one at a time, does not take into consideration the other resistor. This could lead to unpredictable results if the value of the other resistor is high enough because the I in I^2 is a dependent variable – in this case meaning that I depends on the resistance of the other resistor. Studying the equation V = IR = I(R + R) – which as I understand it is true for series circuits – may convince one of this. If I has to be solved first, the method above might not work because in eliminating the I variable from the equation, variables are used that only take into consideration one of the resistors at a time, when I think that the current I actually depends on how both resistors effect I more or less at once. I don’t know if this is right – but if the amount that the metal in the wire obstructs electrons from moving can be though of as resistance, then it might be easier to visualize how the current – which is measure of how much electrons move (according to my interpretation of the first couple of paragraphs of Wikipedia’s article on Ampere), at the end of a wire might be effected by not only one but all areas of resistance before it.

    One thing that is going to be same for both resistors is I. So, the current, I – if I am using the term right, is an independent variable. One does not have to know the mathematics of this to understand it as it is written at Wikipedia’s arProxy-Connection: keep-alive
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    cle on Kirchhoff’s circuit laws - though this is not the simplest explanation that I have seProxy-Connection: keep-alive
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    . Therefore, I might have to be solved for first. We are given two values for P and R in the problem. There is an equation relating these three variables, I, P and R. It is shown above. It can be used for both resistors – and will result in two answers for I. One of these answers – the one that is larger, might be used to calculate a voltage that exceeds the capacity of the resistor circuit described. The other one may also exceed the capacity, but it is safer. Once I is known, the voltage may also be able to be calculated using this known variable and values of the resistors.

    Do you know why V = I(R + R) might be the appropriate formula to use and not a combination of the formulas below - in which RT is the total resistance, R1 is a first resistor resistance value, and R2 is a second resistor resistance value?
    1/RT = 1/R1 + 1/R2
    V = IR

    What I wrote makes sense to me. I worked the problems out in numbers and got the same answer that you said that the book gave. However, it is possible to come to the right conclusion using methods that might not be applicable – and I can not guarantee that I have not made this mistake. However, hopefully I have given you something to think about – even if it means you had to correct my mistakes.
     
    Last edited: Jul 27, 2009
  4. bfickes

    bfickes

    3
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    Jul 22, 2009
    Thanks Jason for all your hard work. I've applied all of your equations and I still don't get the answer the book gives. I don't know how you got the answer but I've come to the conclusion that the answer the book gives is wrong. Here is my reasoning. First, the power rating is only a rating, not power dissipated. Second, in order to solve for voltage you must know current or power dissipated and to solve for current you must know the voltage or power dissipated. None of those variables are given in the word problem. The only variables given are the power rating and resistance. Therefore the only equation that will work is V max which is the square root of the power rating times the resistance. This is the only logical solution. :cool:
     
  5. jasonben

    jasonben

    22
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    Jul 15, 2009
    I agree that the power rating is only a rating – but I still think that it can be used to solve the problem - which I think is solvable. I think that each power rating can be used to determine a highest common value of current for each resistor – which will be one of two values that I think can be solved for. Using, P = (I^2)*R, for one of the resistors, the power rating may actually be equal to power dissipated. This is because we are looking for a highest safest value this is limited by one of the resistors. I think that even this lower value might be considered dangerous – and is why the power rating of resistors may in practice be increased.

    I don’t agree that to find the voltage one has to know the current and power dissipated – if this is what you meant. I’d consider V = IR. Can I be solved for?

    I don’t agree that to find the current one has to know voltage and power dissipated – if this is what you meant. According to P = (I^2)*R, only P and R have to be known. P was discussed above. The value of R is given for each resistor. This will result in two currents. Which one is safe? Maybe the lowest. Why?

    After having solved for the current I, V = I*R can be used.

    I think that this is right. The only way that I can think of it being wrong is if P doesn’t equal (I^2)*(R1 + R2) even though it does equal (I^2)*(RT) – using the same symbols described in an earlier post.

    I tried to explain in my last post why I didn’t think that the formula that you gave would work. I’ll try to simplify my explanation – not because I am sure that I am right, but because I don’t think that I explained it very well. The power rating in the equation I think actually means two different things – the power dissipated and the power rating. These may be the same for one resistor but not the other.

    Basically, there is a circuit. We can apply any voltage to it. According to that voltage, there will be a certain current that flows through the circuit. That current – together with the resistors, will determined power dissipated - and perhaps damage the circuit. Which voltage is best? We don’t want to go too high or low when applying voltage. So, we consider the poProxy-Connection: keep-alive
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    r rating Proxy-Connection: keep-alive
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    roxy-Connection: keep-alive
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    oxy-Connection: keep-alive
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    the resistors – maybe becauseProxy-Connection: keep-alive
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    e don’t want to do any damage. So, how much current? That depends on P and R – which Proxy-Connection: keep-alive
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    e given. They are given for each resistor. We can solve for each resistor and come up with two answers for the current I that will work for each resistor. Will they work for the other resistor as well? Maybe only if only the lower one is used – because power dissipated increases as current increases. Does this make sense according to P = (I^2)*R? So, a lowest current can be selected. Then the formula V = I(R + R) can give the voltage. I think that all this depends on if P = (I^2)*(R1 + R2).

    Basically, there is a circuit. We can apply any voltage to it. According to that voltage, there will be a certain current that flows through the circuit. That current – together with the resistors, will determined power dissipated - and perhaps damage the circuit.. Which voltage is best? We don’t want to go too high or low when applying voltage. So, we consider the power rating of the resistors – maybe because we don’t want to do any damage. So, how much current? That depends on P and R – which are given. They are given for each resistor. We can solve for each resistor and come up with two answers for the current I that will work for each resistor. Will they work for the other resistor as well? Maybe only if only the lower one is used – because power dissipated increases as current increases. Does this make sense according to P = (I^2)*R? So, a lowest current can be selected. Then the formula V = I(R + R) can give the voltage. I think that all this depends on if P = (I^2)*(R1 + R2).


    The reason why I am thinking at this time that it might work is because P = IV, V = IR, and RTotal = R1 + R2.
     
    Last edited: Jul 29, 2009
  6. Georgy

    Georgy

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    Apr 11, 2008
    I thru 1.5-k Ohm 1/8-W resistor is critical, it is thin section at series chain.

    Power > 1/8-W will damage resistor by owerheating.
    Therefore calculate from here is true way only.
    When added together they total 29.5 V ? it more than possible.
    The answer in back of the book is 22.82 V. Yes!
     
  7. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    Yes, no need to get long-winded, just remember they are to be in series and so the current thru them has to be the same - and that has to be the lower of the two currents you can calculate from the Vmax, otherwise the "thinner" one will fry.
    1k 1/4W = 15.8V max = 15.8mA max
    1.5k 1/8W = 13.69V max = 9.13mA max
    0.00913 * 1000 = 9.13V
    9.13V + 13.69V = 22.82V
     
  8. jasonben

    jasonben

    22
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    Jul 15, 2009
    In one of my earlier posts, I said that the equation V = (R*P)^(1/2) could be used for the resistor with a lower rating. Since that time I convinced myself that I was wrong - but that this equation would work for a single resistor and perhaps other series scenarios - but not this one. I tried to explain this. The best explanation that I can think of is that this equation doesn't take into consideration current that is dependent on any of the other resistors in a circuit. I think that the P in V = (R*P)^(1/2) is dependent on current - as has since been clarified. I've been trying to think of a more qualitative way of describing this. The current which influences voltage in both resistors can not be determined only from the variables associated with each resistor independently, but has to consider the effects of both resistors on current, perhaps. But that doesn't take into consideration power. The power dissipated by each resistor is dependent upon the current which is determined by both resistors?
     
    Last edited: Jul 31, 2009
  9. neon

    neon

    1,325
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    Oct 21, 2006
    try this make both .125 w or 1.5k @ .125w and 2k @. .125w total 3.5 k @ .250 watts now solve it easy
     
  10. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    The point of this task is to see if the student can keep more than one thought in his head - and remember Thevenin's theorem: the sum of currents going into a junction must equal the sum of currents going out of it.
    Thus you can't solve this task with one equation, you have to think & make a choice, & then proceed.
    Series connection means equal currents, so you have to forget about the larger current and use only the lower current. This of course makes the voltage across the hi-current resistor lower than it could have been (if it had been on its own).
     
  11. jasonben

    jasonben

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    Jul 15, 2009
    I wasn't very confident when I tried to offer some information that might be useful. I was assuming that the current in a series resistor circuit is dependent on both resistors in the first place - and I am assuming that this is correct. I think that I can solve some problems at this level - but I still don't conceptually understand what P is. In this example - in different equations, it may refer to a maximum amount of something that a resistor can tolerate - if I am using that word correctly - because I know that it has a special meaning when applied to resistors. It may also refer to how much power is actually dissipated. Volts are complicated enough for me to understand. So, unless someone gives me a really insightful explanation, I don't foresee myself understanding P = IV anytime soon. How can I understand P = (I^2)*R? I kind of understand V = IR. I have come across some analogies that compare it to a flow of water in a body of water with an amount of rocks resisting the flow. However, what is I^2? The current I by itself kind of makes sense to me - as a rate of electrons moving past a point or a plane - if that is right. However - and I may be using some words incorrectly, a rate of this would be a number of electrons or something passing per time squared. Why the numerator in this situation is also squared is not intuitively clear to me. I think that what I am asking for is a model of power.
     
  12. Resqueline

    Resqueline

    2,848
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    Jul 31, 2009
    P = (I^2)*R is just a compilation of U=I*R and P=U*I, combined giving P=I*R*I.

    The analogy to water flow is very good. Voltage is pressure (psi) and current is flow (gallons/minute). Power is the product of these two. For a turbine to produce power you need both pressure and flow. A flow alone would stop if you tried to make it do work, unless there's pressure to back it up, and pressure alone does nil.
    A tap is a variable resistor and a pressure reservoir is a capacitor. A length (or a coil) of piping is an inductor (as well as a conductor).

    In a mechanical analogy you can say that torque is what makes a car accellerate (or go up a hill) at all, but it's the power that determines how fast it can do it.

    This understanding is enough for me. I don't know what level of modeling you need.
    I didn't/don't mean to point my finger or anything, just trying to keep it as simple, educational and understandable as possible. I appreciate your ponderings, they are valuable, and I don't see any wrongly used terms.
     
  13. jasonben

    jasonben

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    Jul 15, 2009
    How is pressure and flow related to how a car accelerates?

    Also, what is it about components that may cause them to become dysfunctional under conditions of a lot pressure and flow?
     
    Last edited: Aug 12, 2009
  14. jasonben

    jasonben

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    Jul 15, 2009
    Also, what is it about pressure and flow that can make components dysfunctional?
     
  15. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    Uh-oh..

    Accelleration of a car requires torque and power and is dependent on gearing and all this is an issue greatly debated on car forums, but let's try to keep it simple.

    Imagine you have a steam turbine driven car. Steam & water goes for the same thing in this regard. You open the valve and the steam starts to flow through the turbine. You need pressure to get a flow and both together makes power. Up the pressure and more steam flows, increasing accelleration and speed.
    Now, too much pressure can make the pipes burst (insulation breakdown) or bend the vanes in the turbine. Too much flow and the turbine can overspin and fly apart.

    I find it a little hard to use this analogy for this kind of component behavior though.
    I'm positive you'll find many good sites explaining electronics basics around on the net.
     
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