Last minute circuits question

[Trisphorin]

what happens if you take out one bulb from a parallel circuit? does the other bulb gets brighter or dimmer?
i thought that if you take out one bulb, the equivalence resistance would increase and thus current will decrease. since P=IV, wouldn't the other bulb gets dimmer as result?

also what if you take out a bulb from a series circuit? i read on other websites that doing so would stop the circuit and causes the other bulb to go off. but shouldn't it get brighter as result?

Thanks!

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[justhanging]

what happens if you take out one bulb from a parallel circuit? does the other bulb gets brighter or dimmer?
i thought that if you take out one bulb, the equivalence resistance would increase and thus current will decrease. since P=IV, wouldn't the other bulb gets dimmer as result?

also what if you take out a bulb from a series circuit? i read on other websites that doing so would stop the circuit and causes the other bulb to go off. but shouldn't it get brighter as result?

Thanks!

I was just plugging in some numbers for the first senerio and it seems that the bulb would not change in brightness. It is true that removing a parallel bulb would increase the equivalent resistance of the circuit and thus decrease current but removing it also concentrates all the current to the original bulb rather then splitting it in the original setup where half the current would go to one bulb and the other half to the other.

For the second scenario I think those websites are talking about removing the bulb and not reconnecting the circuit. If the circuit is not complete or is disconnected somewhere then current is going to stop flowing and the bulb would turn off. Similar to adding a capacitor to the circuit. If you do reconnect it then it should increase the voltage on the original bulb and thus increase brightness.

 

[Trisphorin]

Thanks! that was really helpful 🙂

 

[Kimchii]

what happens if you take out one bulb from a parallel circuit? does the other bulb gets brighter or dimmer?
i thought that if you take out one bulb, the equivalence resistance would increase and thus current will decrease. since P=IV, wouldn't the other bulb gets dimmer as result?

also what if you take out a bulb from a series circuit? i read on other websites that doing so would stop the circuit and causes the other bulb to go off. but shouldn't it get brighter as result?

Thanks!

In a parallel circuit the current changes but the voltage doesnt. If you take a bulb out from a parallel circuit, there will be no resistance on that wire making it more likely to have current go through it. This would dim the other two because less current is going towards them. If you use the formula to calculate power in the other bulbs, they would have lower power and less brightness. Yes the overall resistance would increase if you took one bulb out.

In series, if you take a bulb out there would be no more light. The bulb is like a connection, if you take it out how is the current going to flow? Imagine having 1 battery and 1 bulb, if you take the bulb out how would the wires be connected if the bulb connected the wires?

 

[BBB PhD]

OK, simple way to handle this kiddo.
Series Circuits: If you remove anything that breaks the circuit (resistor, inductor, cap, bulb, tv,...) everything else shuts off since no current can flow through the "open circuit".
Parallell Circuit: The bulb that remains will have all the nodal current (the other load is out)--->The "EQUIVALENT PARALLELL resistance decreases with the number of parallell, but the individual branches dictate the current through them.
Am i not clear?

 

[geeyouknit]

Am i not clear?

Reminds me of "Am I not merciful?" from the gladiator. : )

Hijacked question: how does one determine brightness based on math formulas or thinking about resistance/power/current? Thanks.

So doesn't this question depend on whether or not you reconnect the circuit?
If you don't reconnect it
Series - no current flows, everything's off.
Parallel - that one branch disappears; overall resistance increases, overall current decreases, but the same amount goes through as before, thus keeping it same power/brightness (look at Berk's math).

But if you do reconnect it:
Series- total resistance decreases, overall current increases
Parallel - overall resistance again increases, but since this is the path of least resistance, it will have proportionally more current through it.

 

[BerkReviewTeach]

In a parallel circuit the current changes but the voltage doesnt. If you take a bulb out from a parallel circuit, there will be no resistance on that wire making it more likely to have current go through it. This would dim the other two because less current is going towards them. If you use the formula to calculate power in the other bulbs, they would have lower power and less brightness. Yes the overall resistance would increase if you took one bulb out.

In series, if you take a bulb out there would be no more light. The bulb is like a connection, if you take it out how is the current going to flow? Imagine having 1 battery and 1 bulb, if you take the bulb out how would the wires be connected if the bulb connected the wires?

You are correct about series, but not so correct about parallel.

This is a common misconception that is easily cleared up by a real life example. Household circuits are wired in parallel so that all of the outlets and switches are independent of one another. Turning off your computer, unplugging a lamp, or switching on a switch does not change any of the other items in your house drawing current.

Parallel pathways are independent of one another, so removing a parallel light bulb won't change the intensity of any of the remaining lights in parallel.

You can verify this using math in one of two ways. First, parallel circuit elements have the same voltage drop, so they have the same value for I x R. Consider disconnecting a resistor in parallel. That doesn't change the resistance of the remaining resistors or the voltage drop across each remaining resistor, so the current through each remaining resistor cannot change.

You can also solve this by considering three parallel resistors with resistances of 2 ohms, 3 ohms, and 6 ohms connected to a 12 V battery. Equivalent resistance is 1 ohm, so Itotal is 12 amps. The 12 amps of total current splits into 6 amps through the 2-ohm resistor, 4 amps through the 3-ohm resistor, and 2 amps through the 6-ohm resistor. All three experience a voltage drop of 12 V and the sum of the currents (6 + 4 + 2) is 12.

We can disconnect any of the three resistors, so let's consider the 2-ohm resistor. This leaves a 3-ohm resistor and 6-ohm resistor connected in parallel to a 12 V battery. The equivalent resistance is now 2 ohms, so the total current is 6 amps. The 6 amps of total current splits into 4 amps through the 3-ohm resistor and 2 amps through the 6-ohm resistor. Both resistors experience a voltage drop of 12 V and the sum of the currents (4 + 2) is 6 amps. The current through the two remaining resistors is the same whether the 2-ohm resistor is connected or not.

The math is lengthy, but helps to make the conceptual result more believable.

 

[geeyouknit]

You are correct about series, but not so correct about parallel.

The math is lengthy, but helps to make the conceptual result more believable.

Thanks for doing this math, Berk. It was really helpful in clearing that up.

 

[geeyouknit]

Just a side question

What would happen if you removed the resistor in a parellel, but then reconnected the wires? Essentially, you'd have a wire with only resistance of a wire. Would almost all the current go through that wire? This wouldn't decrease the current through other bulbs, right?decrease the brightness of other bulbs? It would just increase the overall current of the circuit?

Thanks

 

[tendiw]

Just a side question

What would happen if you removed the resistor in a parellel, but then reconnected the wires? Essentially, you'd have a wire with only resistance of a wire. Would almost all the current go through that wire? This wouldn't decrease the current through other bulbs, right?decrease the brightness of other bulbs? It would just increase the overall current of the circuit?

Thanks

If it is an ideal wire ALL the current goes through it meaning the other bulbs will go off. If there is a tiny bit of resistance specified by the MCAT, then most current will flow through the wire and only a little through the bulb...making it very dim.
Usually we assume the wire is ideal otherwise specified (or when dealing with actual real life experiments and not theoretical stuff)

 

[BerkReviewTeach]

If it is an ideal wire ALL the current goes through it meaning the other bulbs will go off. If there is a tiny bit of resistance specified by the MCAT, then most current will flow through the wire and only a little through the bulb...making it very dim.

Seriously? After all of the explanations above why is it that you want to convince yourself of your own theory that an independent path will somehow affect another independent path?

It is true that an ideal wire in parallel with another circuit element will absorb all of the current (this is what is known as a short circuit). It will take 100% of the current if it has no resistance at all, and in doing so kill all current in parallel paths. This would be a short-lived circuit as current would flow instantaneously from the cathode to the anode. This is why short circuits usually overload your circuit breaker and cause it to trip.

But there is no such thing as an ideal wire, so let's deal with a real scenario. Adding a wire in paraellel to a light bulb would reduce the overall resistance of the circuit and drastically increase the total current. But the new total current would be the current through the new wire plus the current through the original light bulb. The current through the original light bulb will not change with the addition of a parallel wire and thus its power drain will not change. It will retain the same brightness. This can be proven with a simple experiment.

Seriously, you should sit down with a wire, a light bulb, and 9V battery and experiment. Be careful, because the wire gets very hot. But if you do this experiment it will make it easier to understand (and believe) that parallel additions do not change a resistor. Most of the common misconceptions can be eliminated by a few hands-on setups.