Physics 117 problem about circuits and bulbs

In summary, when the switch is closed in the given circuit, bulb B becomes parallel with the switch, creating a short circuit that bypasses the bulb. This causes bulb B to go out. Meanwhile, the voltage across bulb A increases because it is now receiving all of the current from the battery, making it appear brighter. This is due to the current following the path of least resistance and bypassing bulb B.
  • #1
VictorWutang
13
0
1. In the circuit shown below, describe what happens to the brightness of bulbs A and B when the switch is closed. Explain your reasoning. (no other information is given, just asking about the relative brightness of each bulb)

In the circuit, with the switch open, you have a loop with a battery and bulbs A and B in series. When the switch closes, B becomes parallel with the electrical line that holds the switch. In this case, the electrical line that B is on is inside of the larger loop of the battery, A, and the switch.

Homework Equations

V = IR

The Attempt at a Solution



I thought at first that the bulbs would both slightly dim because the added switch and line would add a small amount of resistance to the line, but I see that A gets brighter and B goes out. Is this because the current takes the path of least resistance and skips over the B bulb?
 
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  • #2
VictorWutang said:
1. In the circuit shown below, describe what happens to the brightness of bulbs A and B when the switch is closed. Explain your reasoning. (no other information is given, just asking about the relative brightness of each bulb)

In the circuit, with the switch open, you have a loop with a battery and bulbs A and B in series. When the switch closes, B becomes parallel with the electrical line that holds the switch. In this case, the electrical line that B is on is inside of the larger loop of the battery, A, and the switch.



Homework Equations

V = IR



The Attempt at a Solution



I thought at first that the bulbs would both slightly dim because the added switch and line would add a small amount of resistance to the line, but I see that A gets brighter and B goes out. Is this because the current takes the path of least resistance and skips over the B bulb?


Welcome to the PF.

I don't see your attachment...
 
  • #3
I do not have a attachment for it, but i think i described it well enough.

Maybe this will help, tried to type it it best as possible. Ignore the asterisks, I just had to put them in there for spacing. The other lines are the wires.

|-----A-----------------|
| ******* | ********|
emf **** B ***** switch
| ******* | ******* |
|------------------------|
 
  • #4
VictorWutang said:
I do not have a attachment for it, but i think i described it well enough.

Maybe this will help, tried to type it it best as possible. Ignore the asterisks, I just had to put them in there for spacing. The other lines are the wires.

|-----A-----------------|
| ******* | ********|
emf **** B ***** switch
| ******* | ******* |
|------------------------|

Hello
Does the diagram look like the image in the link?
http://img684.imageshack.us/img684/404/37304acd6959474e821ad8c.png
regards
Yukoel
 
Last edited by a moderator:
  • #5
yes, it does.
 
  • #6
VictorWutang said:
I thought at first that the bulbs would both slightly dim because the added switch and line would add a small amount of resistance to the line, but I see that A gets brighter and B goes out. Is this because the current takes the path of least resistance and skips over the B bulb?

Yes, this is the reason why B goes out. The wire that has the switch in line with it acts as a "short circuit." It's called that because it gives the current a shorter path back to ground. Instead of having to go through both A and B, the current can now just pass through A and then directly to ground. So B has no current flowing through it: it has been "shorted out" of the circuit (i.e. bypassed).

Now all that is left for you to explain is why A gets brighter. Hint: what happens to the voltage across A when you close the switch?
 

Related to Physics 117 problem about circuits and bulbs

1. How do I calculate the total resistance in a series circuit?

In a series circuit, the total resistance is equal to the sum of individual resistances. This means you simply add up the resistance values of each component in the circuit.

2. What is the relationship between voltage, current, and resistance in a circuit?

According to Ohm's Law, the relationship between voltage (V), current (I), and resistance (R) in a circuit is given by V = I x R. This means that as resistance increases, the current decreases, and as voltage increases, the current increases.

3. How do I calculate the total current in a parallel circuit?

In a parallel circuit, the total current is equal to the sum of individual branch currents. This means you add up the current values of each branch in the circuit.

4. How do I determine the voltage across a specific component in a series circuit?

In a series circuit, the voltage across each component is the same. This means you can measure the voltage across any component in the circuit using a voltmeter.

5. What happens to the brightness of bulbs in a series circuit if one bulb burns out?

In a series circuit, if one bulb burns out, the other bulbs will also go out. This is because the flow of current is disrupted by the broken bulb, causing a break in the circuit. This is one of the disadvantages of using a series circuit for lighting.

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