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Electric circuit

  1. Oct 28, 2007 #1
    1. The problem statement, all variables and given/known data
    1. 3 light bulbs are connected parallel in the circuit ,one of them burns out .How this affects the other bulbs.
    2. We have combination of parallel and series circuit(3 bulbs) and one of them R3 burns out .
    This gives us a series circuit with 2 bulbs .What happens to the light?-----R3------

    2. Relevant equations

    3. The attempt at a solution

    1.I know that in parallel circuit V1=V2=V3.If one bulb burns out the rest will stay on,but I'm not sure if the brightnes will stay the same.
    2.For series circuit I1=I2,current will be the same,voltage will be different.What happens with the light ,is R1( one what was not in the parallel circuit )brighter?Please help me to understand this because I'm getting confused.
  2. jcsd
  3. Oct 28, 2007 #2
    1.why would the brightness stay or not stay the same?
    2.Dont think you are entirely correct here.

    Maybe draw simple circuits. give them arbitrary values like 1 ohm, 2ohms and 3ohms and see what results you get.

    When two resistors are in parallel, their equivalent resistance is lower than the lowest ohm resistor(in that parallel combination).
  4. Oct 29, 2007 #3
    Brightness is related to power. (Increases if power increases.)
    Formula for power, P = (V^2)/R = (I^2)*R = V*I.
    You just have to take care, which thing is remaining constant and what is varying.
    Remember that resistance of an ohmic resistor does not depend upon voltage applied or current flowing. (We assume, generally, ohmic resistor.)
    1st one you did well. As they are in parallel, voltage drop across each of them is equal (and equals the voltage applied). Even if one of them burns out, the voltage drop across the resistors do not change (and continues to be equal to the applied voltage). So, V is constant. Therefore, rest of them continues to glow with same brightness. {From P = (V^2)/R.}
    2nd one is slightly different. As, one of the parallel resistor has burnt out, circuit is still complete and equivalent resistance of the circuit increases. (Satisfy yourself.) Therefore, current entering the circuit decreases. So, R1 (one that was not in parallel) will glow dimmer than before.
    Here is a question for you: Can you show that R2 will glow brighter than before!?
  5. Oct 29, 2007 #4
    Furthermore, if you are talking of total light available:
    1. "the equivqlent resistance will increase,voltage will stay the same, dimmer light.Is this correct?"
    Yes, it is correct. But a word of caution about the current. Current flowing out of the voltage source decreases but not through individual resistors. Brightness of each of the remaining two bulbs is same as they were initially. Please note current through each of the remaining resistor remains same. But initially, there was an extra bulb. Therefore, total brightness decreases. {Power decreases by an amount (V^2)/R3 -- if R3 burns out.}

    2. This one you have got right. Since the equivalent resistance increases, light gets dimmer. {Show it!}
  6. Oct 29, 2007 #5
    Why R2 should be brighter than before? When one bulb burns out in the parallel circuit it does not affect the other bulbs,so it should stay the same.Is that correct.
  7. Oct 29, 2007 #6
    No! It does not affect only voltage drop (and other related things) for the ones in parallel. It affects for the series ones. You can say, in effect, the source voltage for the parallel circuit has changed! Well, dont rack your brain on what I said, rather work out the power delivered by R2. Later on you can try to give a logic to what you obtain!
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