Light bulbs in a circuit


by JJBladester
Tags: bulbs, circuit, light
JJBladester
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#1
Mar2-10, 06:49 AM
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1. The problem statement, all variables and given/known data

Connect three bulbs as shown in the figure. Predict how the brightness of both bulbs A and B will change if you unscrew bulb C. Explain your reasoning.

Image here:
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2. Relevant equations

N/A

3. The attempt at a solution

I Think A and B will both get brighter because there is less resistance in the circuit overall. However, I am not sure if that's true because of the fact that A is in parallel with B and C and B is in series with C.

I know that the equivalent resistance for a bunch of resistors in series is additive (Requiv = R1 + R2 + R3, etc... and that the resistance in a parallel circuit is an inverse relationship.
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Sovos
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#2
Mar2-10, 08:44 AM
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I think this problem could be understood in 2 ways.
Regardless of which way it is worded: At the start, A would be the brightest, since path A offers the least resistance of the two paths in the circuit. B and C would have a brightness less than A, but equal to each other.

Perception 1: When unscrewing C, wouldn't B go out assuming the path along BC is now an open circuit? In this case, the brightness of A would increase as the full voltage of the power supply is now being dropped across A, and the brightness of B would, of course, now be zero.

Perception 2: If the problem simply says to remove C from the the circuit, then the resistance along the 2 paths is now equal. A's brightness would diminish and B's would increase, so that they are now of equal brightness.

The wording of the problem is crucial in this problem. The difference lies in unscrewing a light bulb (creating an empty socket in the circuit) or removing C completely, leaving the circuit intact.
Mindscrape
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#3
Mar2-10, 08:51 AM
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You're right that there will be less resistance in the circuit overall, so you know that there will be more current now, but you have to think about where the extra current is going to go now. Think of the power relations: P=V^2/R and P=I^2 R. Does the voltage change across lightbulb A?

JJBladester
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#4
Mar2-10, 04:35 PM
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Light bulbs in a circuit


Quote Quote by Mindscrape View Post
You're right that there will be less resistance in the circuit overall, so you know that there will be more current now, but you have to think about where the extra current is going to go now. Think of the power relations: P=V^2/R and P=I^2 R. Does the voltage change across lightbulb A?
The voltage doesn't change across A as far as I know. I think the only thing that would change the voltage would be to replace the battery with a more (or less) powerful one.
JJBladester
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#5
Mar2-10, 04:41 PM
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Also... what is it that makes a light bulb get brighter? Is it voltage or current? Why?

I know that voltage is sometimes likened to a fat pipe whereas current is likened to the amount of stuff flowing through a pipe.

If more current means more brightness, what does voltage do?
mgb_phys
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#6
Mar2-10, 05:08 PM
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The brightness of the bulb depends on the power = voltage across it * current through it.
Assuming the power supply can still provide the same voltage, then the only thing that can change the current through the bulb is it's resistance.

Without getting into real world complications (in a real bulb the resistance depends on the temperature which depends on the current, which depends on the resistance - so it gets complicated) the only thing you need to consider is the change in voltage across the bulb.

If you unscrew C does this change the voltage across A?

If you unscrew C - where does the current flow through B?
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#7
Mar2-10, 05:50 PM
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Quote Quote by mgb_phys View Post
The brightness of the bulb depends on the power = voltage across it * current through it.
Assuming the power supply can still provide the same voltage, then the only thing that can change the current through the bulb is it's resistance.

Without getting into real world complications (in a real bulb the resistance depends on the temperature which depends on the current, which depends on the resistance - so it gets complicated) the only thing you need to consider is the change in voltage across the bulb.

If you unscrew C does this change the voltage across A?

If you unscrew C - where does the current flow through B?
Unscrewing C doesn't change the voltage across A.

I found this and it helps a lot...

When considering various circuit arrangements, the old adage that “current follows the path of least resistance” holds true. When comparing bulbs in combination arrangements, it is useful to “trace” the current as it flows around the circuit. As the current reaches places where it can split, we can be sure that more current will pass through the paths of least resistance, making bulbs in these sections brighter than the others. The current will split proportionally to its resistance, meaning that in a parallel branch with two series bulbs in one branch and a single bulb in the other, twice as much current will flow through the branch with the single bulb (66.7% of the total current) since it has one-half as much resistance as the other branch, which receives 33.3% of the current.
Mindscrape
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#8
Mar2-10, 07:58 PM
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Yep that's absolutely right. For example, let's say the lightbulbs are 10 ohms and the battery is 10V. 1A flows through the lightbulb on the left and 1/2A flows through the two on the right. The battery puts out 1.5A total, so the lightbulb on the left gets 2/3 of the current and the lightbulbs on the right get 1/3 of the current. The power through the left is 10W and power through the lightbulbs on the right is 5W (2.5W per bulb).

So now what happens when it is just two 10 ohm resistors in parallel? Assume that the battery can source the proper amount of current to maintain the voltage (this must be why Sovos thinks there should be a slight decrease is because in reality the battery may not be sourcing all that it should).
JJBladester
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#9
Mar3-10, 06:40 AM
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Quote Quote by Mindscrape View Post
The battery puts out 1.5A total...
I thought the battery was only capable of providing a potential difference (V) due to the differing charges on its inner plates. Did you mean 1.5V?

I like your example, Mindscrape... Could you possibly lay out the math step-by-step? FYI this is beyond what the homework asked me, but I'm just interested in how it all works. Thanks in advance!
Mindscrape
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#10
Mar3-10, 11:09 AM
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The battery does whatever it takes to maintain the potential difference (V) that it is supposed to do. So depending on the load, it may put out 100mA of current, 1A of current, or 1.5A of current. You know that the battery maintains that potential difference, and anything in parallel will also have that potential difference.

As far as the math goes, you know by Kirchoff's Voltage Law (KVL) that all the potentials in a loop have to sum to zero. FYI, there is an EM law that says this too, but apparently, and I've never myself found out the difference, the circuit law is for some reason different from the EM law.

Loop 1 (going clockwise, battery around the loop through light bulb A, and note that V=IR):
-10V(battery) + I_1*R (lightbulb A) = 0
-10V + 10*I_1 = 0
I_1 = 10/10 = 1A

Loop 2 (going clockwise, battery around the loop through light bulb B and C):
-10V + 10*I_2 (light B) + 10*I_2 (light C) = 0
20*I_2=10
I_2=1/2
mgb_phys
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#11
Mar3-10, 12:06 PM
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Since the question doesn't describe the power supply, ideal source, current limited PSU or real battery with internal resistance - you have to assume it's a perfect constant voltage source.


The current will split proportionally to its resistance, meaning that in a parallel branch with two series bulbs in one branch and a single bulb in the other, twice as much current will flow through the branch with the single bulb (66.7% of the total current) since it has one-half as much resistance as the other branch, which receives 33.3% of the current.
Yes but the question doesn't ask that - it only asks what happens when you disconnect bulb C.
For an ideal source, removing C doesn't have any effect on A. The voltage is the same, the resistance is the same - so the power and brightness are the same.
JJBladester
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#12
Mar3-10, 05:32 PM
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Quote Quote by Mindscrape View Post
The battery does whatever it takes to maintain the potential difference (V) that it is supposed to do. So depending on the load, it may put out 100mA of current, 1A of current, or 1.5A of current. You know that the battery maintains that potential difference, and anything in parallel will also have that potential difference.

As far as the math goes, you know by Kirchoff's Voltage Law (KVL) that all the potentials in a loop have to sum to zero. FYI, there is an EM law that says this too, but apparently, and I've never myself found out the difference, the circuit law is for some reason different from the EM law.

Loop 1 (going clockwise, battery around the loop through light bulb A, and note that V=IR):
-10V(battery) + I_1*R (lightbulb A) = 0
-10V + 10*I_1 = 0
I_1 = 10/10 = 1A

Loop 2 (going clockwise, battery around the loop through light bulb B and C):
-10V + 10*I_2 (light B) + 10*I_2 (light C) = 0
20*I_2=10
I_2=1/2

Thanks for all of this Mindscape. I didn't know Kirchnoff's law but reading further in my textbook shows that the sum of the current into a junction shall equal the sum of the current out of a junction. Another of his laws states that the sum of all the potential differences encountered while moving around a loop or closed path is zero. I believe it is this second law that you used above to show what the resulting current would be through each loop. With V = IR, we can set the sum of all of the I*R's in a loop to zero and solve for unknowns. Reminds me of equilibrium equations and free body diagrams.

This topic (Kirchnoff's Laws) is actually the next chapter in my Physics II course's reading line-up. It's called "Potential and Field". Can't wait to read more tonight!
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#13
Mar3-10, 05:34 PM
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Quote Quote by mgb_phys View Post
For an ideal source, removing C doesn't have any effect on A. The voltage is the same, the resistance is the same - so the power and brightness are the same.
But... you are removing a bulb, which is in essence a resistor, from a circuit. All equations aside, less resistance would mean more current and more current results in more power and that makes the remaining bulbs brighter... or did I miss something?

By the way, the question is assuming a fallable everyday D cell battery that you'd buy off the shelf and put into a flashlight or child's toy.
mgb_phys
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#14
Mar4-10, 09:43 AM
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The voltage available from a real battery with finite internal resistance will reduce with current but not very much - and of course it depends on the current.

eg. Assuming a D cell with 0.25R internal resistance, if these were 1W lamps using 650mA then halving the total current drawn will only change the voltage by 0.15V = 10%

The question is also ambiguous about what it means to 'remove C' do you mean remove it from the circuit leaving a break - in which case B doesn't light at all - or does it mean replace it with a short, in which case A and B are equal.
JJBladester
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#15
Mar5-10, 08:57 AM
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Bulb C (as well as A and B) is a mini-bulb and it sits in a tiny socket. When you remove C, the socket remains and current can still move along the circuit, just without the resistance of having C plugged in.


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