Physics Problem of the Week: 01/03/2004


Physics Problem of the Week: 01/03/2004

Well, we're on to electricity and circuits to draw this circuit out.hrm.

Okay, well I'll explain it first.

Four lightbulbs of the same wattage are connected on a circuit so that one B1 (bulb1) is in series with the battery and the other three are all parallel.

Attempt to draw:

Okay, so then, you bulb3 burns out completely, what happens to B1, B2, B4:

a) B1 gets brighter, B2 and B4 get brighter
b) B2 gets dimmer, B2 and B4 get dimmer
c) B1 gets brighter, B2 and B4 get dimmer
d) B1 gets dimmer, B2 and B4 gets brighter
e) None of the above

if you answer, please include an explanation.


Try this:


Diabloii.Net Member
Ok, Bulb one decreases, bulb 2 and bulb 4 remain the same intensity. Here is why:

When you remove bulb 3 from the circuit, there is less total current being drawn out of the battery. This means that bulb 1 disspates less power (P=I^2*R), and is dimmer.

Bulb 2 and 4 however, see the same voltage drop as they did before across bulb 1 (they still have the same voltage) and each branch draws the same current (since nothing in those branches changed, they are in parallel) and thus remain the same brightness

More mathematical:

Situation 1: All 3 bulbs: you'd have 3I flowing across B1, and 1I across B2, B3, and B4.

Situation 2: Bulbs 1, 2, 4. You'd have 2I flowing across B1, and 1I across B2 and B4.

So the answer is E, none of the above, because 2 and 4 remain the same brightness.


Diabloii.Net Member
battery puts out power, lights don't draw it. so bulb 1 stays the same, other two bulbs get brighter since I isn't split up over as many.

gogo A in physics


Diabloii.Net Member
heh, we are both wrong, but its a simple thing really.

Dantose, light bulbs are resistors, they just give off their power in terms of light, and we neglect heat in most cases.

Now, heres what happens:

In the first situation, you can minimise the resistor network (lightbulbs are resistors) to figure out what the current coming out of the battery is. You do this and you come up with a resistor of 4*R/3
Ohms law: V=I*R. V= I*4*R/3
Solving for I:
I= 3V/4R

That is the current across B1. Since the other 3 bubls are the same resistance, they get equal current, or V/4R.

Now if we remove one resistor from the parallel network, you get an equivilant resistor of 3R/2 By ohms law again:
I=2V/3R This is LESS Current than before, meaning that B1 gets dimmer.
But again, B2 and B4 are the same, meaning they get equal current, or I=V/3R, a LARGER amount of current than before, meaning they get brighter.

Calculating Brightness:

You can figure the brightness by the power dissipated. For a resistor, or a lightbulb, the easiest way to do this is P=I^2*R. In the first case:

B1: P=9V^2/16R
B2: P=V^2/16R
B3: P=V^2/16R
B4: P=V^2/16R

Without B2:

B1: P= 4V^2/9R
B2: P= V^2/9R
B4: P= V^2/9R

The easiest thing to do to get a hard number is go with a 1V batter, 1ohm resistor. For each of the bulbs above:

First case:

B1 = 9/16 Watt
B2 = B3 = B4 =1/16 Watt

Second Case:
B1 = 4/9 Watt
B2 = 1/9 Watt

So the end answer you want is D, that B1 gets dimmer, and B2 and B4 get brighter. Think of it this way, when you plug a vaccuum cleaner into the wall, you pull more current, and the other lights, that are in parallel, get dimmer.


I think it's a), they all get brighter:

a) B1 gets brighter, B2 and B4 get brighter.
First of all, with all four bulbs working, let's name the voltage across B1 in series as V1 and the voltage across the parallel resistors as V2. V1 + V2 is equal to Vtotal. We now take the same voltages but the parallel section only have two bulbs left due to to B3 burning out, this is now V2' and the voltage across the series bulb is known as V1'. V1' + V2' is also equal to Vtotal, therefore:

V1 + V2 = V1' + V2'

Now, we calculate the voltages with I being the current with all four bulbs and I' the current with B3 burned out:

I * r + I * (r/3) = I' * r + I' * (r / 2)

The r is canceled out and we factorize the I out:

I(1 + 1/3) = I'(1 + 1/2)

We simplify:
I * 4/3 = I' * 3/2

I = (9/8)I

And therefore:

I < I'

Which means that the current before B3 burned out is inferior to the current after B3 burned out. Thus, the current running across V1' is greater than V1 and B1 is now brighter. Since V1 equals V2 and V1' equals V2', V2' has a greater current than V1 and is only split amongst two branches, which means B2 and B4 are now much brighter.

Go go dad w/ physics major.


Diabloii.Net Member
Im pretty sure its A, i couldnt see the image but from the description, all lit bulbs should be lit evenly no matter how many burn out.

Edit: i have to say all that math an figuring is nice but its a very basic principle, which if you know it will take all of 2 seconds to figure out...that said its been years since i dealt with that stuff. the math helps if you dont get it, but is completely a waste of time otherwise.


Diabloii.Net Member
I stand by my circuit analysis. That and I've seen this done in a lab and I know the basic result, moreso what happens to bulb 1


Diabloii.Net Member
ShadoweOrbe said:
I stand by my circuit analysis. That and I've seen this done in a lab and I know the basic result, moreso what happens to bulb 1
Yes, answer A is correct. You didn't really explain it clear so I'll try to sum it up.

For this problem, voltage remains constant. The individual resistance of each bulb is also constant (duh).

We could used letters all day but to make things easier, lets make b1=b2=b3=b4= 10 ohms and V = 6 volts

Situation 1, the 3 bulbs in parallel give you a total resistance of 10/3 ohms. I did this by 1/10 + 1/10 + 1/10 = 1/total resistance

This gives the total circuit a combined resistance of 10 + 10/3 = 40/3 ohms
By using V=IR, we get 6=I*(40/3)
I=(3*6)/40= 0.45 amps

Situation 2 we get two 10 ohm resistors in parallel and the total resistance of the two is 5 ohms. The total resistance of the circuit is 15 ohms.
Using V=IR, we get 6=I*15
I=6/15= 0.4 amps

There are 2 key to understanding this problem.

First, the overall resistance of the first circuit is actually less then the overall resistance of the second circuit. This is due to the principle of resistances in parallel.

Second, the same current is flowing through EVERY component connected in series with the cell. Clearly, we get more current flowing through B1 in the first situation since the overall resistance of the circuit is smaller.


Diabloii.Net Member
Actually, you provd my point, Wuhan_Clan. You proved there is more current flowing across bulb 1 in situation 1 than two, hence it dims in situation 2. We can use your numbers if you'd like. By a current divider, all 3 bulbs in the first situation are going to see the same current, since they have equal resistance:
Bulb 1:

Situation 1:
P=.45^2*10 = 2.025W

Situation 2:
P=.4^2*10 = 1.6 W

Bulb 2, 3, 4
Situation 1: .45A/3 = .15A per bulb

P= .15^2*10 = .225W
Bulb 2, 3
Situation 2: .4A/2 = .2A.

P= .2^2*10 = .4W

I.e. Bulb 2 and 4 get Brighter. So the answer is D. I can check it in PSpice if you'd like, but its going to come out the same way. Again, I stand by my circuit analysis, and my electrical engineering major ^^

Again, think of a house circuit. When you turn on a vaccuum cleaner, the lights dim. This is because you hooked in somethign with a great current draw (less resistance) up to the parallel circuit (lights in a house are hooked in parallel) this pulls current and dims the lights. you take out this resistance, and the lights brighten up again.

As far as bulb 1 goes, as a general rule the more things hooked in parallel the more current pulled out of the battery. so less things in parallel = less current across B1 = less power dissapated.

edit: Here, PSpice proves it:

notice, more current comming across B1 in situation 1 than 2, hence it dims
more current commign across the 2 bulbs in parallel in situation 2, hence they brighten.


Diabloii.Net Member
WTF? Revived thread?

Now that I look at my post, I realized I put A instead of D. My working was done to prove D but in my first sentence, I put down A for some reason. Oh well, at least I know my math is right.


Diabloii.Net Member
well was close on my answer, just forgot to factor in the incresed total resistance dimming the one in series