arindameducationusc
  • arindameducationusc
Electric Circuit Question
Physics
chestercat
  • chestercat
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arindameducationusc
  • arindameducationusc
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arindameducationusc
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  • arindameducationusc

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arindameducationusc
  • arindameducationusc
arindameducationusc
  • arindameducationusc
arindameducationusc
  • arindameducationusc
arindameducationusc
  • arindameducationusc
Answer is little confusing... so I asked this question......
Empty
  • Empty
So is this in series or parallel? How do resistors add in this situation?
UsukiDoll
  • UsukiDoll
How would I know? I'm bad at science problems like these >:(
arindameducationusc
  • arindameducationusc
No problem..... @UsukiDoll Sorry about that.....
arindameducationusc
  • arindameducationusc
Parallel ofcourse @Empty
UsukiDoll
  • UsukiDoll
@arindameducationusc #forgiven
arindameducationusc
  • arindameducationusc
@Empty, Answer and why?
arindameducationusc
  • arindameducationusc
Astrophysics
  • Astrophysics
Mhm been a while since I've done E&M so since it's parallel then each component has the same voltage yes?
arindameducationusc
  • arindameducationusc
yes
arindameducationusc
  • arindameducationusc
so, D?
arindameducationusc
  • arindameducationusc
@Astrophysics Light Intensity depends on Voltage or Current?
Astrophysics
  • Astrophysics
Ideally since every emf has some internal resistance, we know the voltage then depends on the current, and in a parallel circuit to get the total current we would have to add it all up, so you have I1+I2+I3 = total I
Astrophysics
  • Astrophysics
So I'm not sure if this is the right thinking as I haven't done this in a long while xD, but what happens if the current is gone in one of the bulbs?
arindameducationusc
  • arindameducationusc
Okay I have the answer and reason... but I am not understanding it... You can think, recall and try explaining...
arindameducationusc
  • arindameducationusc
D,, If each of the identical bulbs has resistance R, then the current through each bulb is E/R. This is unchanged if the middle branch is taken out of the parallel circuit. (What will change is the total amount of current provided by the battery).. //This is the explanation given
Empty
  • Empty
Sorry I was answering another question. When the middle bulb goes out, the resistance goes to infinity for that particular bit. So remember in parallel resistances add like this: \[\frac{1}{R_T} = \frac{1}{R_1}+ \frac{1}{R_2}+ \frac{1}{R_3}\] So if the middle one \(R_2\) goes out, we have \(R_2 = \infty\) so when we plug it in, \(\frac{1}{\infty} = 0\) so we get the new total resistance: \[\frac{1}{R_{T'}} = \frac{1}{R_1}+ \frac{1}{R_3}\]
arindameducationusc
  • arindameducationusc
Light Intensity depends on Voltage or Current? @Empty
Astrophysics
  • Astrophysics
Is the answer suppose to be D?
arindameducationusc
  • arindameducationusc
Empty
  • Empty
The voltage will remain the same, so the current is what changes to make up for it, like your answer says. Light intensity depends on voltage apparently. I feel like I should know why but I don't.
Astrophysics
  • Astrophysics
Exactly!
arindameducationusc
  • arindameducationusc
hmmmm... Guys I am confused.... What should I do? Leave this question for now?
Astrophysics
  • Astrophysics
I was just writing, since the emf is equal to the potential difference across the circuit when there is no current, we know the potential difference (voltage) will remain the same, so if the middle bulb burns out, only the current will be affected, so if the light intensity is then only depended on the voltage it wouldn't really matter if the current increases or decreases as you will have the same voltage...I hope I said that right and it makes sense haha.
arindameducationusc
  • arindameducationusc
That one I am understanding.... Intensity depends on Volatage.... hmmmm. this one bothers me.... Its okay Thank you for trying..... @Astrophysics @Empty
arindameducationusc
  • arindameducationusc
If I get why, I will message you both....
radar
  • radar
Theoretically the answer is D, but that is assuming the internal resistance of the battery is 0, when in the real world, that ain't so. With the bulb filament opening the current is reduced, this results in the voltage across the remaining two bulbs to increase slightly. Why!, you may ask, because the voltage drop across or loss across the internal resistance is decreased due to this decrease in current.
radar
  • radar
So C is a consideration for someone astute to real world electrical problems.

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