Which lamp is dimmed in the circuit?

[Pouyan]

Homework Statement

In the picture below we see a circuit with four lamps. The question is which of these lamps shines weaker

Relevant Equations

Kirchhoff's circuit laws

The true answer is B. But I don't understand why!
I know:
Kirchhoff's circuit laws : ∑I_n=0

If we assume that a current that goes from plus to minus, before it passes through lamp B, I know that according to Kirchhof's laws, part of the current will pass through the bottom path where there is no lamp there, and another part through the lamp B. Then all of the current wants to go and then split to the next intersection ie part of the current wants to pass through the lamps A and D and another part through the lamp C.

The one I do not understand in this figure is B parallel with the three lamps (A, D, C) or is it in series? Why do we say that B shines weaker?

 

[mitochan]

Say $I_A$ is current through A
I_A=I_D
I_B+I_O=I_A+I_C
where $I_O$ is current through wire parallel to B with no lamp.

I_O >> I_B
because of scarce resistance of wire.

 

[Pouyan]

Say $I_A$ is current through A
I_A=I_D
I_B+I_O=I_A+I_C
where $I_O$ is current through wire parallel to B with no lamp.

I_O >> I_B
because of scarce resistance of wire.

Thank you! But is B in serie with A,D,C or is in parallell ?

 

[haruspex]

Thank you! But is B in serie with A,D,C or is in parallell ?

B, as such, is neither. The system consisting of B and the wire that bypasses it is in series with the system consisting of A, D and C.
If the bypass wire had zero resistance, why would any current go through B?

 

[phinds]

I know that according to Kirchhof's laws, part of the current will pass through the bottom path where there is no lamp there, and another part through the lamp B.

It is unfortunate that you "know" that, since it is wrong. Assuming an idea circuit with no resistance in the wires, why would any of the current pass through B ?

 

[Pouyan]

It is unfortunate that you "know" that, since it is wrong. Assuming an idea circuit with no resistance in the wires, why would any of the current pass through B ?

you mean that the current through B is negligible? or in other words it is so small that we can ignore it?

 

[phinds]

you mean that the current through B is negligible? or in other words it is so small that we can ignore it?

No, I mean it is ZERO. Do you understand what a short circuit is, what it does?

 

[Delta2]

No, I mean it is ZERO. Do you understand what a short circuit is, what it does?

It can't be absolute Zero, you need a short circuit wire with perfect zero resistance to achieve that.

 

[phinds]

It can't be absolute Zero, you need a short circuit wire with perfect zero resistance to achieve that.

Which is why I specified "in an idea circuit / no resistance in the wires" which is absolutely the standard in beginning circuits classes.

 

[kuruman]

Homework Statement:: In the picture below we see a circuit with four lamps. The question is which of these lamps shines weaker
Relevant Equations:: Kirchhoff's circuit laws
The true answer is B. But I don't understand why!

The short answer is that B is shorted.

 

[phinds]

The short answer is that B is shorted.

Good pun but he doesn't seem to understand what a short IS.

 

[mitochan]

But is B in serie with A,D,C or is in parallell ?

Kirchhoff's law does not care they are series, parallel or neither. That is the reason why it is used.

 

[Delta2]

you mean that the current through B is negligible? or in other words it is so small that we can ignore it?

It depends on the resistance of the wire that is parallel with lamp B. The smaller that resistance, the smaller the current through B. If you ask why, it is what we get if we do the math using Kirchhoff's laws.

 

[phinds]

It depends on the resistance of the wire that is parallel with lamp B. The smaller that resistance, the smaller the current through B. If you ask why, it is what we get if we do the math using Kirchhoff's laws.

Why do you keep bringing up resistance in the wire when he is clearly a rank beginner and should learn ideal circuits at this point, particularly since he clearly does not understand them yet.

 

[kuruman]

Good pun but he doesn't seem to understand what a short IS.

Then let's try the incremental approach.
Add a fifth bulb E, with resistance identical to that of bulb B in the straight wire, ($R_E=R_B$.) Say current $I_{tot.}$ is the total current provided by the battery. The current in the two bulbs will be equally split because the total current has no reason to prefer one resistance over the other; the two bulbs will be equally bright because they have the same current and resistance.

Now make the two bulbs different, ($R_E \ne R_B$.) They have the same voltage across them because they are in parallel. Since power is $P=IV$, the bulb with the higher current will dissipate more power at the expense of the other bulb. So if $R_E$ is much, much smaller than $R_B$, which is the case with shorts or straight wires, it will draw almost all of the current leaving almost nothing for B.

Mathematically, $$I_E~R_E=I_B~R_B~ \Rightarrow~I_E=\frac{R_B}{R_E}I_B=\frac{R_B}{R_E}(I_{tot.}-I_E).$$If you solve for $I_E$, you get $$I_E=\frac{R_B}{R_B+R_E}I_{tot.}$$ The smaller you make $R_E$, the larger fraction of $I_{tot.}$ is shunted to $R_E$ leaving a smaller fraction to $R_B.$ In the limit $R_E \rightarrow 0$, $I_E \rightarrow I_{tot.}$.

Current will take the path of least resistance.

 

[Delta2]

Why do you keep bringing up resistance in the wire when he is clearly a rank beginner and should learn ideal circuits at this point, particularly since he clearly does not understand them yet.

Though he might be a beginner, he is mostly right at what he says at post #6 and I think he deserves to know why he is right.

 

[phinds]

Though he might be a beginner, he is mostly right at what he says at post #6 and I think he deserves to know why he is right.

A reasonable point but since he doesn't understand basic circuitry I still think he should focus on ideal circuits which is where we all start.

 

[Delta2]

This problem isn't meant to be an ideal circuit, the wire must have some resistance otherwise it wouldn't ask which bulb light is weaker but which bulb doesn't work at all.

 

[phinds]

This problem isn't meant to be an ideal circuit, the wire must have some resistance otherwise it wouldn't ask which bulb light is weaker but which bulb doesn't work at all.

I disagree but at this point we're just arguing about how many angels can dance on the head of a pin.