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arindameducationusc

  • one year ago

Electric Circuit Question

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  1. arindameducationusc
    • one year ago
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    See attachment

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  2. arindameducationusc
    • one year ago
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    @Michele_Laino

  3. arindameducationusc
    • one year ago
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    @UsukiDoll

  4. arindameducationusc
    • one year ago
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    @Empty

  5. arindameducationusc
    • one year ago
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    @Astrophysics

  6. arindameducationusc
    • one year ago
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    @sweetburger

  7. arindameducationusc
    • one year ago
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    Answer is little confusing... so I asked this question......

  8. Empty
    • one year ago
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    So is this in series or parallel? How do resistors add in this situation?

  9. UsukiDoll
    • one year ago
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    How would I know? I'm bad at science problems like these >:(

  10. arindameducationusc
    • one year ago
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    No problem..... @UsukiDoll Sorry about that.....

  11. arindameducationusc
    • one year ago
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    Parallel ofcourse @Empty

  12. UsukiDoll
    • one year ago
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    @arindameducationusc #forgiven

  13. arindameducationusc
    • one year ago
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    @Empty, Answer and why?

  14. arindameducationusc
    • one year ago
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    @wio @zepdrix

  15. Astrophysics
    • one year ago
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    Mhm been a while since I've done E&M so since it's parallel then each component has the same voltage yes?

  16. arindameducationusc
    • one year ago
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    yes

  17. arindameducationusc
    • one year ago
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    so, D?

  18. arindameducationusc
    • one year ago
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    @Astrophysics Light Intensity depends on Voltage or Current?

  19. Astrophysics
    • one year ago
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    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

  20. Astrophysics
    • one year ago
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    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?

  21. arindameducationusc
    • one year ago
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    Okay I have the answer and reason... but I am not understanding it... You can think, recall and try explaining...

  22. arindameducationusc
    • one year ago
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    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

  23. Empty
    • one year ago
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    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: \[\frac{1}{R_{T'}} = \frac{1}{R_1}+ \frac{1}{R_3}\]

  24. arindameducationusc
    • one year ago
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    Light Intensity depends on Voltage or Current? @Empty

  25. Astrophysics
    • one year ago
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    Is the answer suppose to be D?

  26. arindameducationusc
    • one year ago
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    Yes @Astrophysics

  27. Empty
    • one year ago
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    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.

  28. Astrophysics
    • one year ago
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    Exactly!

  29. arindameducationusc
    • one year ago
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    hmmmm... Guys I am confused.... What should I do? Leave this question for now?

  30. Astrophysics
    • one year ago
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    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.

  31. arindameducationusc
    • one year ago
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    That one I am understanding.... Intensity depends on Volatage.... hmmmm. this one bothers me.... Its okay Thank you for trying..... @Astrophysics @Empty

  32. arindameducationusc
    • one year ago
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    If I get why, I will message you both....

  33. radar
    • one year ago
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    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.

  34. radar
    • one year ago
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    So C is a consideration for someone astute to real world electrical problems.

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