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Homework Help: Brightness of bulb question

  1. Jun 28, 2012 #1
    1. The problem statement, all variables and given/known data
    20 bulbs are wired together in a series position.
    A) what happens if 10 more bulbs are added in series?
    B) the bulb was modified such that an additional resistor is placed across the filament (parallel to the filament). The value of the resistor R is much larger than the resistance of the filament. How does the brightness change?

    2. Relevant equations

    V=RI P=VI

    3. The attempt at a solution
    A) brightness decreases as the voltage is becomes smaller across each bulb hence since P=V^2/R where R is the constant resistance of each bulb, the power drops and subsequently the brightness drops.

    I'm more concerned about part B as I was marked wrong even though I seem right. Here is my answer: although the effective resistance of the bulb decreases, as there is exactly 20 of them hence the voltage across each of the bulb is still the same. And since each bulb contains a parallel branch of the filament and the resistor thus, the filament still receives the same voltage despite their difference in resistance. Since the resistance of the filament remains the same, that means the power developed by the bulb also remains the same. Thus the brightness of each bulb also remains the same.

    I don't get how this is wrong. I asked my teacher about it but she seemed uncertain about it. At first she said that the power developed as a whole is greater so the brightness increases. I explained that the resistor doesn't give out light but only heat (she agreed with that). So I explained what I wrote above and her, with a totally different answer from before, said that since the question says that the resistance of the resistor is a lot higher that means most of the current will past though the filament (short circuit in essence). But the sudden change in answer makes me feel worried. Could she be embarrassed to admit her mistake because this is my mid years paper and many students did it so it will be harder to change all of our marks or am I really wrong?

    Thanks for the help!
  2. jcsd
  3. Jun 28, 2012 #2


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    Your teacher's argument that the power dissipated is nearly the same is true, because if the parallel resistance is much larger than the filament resistance, then to a good approximation, all of the current still goes through the filament, and none through the shunt resistor.

    I can show you that quantitatively. Let's just consider what happens to ONE bulb with a voltage V across it. The bulb has been modified in the way described in the problem. Let the filament resistance be r, and the parallel (shunt) resistance have a value of R. In other words, r is the small resistance and R is the big resistance.

    Since the two resistors are in parallel, the current from the supply has to branch out (split) before the two resistors and then meet up again "downstream" of them. So the total current in the circuit is sum of the currents through each of the two resistors. The current through each resistor is given by Ohm's law, and since both of them have the same voltage across them:

    $$I_\mathrm{tot} = \frac{V}{r} + \frac{V}{R} = V\left[\frac{1}{r}+\frac{1}{R}\right] = V\left[\frac{rR}{r+R}\right]^{-1}$$

    The thing in square brackets is just the effective resistance of two resistors in parallel.

    So P = VI

    $$P = V^2 \left[\frac{rR}{r+R}\right]^{-1} = V^2\left[\frac{r+R}{rR}\right]$$

    Now, as you can see, in the limit where R >> r, we can approximate that ##(r+R) \approx R##. (For example, if R = 1000 ohms and r = 1 ohm, then 1001 ohms is approximately 1000 ohms). With this approximation, we get:

    $$P \approx V^2\left[\frac{R}{rR}\right] \approx \frac{V^2}{r}$$

    In this limit, the power dissipated is approximately the same as it was before, with just the filament.
    Last edited: Jun 28, 2012
  4. Jun 29, 2012 #3
    Wait, so am I right that the power developed by the bulb is the same is she right that the power developed by the bulb is greater? Thanks! Btw is the shunts resistance the resistance of the additional resistor?
  5. Jun 29, 2012 #4


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    Hi sgstudent. Does the question refer to adding a parallel resistor across one and only one of the 20 bulbs in the string? I suspect that it may, so you should clarify that with your teacher.

    If the modification is to all of the bulbs, each and every one, and identically, then of course your answer is correct, there will be no change in any bulb's brightness, and for the reason you so eloquently put forward.
  6. Jun 29, 2012 #5


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    I'm puzzled why you're asking this, because the answer is there in the math above. If you look at my (full, not approximate) expression for the power, you see that it just reduces to P = V2/R + V2/r. In other words, the total power dissipated is just the sum of the power dissipated in each resistor. We assumed that R was so large that we could ignore the first term.

    So, in the case where you add a large resistor in parallel, the total power is approximately the same as, but slightly larger than the power dissipated in the case where there is just the filament.

    Yes, it's common refer to a resistor that you add in parallel like that as a "shunt resistor." A shunt is like a bypass or an additional path for current to flow.
  7. Jun 30, 2012 #6
    Oh sorry I didn't read the last part properly. In the question, a picture was given for the bulb - there is two connecting wires to the resistor with a parallel branch for the filament above it. Also, will it be okay for me to edit some stuff to show this to my teacher? Thanks again guys

    Oh when You said my teachers approximation was right are you referring to me? Because my teacher's explanation was only about have more power from the whole bulb as you said V^2/R+V^2/r (but the V^2/R part doesn't contribute to brightness) after I said that, she said that because R is very high so there is a short circuit and almost none passes through the shunt resistor. But either way the P across the filament still remains the same right?
    Last edited: Jun 30, 2012
  8. Jun 30, 2012 #7


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    Yes, the power used up by the *bulb* is not affected by the change. Therefore, the brightness of the bulb will be exactly the same as before. This is not an approximation, it's exact.

    The total power used up will be slightly larger in the case with the shunt resistor, because some extra power will be used up in the shunt resistor. However, since the shunt resistor is large, it will draw only a tiny current, and so we can say that this extra power used up is approximately 0. Also, you are right that the power dissipated in the shunt resistor will be only in the form of heat (it will emit infrared light, but not visible light).
  9. Jul 1, 2012 #8
    Thank you cepheid I will be editting some stuff in order to show my teacher this will that be okay for the forum?
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