## arindameducationusc one year ago Electric Circuit Question

1. arindameducationusc

See attachment

2. arindameducationusc

@Michele_Laino

3. arindameducationusc

@UsukiDoll

4. arindameducationusc

@Empty

5. arindameducationusc

@Astrophysics

6. arindameducationusc

@sweetburger

7. arindameducationusc

8. Empty

So is this in series or parallel? How do resistors add in this situation?

9. UsukiDoll

How would I know? I'm bad at science problems like these >:(

10. arindameducationusc

No problem..... @UsukiDoll Sorry about that.....

11. arindameducationusc

Parallel ofcourse @Empty

12. UsukiDoll

@arindameducationusc #forgiven

13. arindameducationusc

14. arindameducationusc

@wio @zepdrix

15. Astrophysics

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

yes

17. arindameducationusc

so, D?

18. arindameducationusc

@Astrophysics Light Intensity depends on Voltage or Current?

19. 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

20. 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?

21. arindameducationusc

Okay I have the answer and reason... but I am not understanding it... You can think, recall and try explaining...

22. 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

23. 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: $\frac{1}{R_{T'}} = \frac{1}{R_1}+ \frac{1}{R_3}$

24. arindameducationusc

Light Intensity depends on Voltage or Current? @Empty

25. Astrophysics

Is the answer suppose to be D?

26. arindameducationusc

Yes @Astrophysics

27. 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.

28. Astrophysics

Exactly!

29. arindameducationusc

hmmmm... Guys I am confused.... What should I do? Leave this question for now?

30. 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.

31. arindameducationusc

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

32. arindameducationusc

If I get why, I will message you both....

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.