Physics Help Forum Parallel circuits

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 Dec 10th 2009, 09:17 AM #1 Junior Member   Join Date: Sep 2009 Posts: 22 Parallel circuits Hello Now, suppose there are one light bulb connected in a complete circuit, later we add one more light bulb to the circuit, in which the bulb is identical to the first one, such that the two bulbs are in parallel. In this case, I think the individual brightness of the two bulbs will decrease and it is very logic to think that --- as more light bulbs are added to the circuit in parallel, the brightness of EACH bulb will decrease... it is very logic.. but why as I use formula to calculate the power (which is related to brightness), the story is different? P=VI and then P=V sqaure over R no matter how many light bulbs are added in a parallel way, the voltage across each bulb should always equal to the cells, which resistance is the intrinsic properties of the bulb that is not affected by the number of bulbs connected in parallel. V and R of individual light bulb unchanged with added light bulb. and so Power of each bulbs (and thus brightness) should remain unchanged with added number of light bulbs in parallel, with the fixed number of cells. I think it is quite not logical, as the number of light bulbs increases, I think they will all go less bright. What's problem behind /
 Dec 10th 2009, 06:26 PM #2 Senior Member   Join Date: Apr 2008 Location: HK Posts: 886 One thing to note is that although the brightness of each light bulb in parallel remains the same in a parallel circuit but the power dissipated increases with the number of light bulbs. We know that power dissipated in a light bulb is P=V^2/R, for a better understanding, it should be P=NV^2/R where N is the number of light bulbs. For the same number of cells, more power is drawn for more light bulbs. Think of the case that the more light bulbs, the more frequent you have to replace the batteries. __________________ Good results were achieved and the new task is to become a good doctor.
 Dec 11th 2009, 12:46 AM #3 Physics Team   Join Date: Feb 2009 Posts: 1,425 What you say will happen if the total current drawn is close to the rated capacity of the source say a cell. In this condition it is very obvious. Even otherwise actual cells have some internal resistance. The output voltage of the cell is the rated voltage minus the drop across the internal resistance. As more bulbs are added in parallel, more current is drawn and hence the output voltage also drops leading to lower brightness. Let a 10V cell have 0.1 ohm internal resistance. Let each bulb draw say 0.1 amp . When the first bulb is connected, the cell o/p voltage will be 10 - (0.1 x 0.1) = 9.99 V. When the tenth bulb is added in parallel, the o/p voltage will be 10 - (0.1 x 10) = 9 V. Since the power depends on the square of the voltage the effect is notiecable.
Dec 13th 2009, 12:45 AM   #4
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 Originally Posted by physicsquest What you say will happen if the total current drawn is close to the rated capacity of the source say a cell. In this condition it is very obvious. Even otherwise actual cells have some internal resistance. The output voltage of the cell is the rated voltage minus the drop across the internal resistance. As more bulbs are added in parallel, more current is drawn and hence the output voltage also drops leading to lower brightness. Let a 10V cell have 0.1 ohm internal resistance. Let each bulb draw say 0.1 amp . When the first bulb is connected, the cell o/p voltage will be 10 - (0.1 x 0.1) = 9.99 V. When the tenth bulb is added in parallel, the o/p voltage will be 10 - (0.1 x 10) = 9 V. Since the power depends on the square of the voltage the effect is notiecable.
Actually I don't quite understand what you said

How about if I assume there is no internal resistance for the cell, and only the resistance of parallel light bulb is considered? (so voltage across each bulbs is the same as that of the voltage of cell) What is the explanation in this case?

and how can u say that if one bulb in series draw 0.1Ampere then two bulbs is 0.2Ampere, and thus 10 bulb is 0.1Ampere, I would like to know
the explanation on this

Thanks you

 Dec 13th 2009, 10:23 PM #5 Physics Team   Join Date: Feb 2009 Posts: 1,425 If we take the internal resistance to be zero, that is an ideal cell. In this case its current capacity is also treated as infinite. If any number of lamps are connected in parallel to such a cell, it continues to give the required current and the same output voltage hence the bulbs will not dim. Bu we observe dimming in practice and that is because no cell is really ideal. We are assuming that the light bulbs are identical i.e. they have the same resistance and behaviour. Hence each bulb connected across a cell will continue to draw the same additional current. Actually i have simplified things a bit to facilitate understanding. Actually when we connect 1 resistor itself the voltage drops. So when we connect two it drops further. Thus the current passing thru each bulb now will be less than that passing thru the bulb in the first case as the voltage is lower. As for this, and how can u say that if one bulb in series draw 0.1Ampere then two bulbs is 0.2Ampere, and thus 10 bulb is 0.1Ampere, I would like to know There is a typo what meant to say is that since each bulb draws 0.1 amps 10 will draw 1 amp and hence 10 - (0.1 x 10) = 9 V. should have been 10 - (0.1 x 1) = 9.9 V. Last edited by physicsquest; Dec 13th 2009 at 10:32 PM.
Dec 30th 2009, 08:49 PM   #6
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 Originally Posted by werehk One thing to note is that although the brightness of each light bulb in parallel remains the same in a parallel circuit but the power dissipated increases with the number of light bulbs. We know that power dissipated in a light bulb is P=V^2/R, for a better understanding, it should be P=NV^2/R where N is the number of light bulbs. For the same number of cells, more power is drawn for more light bulbs. Think of the case that the more light bulbs, the more frequent you have to replace the batteries.
HI THERE, Pl correct me if i m wrong. when more bulbs are added to a parallel circuit, more current is drawn from the source,therefore,the brightness does not decrease, but we have to replace the battery more often.

1 and 1 makes 11

 Dec 31st 2009, 12:36 AM #7 Physics Team   Join Date: Feb 2009 Posts: 1,425 The brightness does decrease depending on the internal resistance of the cell as stated in the earlier post. When more current is drawn, the drop across the int resistance of the cell is increased leading to alower output voltage. Thus, the voltage across 4 bulbs in parallel is less than the voltage across 3 bulbs in parallel Of course you have to change the battery more often.
 Dec 31st 2009, 10:02 AM #9 Senior Member   Join Date: Sep 2009 Location: india Posts: 409 i really loved that post. it was sensational. however, will u please explain me why/how the battery has an upper limit to current ie. how it realises that "hey, i cannot supply electrons faster than this"

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What happens to the light intensity of its lamp in a parallel circuit when more lamp are added to the circuit

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