Question on lightbulb brightness in circuit analysis

In summary, when someone closes the switch in an ideal situation, the current will split itself and some will go into the new pathway and some will go through the switch. The switch will short out the current and all of the current will flow through the new pathway.
  • #1
RoboNerd
410
11

Homework Statement



question 66.jpg

Homework Equations


I know that power of a lightbulb/resister is I^2 * R

The Attempt at a Solution



OK.

Here's my attempt:

Let's consider the situation where we do not close the lightbulb and assume that the wires and switch are ideal because they have no resistance. I have a Total Itot current flowing through the circuit loop, and everyone is happy.

Now comes along some random guy and he closes the switch. The total current in this case will be flowing from the positive side of the battery to the first lightbulb, go through it, and then reach the junction.

My impression was that the current will then split itself, with some going into the branch of bulb2 and some going through the switch before the two sub-currents meet up and then head back to the battery's terminal.

The equivalent resistance of the initial configuration without the switch will be 2*R (R in this case is the resistance of each of the batteries), but connecting the switch with the ideal resistance of zero will have the Req also equal 2*R.

Thus, the total current in the circuit would remain the same as ElectromotiveForce = I * Requivalent, and Requivalent is going to remain the same.

However, due to the loss of some current going through the switch, the current going through bulb2 would decrease, and thus by Power = I^2 * R, it's power (brightness) would decrease while the current through bulb1 which equals Itotal will remain the same.

The explanation that I read says that the switch will short out the current, with all of the current flowing through the new pathway created by the closed switch, and none will flow through bulb2, making it go out.
With that, the circuits total resistance decreases, so more current flows through bulb1 making it brighter, so we would have B as the answer.

Why is it the case that all of the current goes through the pathway caused by the switch with even a tiny portion of the current not going through bulb2 at all?

Thanks in advance.
 
Physics news on Phys.org
  • #2
RoboNerd said:
Why is it the case that all of the current goes through the pathway caused by the switch with even a tiny portion of the current not going through bulb2 at all?
The potential difference across the closed switch must be zero since it has no resistance (ideal wires and switch). It's in parallel with bulb2, so what's the potential difference across bulb2? What's the rule for the potential across parallel components?
 
  • Like
Likes RoboNerd
  • #3
Parallel components must have the same potential difference so bulb2 must have a potential difference of zero also.

Wow. That explains a lot. Thank you very much for the help!
 
  • Like
Likes CWatters and gneill

FAQ: Question on lightbulb brightness in circuit analysis

How does the brightness of a lightbulb change in a circuit analysis?

The brightness of a lightbulb in a circuit analysis is affected by the amount of current flowing through the circuit. The higher the current, the brighter the lightbulb will be.

What factors influence the brightness of a lightbulb in a circuit analysis?

The main factors that influence the brightness of a lightbulb in a circuit analysis are the voltage of the power source, the resistance of the circuit, and the type of lightbulb being used. These factors determine the amount of current flowing through the circuit and therefore affect the brightness of the lightbulb.

How does a lightbulb's resistance affect its brightness in a circuit analysis?

The resistance of a lightbulb is directly proportional to its brightness in a circuit analysis. This means that as the resistance increases, the brightness decreases. This is because a higher resistance limits the amount of current that can flow through the circuit and reach the lightbulb, resulting in a dimmer light.

Why does a lightbulb get brighter in a series circuit compared to a parallel circuit?

In a series circuit, all of the current flowing through the circuit must pass through each component, including the lightbulb. This means that the lightbulb will receive a higher current, resulting in a brighter light. In a parallel circuit, the current is divided between different branches, so the lightbulb will receive less current and therefore be dimmer.

How does the type of lightbulb affect its brightness in a circuit analysis?

The type of lightbulb being used can have a significant impact on its brightness in a circuit analysis. Different types of lightbulbs have different resistances and therefore require varying amounts of current to produce the same level of brightness. For example, an LED lightbulb will be much brighter than an incandescent bulb using the same amount of current. Additionally, different types of lightbulbs may have different voltage requirements, which can also affect their brightness in a circuit.

Back
Top