Question

Electric Circuit Question

Answer 1

See attachment

Answer 2

@Michele_Laino

Answer 3

@UsukiDoll

Answer 4

@Empty

Answer 5

@Astrophysics

Answer 6

@sweetburger

Answer 7

Answer is little confusing... so I asked this question......

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

Parallel ofcourse @Empty

Answer 12

@arindameducationusc #forgiven

Answer 13

@Empty, Answer and why?

Answer 14

@wio @zepdrix

Answer 15

Mhm been a while since I've done E&M so since it's parallel then each component has the same voltage yes?

Answer 16

yes

Answer 17

so, D?

Answer 18

@Astrophysics
Light Intensity depends on Voltage or Current?

Answer 19

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

Answer 20

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?

Answer 21

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

Answer 22

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

Answer 23

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:


\[\frac{1}{R_{T'}} = \frac{1}{R_1}+ \frac{1}{R_3}\]

Answer 24

Light Intensity depends on Voltage or Current? @Empty

Answer 25

Is the answer suppose to be D?

Answer 26

Yes @Astrophysics

Answer 27

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.

Answer 28

Exactly!

Answer 29

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

Answer 30

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.

Answer 31

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

Answer 32

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

Answer 33

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.

Answer 34

So C is a consideration for someone astute to real world electrical problems.