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arindameducationusc
 one year ago
Electric Circuit Question
arindameducationusc
 one year ago
Electric Circuit Question

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arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0See attachment

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0@UsukiDoll

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0@Astrophysics

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0@sweetburger

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0Answer is little confusing... so I asked this question......

Empty
 one year ago
Best ResponseYou've already chosen the best response.1So is this in series or parallel? How do resistors add in this situation?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0How would I know? I'm bad at science problems like these >:(

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0No problem..... @UsukiDoll Sorry about that.....

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0Parallel ofcourse @Empty

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0@arindameducationusc #forgiven

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0@Empty, Answer and why?

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0@wio @zepdrix

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Mhm been a while since I've done E&M so since it's parallel then each component has the same voltage yes?

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0@Astrophysics Light Intensity depends on Voltage or Current?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Ideally 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
 one year ago
Best ResponseYou've already chosen the best response.0So 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
 one year ago
Best ResponseYou've already chosen the best response.0Okay I have the answer and reason... but I am not understanding it... You can think, recall and try explaining...

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0D,, 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
 one year ago
Best ResponseYou've already chosen the best response.1Sorry 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
 one year ago
Best ResponseYou've already chosen the best response.0Light Intensity depends on Voltage or Current? @Empty

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Is the answer suppose to be D?

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0Yes @Astrophysics

Empty
 one year ago
Best ResponseYou've already chosen the best response.1The 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.

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0hmmmm... Guys I am confused.... What should I do? Leave this question for now?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0That one I am understanding.... Intensity depends on Volatage.... hmmmm. this one bothers me.... Its okay Thank you for trying..... @Astrophysics @Empty

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0If I get why, I will message you both....

radar
 one year ago
Best ResponseYou've already chosen the best response.0Theoretically 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
 one year ago
Best ResponseYou've already chosen the best response.0So C is a consideration for someone astute to real world electrical problems.
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