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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.
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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  Junior Member
Join Date: Sep 2009
Posts: 22

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  Member
Join Date: Nov 2008 Location: PAKISTAN
Posts: 79

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, 09:02 AM

#8  Senior Member
Join Date: Dec 2009
Posts: 209
 This is what happens when you attach loads to a source of potential energy. Hi there, So you have a source of potential energy that can shove electrons along a wire through a set of bulbs and then back to the other end of the source. This source of potential energy is what you call the battery for the circuit. It is a potential source of energy because it has the potential when applied to a load to deliver energy to that load where it can perform work upon that load, it may be transformed, used to trigger another event, converted into other forms of energy such as light and heat energy; in all of these cases the energy delivered to the load is done so over some time period. Now energy delivered over some period of time means that a particular amount of power was delivered to that load (where power is just energy per unit time). Now your battery has an upper limit in how much power it can supply to any number of loads attached to its leads. Let's call this upper power limit Pu (neglecting internal power loses, but I'll get back to this area in a little bit). Now, as long as your battery can supply the current needed for each little load, then no the bulbs do not diminish in intensity. They only diminish in intensity if they do not get the current they want/need to illuminate fully. You're limited in the amount of current that you can supply to all the bulbs by the amount of available power that the battery can deliver. By placing another bulb in parallel with the one(s) you already have in the circuit you open up a new path for electrons to go through, therefore as far as the source is concerned it "sees" a lower resistance attached to its leads because you have given the source another path to let charge flow through, so more charge leaves the source and heads down this new path, along with all the other charges that were flowing before you added another bulb. At some point after adding enough bulbs the battery "says, hey I can't give you anymore current, I'm throwing out electrons as fast as I can and I can't do it any faster or in any greater quantities then I am now doing". The moment you add another bulb to this circuit in parallel with the others, you've now "loaded" the supply to the point where not only can't it deliver the current needed to get the bulb to light up with the same intensity as all the others have, but in adding this extra bulb, some of the current that was going through all the other bulbs in the circuit gets shoved into the new bulb that you've added. Since the source can no longer increase the amount of current within the circuit, the bulbs have to "share" that available current amongst themselves. This will decrease the intensity of light coming from the bulbs as now the amount of power being dissipated or rather being dissipated and converted into light energy by each bulb is slightly less than it was before. Also Ohms law must hold true at all times during circuit operation, so as you add more and more bulbs to the circuit you keep decreasing the resistance seen by the source, which in turn will keep increasing its current output until it can longer deliver anymore current than it is doing at this point in time, but if you keep adding bulbs and therefore keep lowering the resistance, if no more current can flow, and power and energy are conserved values then something has to give in order to satisfy Ohms law. What gives is the voltage drop across the bulbs, here's why. If your delivering the most current the battery will deliver to a load then the equations can be written as, Pu = V*Imax and V = Imax*R. Looking at the second equation one can see that I can get no larger, it is at its maximum value, but R can become smaller and does as you add more bulbs to your circuit, therefore the product of some current and a lowering resistance means the voltage drop across the bulbs must also decrease in order to keep Ohms law true. No matter how many bulbs you add you can never get the resistance seen by the battery to be equal to zero, you would need an infinite number of bulbs in parallel to get a zero resistance load; finite resistance with a current flow means a finite voltage will exist. Basically what is happening is that as you add more bulbs to your circuit the resistance seen by the source decreases to the point where the internal resistance of the source starts to compete and then become larger than that seen by the source from the attached load(s), so the voltage that keeps dropping as you add more and more bulbs, is actually being dropped inside the battery across its internal resistance, which is typically small compared to the load and therefore is often ignored. So initially, assuming your battery can supply enough power for say 10 bulbs, then every bulb will be of the same brightness as all the others from 1 bulb all the way up to 10 bulbs, after which point, the supply can no longer supply the needed current for an eleventh bulb, therefore the voltage across the battery terminals drops by some amount to keep Ohms law true (V = I*R), and the brightness of all the bulbs drops by some small amount as each of the first 10 bulbs now has to give away a little bit of its current to the eleventh bulb such that each bulb now has the same current flowing through all of them. Another way to look at this is your battery has the potential to deliver some maximum amount of power to a load. If that load is a number of small bulbs each of which needs only a fraction of the available power to fully illuminate then they will take the power they need and illuminate fully. At some point after attaching enough bulbs the battery no longer has any additional power it can provide for extra bulbs, so it instead takes some of the power that is being delivered to all the other bulbs in the circuit and keeps "spreading" out its available power to all the bulbs within the circuit. So the more bulbs you add, the less power available for each bulb. For the individual bulbs, probably the best way to find out how much power an individual bulb is using is to measure the voltage drop across the bulb, or the terminals of the battery since the bulbs are in parallel with the battery leads, these voltages must be the same, and I would probably just measure the total current flow from the battery and divide it by the number of bulbs within the circuit and say on average that much current must be flowing through each bulb, only because it is easier to put an ammeter in line with one of the wires attached to one of the terminals of the battery than it is to try and stick an ammeter inside the path of an individual bulb. With these two values you can say that the average power dissipated by each bulb is (Vbattery*Itotal)/(number of bulbs). If you want the exact power being used by any bulb in particular you would have to measure the current flow through that particular bulb and multiply it by the voltage drop across that bulb; I'm quite sure you will find, as long as all the bulbs are basically the same, that the power used by any bulb in particular will be very very close to what each bulb uses on average. As far as the internal resistance of the battery goes, you should only worry about it when your supply is "maxed" out in terms of its power delivering capability. As long as you keep the number of bulbs within your circuit under a certain number such that the battery is able to deliver the desired current to each bulb so they fully illuminate, then why care about the power lost to the internal resistance of the battery. If you add on more bulbs to the point where you notice them to be dimmer than at an earlier time when there were less bulbs attached, then you are dropping more power inside the battery than you should be doing, and in fact if you go overboard and attach a whole bunch of bulbs to your circuit then the voltage seen at the battery terminals will most likely drop to a very small value as most of the voltage is now being dropped across the internal resistance of the battery; this is not a good thing, because now you have a maximum current flow coming from the battery whose voltage potential is being almost entirely dropped within the battery itself; this causes a far larger power dissipation inside the battery (P=V*I) that may be unsafe and will cause the battery to become very hot; because in a sense you are basically shorting the terminals of the battery together (as can be done with a short piece of conductive wire) by attaching so many loads in parallel. If you did this a maximum current would flow through the wire from one terminal into the other. Since power = V*I and most of the V is dropped inside the battery, with a maximum I flowing, then almost all of the available power the battery can deliver to a load is actually dissipated within the battery itself and very little is dissipated by the load (a short piece of wire in this case). Therefore the battery gets hot, no useful work is done by the battery and therefore power is simply being wasted. Many Smiles, Craig 
 
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"

 
Dec 31st 2009, 10:30 AM

#10  Senior Member
Join Date: Dec 2009
Posts: 209
 One more thing to note Hello again, I should have pointed out that as more bulbs are added to the circuit in parallel, the more current that is drawn from the source and hence the greater the internal voltage drop inside the battery. Again, though, if the supply is capable of supplying more current then simply turn up the dial on the voltage source ever so slightly such that the voltage seen between the terminals is once again what it was before another bulb was added; if not, the additional voltage that is dropped within the source, while it would be a small increase in the internal voltage drop, it would nonetheless take away from the voltage seen between the leads of the supply to the circuit. So even though lowering the load resistance alone, if the supply has the capability, will source more current to the load, the product of this increased current and the resistance of the bulb will be slightly less than that before the extra bulb was added, because that slight difference is lost to the sources internal resistance and is dropped across it leaving less potential for the loads; hence decreasing power used by the bulbs. If the source voltage can be increased by the same amount that was lost due to the larger internal voltage drop, then each bulb gets the power needed to shine with the same full intensity they are made to emit light at. I was assuming that would be a known thing to do on your part if you wanted the brightness to remain unchanged as you added more bulbs, so I didn't mention it, but after giving it some thought, I decided if I didn't say something about it you may be under the impression from what I wrote that simply adding more bulbs to a source that can deliver the current needed to make the bulbs shine brightly means that in adding more bulbs they will all still shine with the same intensity as before more bulbs were added; and this is not true. Adding more bulbs decreases the load resistance seen by the source therefore allowing the source (if capable) to deliver more current to the load, but this increased current must first flow through the internal resistance of the source which will cause a larger internal voltage drop leaving less voltage to be dropped across the bulbs. Therefore one needs to increase the source voltage by the same amount it decreased by in order to keep the bulbs shining at the same brightness as before. Everything I had said about loading the source still holds true, in that at some point after adding enough bulbs the resistance decreases to the point where the voltage cannot be increased any further and the supply cannot source any more current than it is doing at that time. Furthermore if you then keep adding more bulbs, the internal source resistance no longer seems so small compared to the load resistance and therefore more and more voltage is dropped within the source leaving less for the loads. At some point with enough bulbs attached, the voltage across the bulbs will be small compared to that dropped across the internal source resistance and with the maximum amount of current flowing through the circuit at this point, the power being dissipated inside the battery is far larger than the amount of power that is being delivered to the bulbs. Therefore the bulbs may just barely shine and the supply would probably get too hot to touch (without getting burned). It's not so much that I assumed that you knew you would have to do this in order to keep the brightness the same as more bulbs are added, but it being such basic circuit theory I know that it has to be done and would automatically do so to keep the brightness the same. I wasn't thinking about whether or not you knew to do so. It's kind of like as if I lent you my car, you would assume that I left gas in the tank or you wouldn't be able to go anywhere; you don't ask me if there's gas in the car you simply know there's going to be gas in the car, as I simply know that for the brightness to remain the same in the parallel load circuit, the voltage supply would have to be adjusted for every bulb attached until a point is reached when that voltage can no longer be increased and no further current can flow, at which point if you keep adding bulbs to the circuit the voltage at the terminals starts to drop as more is dropped within the source than across the bulbs and the current flow remains constant at its maximum value. For not mentioning that, I apologize; just because I know that as the current from a supply is increased within a circuit the voltage seen at the terminals of that supply has dropped by a small amount due to the increased voltage drop within the supply, doesn't mean that you know that. If you did, great, then just ignore everything I've just written in this reply, if you didn't then now you do. Craig
Last edited by clombard1973; Dec 31st 2009 at 11:45 AM.

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