Math

[sensei_tanker 3,098]

I'm using TO Forums as a last resort, but someone please help me with this basic physics concept:

Here I drew a circuit with a switch, and there is both a light bulb and a capacitor in parallel to the battery. The problem asks how would the brightness of the lightbulb change overtime when the switch is closed.

 

I'm extremely conflicted over two answers.

One way I think about this is since the capacitor "steals" all of the current away from the light bulb when it begins charging, the bulb should not emit any light at the beginning. And then as the capacitor becomes fully charged, all of the current will be "returned" to the light bulb and the light bulb would gradually get brighter until it reaches its maximum brightness. 

Another way I think about this is since brightness is associated to power of the lightbulb and power is just V^2/R, the lightbulb should remain at the same brightness the entire time. This is because the battery is always in parallel to the battery, and therefore would have the same voltage drop as the battery. Therefore, the voltage drop across the battery is the same regardless of whether the capacitor is charging up, so the brightness of the light bulb is the same. 

 

Could someone please help explain to me which one is the correct answer and why the other answer is incorrect?

 

[LOLKILLERTOTHEDEATH 1,921]

8 minutes ago, sensei_tanker said:

Spoiler

I'm using TO Forums as a last resort, but someone please help me with this basic physics concept:

Here I drew a circuit with a switch, and there is both a light bulb and a capacitor in parallel to the battery. The problem asks how would the brightness of the lightbulb change overtime when the switch is closed.

 

I'm extremely conflicted over two answers.

One way I think about this is since the capacitor "steals" all of the current away from the light bulb when it begins charging, the bulb should not emit any light at the beginning. And then as the capacitor becomes fully charged, all of the current will be "returned" to the light bulb and the light bulb would gradually get brighter until it reaches its maximum brightness. 

Another way I think about this is since brightness is associated to power of the lightbulb and power is just V^2/R, the lightbulb should remain at the same brightness the entire time. This is because the battery is always in parallel to the battery, and therefore would have the same voltage drop as the battery. Therefore, the voltage drop across the battery is the same regardless of whether the capacitor is charging up, so the brightness of the light bulb is the same. 

 

Could someone please help explain to me which one is the correct answer and why the other answer is incorrect?

 

You may have posted this in the wrong thread. In either case, consider asking @Person_Random as he is interested in Physics and is quite well versed in it too.

[Person_Random 4,991]

On 5/10/2020 at 8:49 AM, sensei_tanker said:

I'm using TO Forums as a last resort, but someone please help me with this basic physics concept:

Here I drew a circuit with a switch, and there is both a light bulb and a capacitor in parallel to the battery. The problem asks how would the brightness of the lightbulb change overtime when the switch is closed.

 

I'm extremely conflicted over two answers.

One way I think about this is since the capacitor "steals" all of the current away from the light bulb when it begins charging, the bulb should not emit any light at the beginning. And then as the capacitor becomes fully charged, all of the current will be "returned" to the light bulb and the light bulb would gradually get brighter until it reaches its maximum brightness. 

Another way I think about this is since brightness is associated to power of the lightbulb and power is just V^2/R, the lightbulb should remain at the same brightness the entire time. This is because the battery is always in parallel to the battery, and therefore would have the same voltage drop as the battery. Therefore, the voltage drop across the battery is the same regardless of whether the capacitor is charging up, so the brightness of the light bulb is the same. 

 

Could someone please help explain to me which one is the correct answer and why the other answer is incorrect?

 

Okay, I've just been pinged. Looking through this to endure a painful hour of online church

So... let's start at when the switch is just closed. The capacitor will begin charging, and at that moment, it acts as a wire with negligible resistance. The current then follows the path of least resistance, so the capacitor will steal all the current, leaving the lightbulb with zero current. Here, we can think of power as P = I²R for now (I'm currently still thinking about the voltage one right now, my mom explained it to me during AP Review but I can't remember right now). As R remains constant, I is zero, so the power is zero, therefore the lightbulb is completely dimmed.

As the capacitor charges, the capacitor gains more charge (and more effective resistance). This diverts some current to the lightbulb. Now, power is (some nonzero number)²R, which is greater than 0²R, which is zero, so the brightness goes up.

When the capacitor is fully charged, no more charge go through the parallel plates, which results in the capacitor having an infinite effective resistance. Since the current chooses to go the path of least resistance, all the current goes to the lightbulb, which gives the lightbulb its maximum current. Because P=I²R, it attains its maximum brightness at this point.

Also, about the V²/R concept, I believe now I have an answer. So in this case, the voltage drop is also equivalent to IR (current multiplied by resistance). When the current is zero, the effective voltage drop is zero; therefore, the power and effective brightness is also zero.

 

[spiderman1000 737]

15 hours ago, Person_Random said:

 

Man you are good at physics like my brother. (im good at too but he is better (since he is older))

[ThirdOnion 4,898]

21 hours ago, Hazeral said:

 

When you try to write a program in the fewest lines possible

[spiderman1000 737]

On 5/12/2020 at 8:14 AM, Hazeral said:

let tokens = m.split('\n\n').slice((i > 1) ? (5 * i) - 1 : (5 * i), ((i + 1 == Math.ceil(m.split('\n\n').length / 4)) ? (m.split('\n\n').length - c) + 10 : ((i > 1) ? ((4 * (i + 1) + (i * 2))) : (4 * (i + 1) + i))));

why doesnt this work

Can you pls tell me which comp language is this? if its python, then maybe i can help. (unless the improved version works. then this comment is awkward)

[Troll 8]

Ohh boy 

[spiderman1000 737]

On 5/15/2020 at 6:30 PM, Hazeral said:

No worries, I posted that as a joke 

It's JavaScript, and it actually did work, just very slowly. I've replaced the code with a more efficient alternative already.

Ik it is a joke. (well kinda an inside joke for people who know programming.) and i guess it wouldnt have worked on python if it all would have been in one line. BTW can you pls post the result of this program? im kinda curios to see what this program means and does.