series circuit

Discussion in 'Homework Help' started by GMChandio, Feb 26, 2015.

1. GMChandio Thread Starter New Member

Feb 26, 2015
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three bulbs of ratings 60W, 80W and 100W are connected in series to work on 240V. Which bulb will burn most brightly?

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2. Veracohr Well-Known Member

Jan 3, 2011
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Your bulbs are connected in series, yet you have calculated different currents for each. Revisit Ohm's Law and series circuits.

3. crutschow Expert

Mar 14, 2008
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4,338
Okay you've calculated the resistance of each bulb when it's connected by itself to 240V.
Now put them in series and calculate which one dissipates the most power (that one will be the brightest) when 240V is applied to the string.
(The resistances of real bulbs will actually go down when each is operating at a lower voltage in series but we don't have to worry about that for the purposes of this question).

4. nDever Active Member

Jan 13, 2011
154
5
You say that the bulbs are connected in series to a 240V source, but your work specifies a voltage of 220V.

5. WBahn Moderator

Mar 31, 2012
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You don't saw what voltage the three bulbs are individually rated at. If this is all the information that is given, you need to make some assumptions because I can pick voltage ratings for the three bulbs such that any one of the three would be the brightest when all are put in series. What's the simplest assumption that makes the problem solvable as given?

6. Dodgydave AAC Fanatic!

Jun 22, 2012
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The 60 W bulb will be brighter, as it has the largest resistance,so will have the largest voltage across it,

7. WBahn Moderator

Mar 31, 2012
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Just having the largest voltage across it isn't a sufficient reason. After all, if they are all in parallel then they all have the same voltage across them and then the one with the least resistance will dissipate the most power. The argument needs to exploit the fact that they are in series.

8. GMChandio Thread Starter New Member

Feb 26, 2015
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So what do you suggest should be the answer?

Feb 26, 2015
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10. WBahn Moderator

Mar 31, 2012
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But how can that be? First off, 40 is just a number. You need to pay attention to units. Assuming you mean 40W, you don't HAVE a 40W bulb to choose from. You have three options -- the 60W, the 80W, or the 100W bulb.

11. WBahn Moderator

Mar 31, 2012
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Depends on what assumptions you make.

If I assume that the 60W bulb is intended for 12V operation, that the 80W bulb is intended for 120V operation, and the 100W bulb is intended for 240V operation, I'll get a different answer than if I assume it the other way around. That's why I asked you what the simplest assumption is that makes the problem workable. As it stands, unless you make SOME assumption, the problem has no definite solution.

12. GMChandio Thread Starter New Member

Feb 26, 2015
28
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Sorry, by mistake I took the wrong value.

13. Dodgydave AAC Fanatic!

Jun 22, 2012
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If all the bulbs are rated for the same voltage, the lowest wattage bulb will be the brightest in series.

14. djsfantasi AAC Fanatic!

Apr 11, 2010
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I'm a little confused. Could you explain the error in my math?

First, all the bulbs are in series, so that tells something about the current and voltage across each of them. Calculating the resistance based on their wattage gives us these general formulas.

$
W=I^{2}R

60=I^{2}R_{60}
80=I^{2}R_{80}
100=I^{2}R_{100}
$

Note that since the bulbs are in series, current can be treated specially in this analysis, and the result is that we can see that Watts are a certain type function of R. So what can we say about the following relation?

60 < 80 < 100
$
R_{60} ? R_{80} ? R_{100}
$

Note: Could not to save my life, get this to work with Tex!

Last edited by a moderator: Mar 2, 2015
15. WBahn Moderator

Mar 31, 2012
20,076
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Please let's remember that this is Homework Help and not Homework Done For You. The idea is to guide the OP into finding the answer for themselves, not simply giving them the answer.

To the OP:

Q1) If the bulbs are in series, what do we know about the current through them?

Q2) If two different-valued resistors have the current relationship from Q1, which one dissipate the most power?

Q3) If two different-wattage bulbs are rated at the same voltage, which one has the lower (or higher) resistance?

16. MrAl Distinguished Member

Jun 17, 2014
3,625
763
Hello,

Amazingly, the one that came out the brightest was the 80 watt bulb.

This is really a static nonlinear problem. The vacuum lamp characteristics follow a power
law, and will deviate from this for wide changes in operating point.

Using common approximations for incandescent lamps, i calculate the light output from all
three lamps, and the results are normalized:
60 watt lamp: 4.06
80 watt lamp: 9.51
100 watt lamp: 6.01

These are the relative light outputs from all three bulbs in this series circuit, and
because the 80 watt lamp has the most output it will show up as the brightest from any
angle chosen for viewing all three bulbs. The results are expressed in units of "relative
lumens".

Assumptions and reason for them:
1. All the bulbs are 240v incandescent vacuum bulbs. Usually bulbs run at the stated
voltage.
2. All the bulbs have the same luminous efficacy. Same construction.
3. All the bulbs have the same construction, shape, size, etc. They are usually similar
enough even for different powers.
4. The bulbs behave like most incandescents in that they have a power law over a portion
of their behavior.
5. All the bulbs are viewed at the same solid angle and distance for brightness
comparisons.
6. If any of these assumptions change the result could change drastically. For example,
if the 60 watt bulb was made to be used normally for a 50 volts source instead of 240, it
could shine much brighter.

Solution:
As noted, this was done as a static non linear problem. The bulbs current, voltage, power, and
efficacy are all non linear because they follow a basic power law over at least part of
their total behavior.
The key to solving this is the fact that the current in all three bulbs is exactly the
same.
Given the function that relates voltage to current we can solve for the (as yet unknown)
voltage for each bulb. We can then form a sum
Vs=v60(i)+v80(i)+v100(i)
Vs is the source voltage of 240v. The three functions on the right are actually solved
for the current because all three currents are equal.
Once we get the current, we can then go back to the function that relates voltage to
current and solve for the voltages We can then solve for the power (or look for a
function that can give us the power from the current and then calculate the power).
Now we have the current and voltage for each bulb, and the power for each bulb.
Next, using a function that relates the efficacy to the current (or voltage) we solve for
the normalized efficacy, then multiply by the watts for each bulb to get the relative
light output for each bulb.
Assuming each bulb is viewed at the same angle, we can then determine which bulb appears
brightest.
We can look into these functions deeper if you like. They are not that hard to solve,
just some algebraic manipulation.
It should also be noted that the solution led to an unusually low current for all the
bulbs so the functions used probably do not hold exactly. For a more exact solution we'd
have to find better functions than the ones given in my engineering handbook. We can do
this too. The solution will follow the same procedure: solve the functions for the
unknown voltages, sum all the voltages and solve for the current, then plug back to solve
for the actual voltages, then calculate the watts, then the lumens per watt, then the
lumens.

We might also have to look at the bulb dynamics. For example, what if the 60 watt bulb
heats up much faster than the other two. It may drop much more voltage than in the static
case before the other two get hot, and that would make it brighter than calculated. This
is interesting because the thermal mass will definitely be larger for the 80 and 100 watt
units meaning they will heat up slower. We would only be able to tell if we used a more
sophisticated model or alternately apply full voltage to all the bulbs through a fast
switching network, then quickly switch them back in series. This way they would not be
connected in the circuit until they were all hot already.
Anyone up for the test, it's not that difficult? We could use 120v bulbs instead and a
120vac supply. All the bulbs should come from the same manufacturer.

Last edited: Feb 28, 2015
17. JoeJester AAC Fanatic!

Apr 26, 2005
3,663
1,530
My calculations shows the power for the new configuration .... I think I'll let the TS figure out which represents which one is the brightest.

18. tjohnson Active Member

Dec 23, 2014
625
122
@djsfantasi: Just enclose it in {tex} tags, as explained in this post.
Code (Text):
1. [tex]W=I^{2}R
2. 60=I^{2}R_{60}
3. 80=I^{2}R_{80}
4. 100=I^{2}R_{100}[/tex]
5.
6. [tex]60 < 80 < 100
7. R_{60} ? R_{80} ? R_{100}[/tex]
EDIT: This post is even more comprehensive in explaining how to use LaTeX. I didn't mention it earlier because for some odd reason I couldn't see any posts in it when I looked at it before.

Last edited: Feb 28, 2015
19. MrAl Distinguished Member

Jun 17, 2014
3,625
763
Hello again,

I had to redo the calculations in post #16 due to an error. Amazingly, the one that came out the brightest was the 80 watt bulb !

This makes sense when you think about it, because the 60 watt bulb doesnt shine as bright as the others even at full power, yet the 100 watt bulb is operating way under current, so the one that wins is the one in between where it gets a reasonable current and voltage relative to what it normally has, even though not what it really needs for full brightness.

To repeat the relative brightnesses:
60 watt lamp: 4.06
80 watt lamp: 9.51
100 watt lamp: 6.01

The 80 watter is highest. This was quite an amazing result.

Still interesting would be a time model for the bulb filaments to see if the 60 watter heats up fast enough to steal all the voltage before the others get a chance to heat up significantly.

20. WBahn Moderator

Mar 31, 2012
20,076
5,666
While that may be the case, I suspect that the homework problem was intended to use a far simpler model, probably one in which the resistance of the bulb filaments is completely temperature independent. Not realistic, but I don't think that was the point of the problem.