Intus Games 01: Circuit Current

[Cygna]

Intus Games Episode 01

Circuit Current

Question open to all Ordo members only. This one's worth some serious L$. Since this problem is fairly easy only L$2000 is in order. Later problems will be worth L$3000. I'm probably going to end up paying it to Waffle, though. Make sure to show all your work and commentary by posting in this thread. Remember, you can discuss the problem in the thread, but it's a contest: there's only one winner.

Hints: Ohm's Law, V=IR

Question

Rank the brightness of the identical bulbs in the following circuit.

[Agares Tretiak]

A, B, D, E, C ?

[Jester Spearmann]

B and C will be dimmer since it's a series circuit, D and E should be the same brightness since it is a parallel circuit.... but probably dimmer than A

[Cyphre Iredell]

D and E is sort of a trick it appears, since they are both technicalyl in the same parallel circuit as A. Where B and C are a series circuit within the parallel. I can't remember which does what though; one decreases voltage while the other decreases amperage or something?

Edit: I think Agares has it right, at A, B, D, E, C, if in fact the negative lead on the left reaches C last due to subsequent parallel circuits terminating last on the positive side, and the series circuitry causing the most resistance to C altogether.

[Niiya Narayan]

Cyph had it the first time... A,D and E will be the three brightest as they're in parallel decreasing that side of the circut's resistance whereas B and C are in series thus doubling the resistance.

so: A+D+E, B+C?

[Agares Tretiak]

Please note, I'm only a layman when it comes to this sort of thing. But I am trying to remember my basic science, and expanding the concepts to fit this problem.:(

My reasoning is that A would have both the shortest path (Which doesn't make a huge difference since it seems to be a very local circuit) and the least resistance due to being the only bulb on it. That was apparent.

I felt that B would also be fairly bright, because it would receive the same amount of initial energy, but the following bulb would suffer because of that. I don't know if this would actually lower BOTH bulbs brightness, so I decided B would probably be brighter. D and E are on a parallel circuit, would draw the same amount of energy without progressively losing it, but still not be as bright as A or B due to distance (which I felt though maybe very small indeed, would still matter *shrug*) To me, this all indicates that C is the loser in the energy department.

Also, is the circuit running right to left, or left to right?

By the way, is this a direct or alternating current or does that even matter? I'm assuming tyhis is on a board a few feet across, rather than several miles in distance.

[Cygna]

Niiya's the winner. Explanation here in a bit.

[Cygna]

A, B, D, E, C ?

A ≤ B

(A is not greater than B)

B and C will be dimmer since it's a series circuit, D and E should be the same brightness since it is a parallel circuit.... but probably dimmer than A

D = E ≥ A

(D and E are not less than A)

D and E is sort of a trick it appears, since they are both technicalyl in the same parallel circuit as A. Where B and C are a series circuit within the parallel. I can't remember which does what though; one decreases voltage while the other decreases amperage or something?

Edit: I think Agares has it right, at A, B, D, E, C, if in fact the negative lead on the left reaches C last due to subsequent parallel circuits terminating last on the positive side, and the series circuitry causing the most resistance to C altogether.

Lalala, good observation Cyphre. R_D and R_E are simply in parallel with R_A and R_BC. What a mean tricky drawing :o

Cyph had it the first time... A,D and E will be the three brightest as they're in parallel decreasing that side of the circut's resistance whereas B and C are in series thus doubling the resistance.

so: A+D+E, B+C?

Right answer. Although wrong notation, I'll accept it.

Correct notation: A = D = E > B = C

Niiya's notation implies nothing.

Please note, I'm only a layman when it comes to this sort of thing. But I am trying to remember my basic science, and expanding the concepts to fit this problem.:(

My reasoning is that A would have both the shortest path (Which doesn't make a huge difference since it seems to be a very local circuit) and the least resistance due to being the only bulb on it. That was apparent.

I felt that B would also be fairly bright, because it would receive the same amount of initial energy, but the following bulb would suffer because of that. I don't know if this would actually lower BOTH bulbs brightness, so I decided B would probably be brighter. D and E are on a parallel circuit, would draw the same amount of energy without progressively losing it, but still not be as bright as A or B due to distance (which I felt though maybe very small indeed, would still matter *shrug*) To me, this all indicates that C is the loser in the energy department.

Also, is the circuit running right to left, or left to right?

By the way, is this a direct or alternating current or does that even matter? I'm assuming tyhis is on a board a few feet across, rather than several miles in distance.

Path length does not effect current. Recall that, in an electrical circuit, electrons are always populating the circuit, even if no emf(electromotive force) is connected. All the emf does is "push" the electrons around the circuit. An analogy to this situation: Imagine having a square track that is completely populated with spherical balls that can not pass eachother (only travel in one at a time). If you were to push one ball, where the battery/emf/voltage is located, all the balls will move the same speed in the entire track (first ball pushes the next which pushes the next etc. at exactly the same time).

Next stanza, see explanation below.

Circuit doesn't run anywhere. Current does. Current in this circuit moves from the positive end of the emf (ε) source (the longer line of the two), through the circuit, coming back in the same amperage as it left, at the negative end of the emf.

Current is the derivative of charge in respect to time. Or more simply, charge per unit (single) time. I is in units of Ampere. Ampere is equal to the same unit called Coulomb Time^-1, or Coulomb per time. Coulomb is the standard unit of charge. An electron (e^-) has a charge of ~1.6x10^-19 Coulomb, for example.

I = (dQ/dt)

Current goes path of least resistance. Here's how current works in a circuit:

Series:

I₁ = I₂ = I₃ = ∙∙∙

Parallel:

I_circuit = ∑ I_i (from 1 to ∞) (or I₁ + I₂ + I₃ + ∙∙∙)

B is in a series configuration with C, and resistance in series is the sum of the two.

R_B + R_C = R_BC

R_BC = R_A = R_D = R_E

The board can be of any size. As I said before, consider that the circuit is already populated with charged particles, and by adding an emf (voltage V potential gradient) just moves the charged particles in circuit. Alternating and direct current doesn't matter in this particular case. Either way properties will be the same.

So anyway, from Ohm's Law,

I_A = I_D = I_E = (V_ADE/R_ADE)

I_B = I_C = (V_BC/2R_{B or C})

The more current, the brighter the bulb. So Niiya was correct in saying it'll be double the resistance in series.

This was good fun, everyone learned some stuff, and moar L to come.

[Agares Tretiak]

Right. NExt, do a history analysis.

[Cyphre Iredell]

Oh god, theres a bunch of weird characters and numbers in Intus' post and my brain ceased functioning o_o

[Lestat Umarov]

*Sobs cause he doesn't understand any of this.*

[Sekonda Huet]

*Hemorages quietly in the corner.*