Electricity: Basic circuit question with a switch

  • #1
Femme_physics
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This is a question that was on my test

Homework Statement





http://img269.imageshack.us/img269/5125/circuitthingy.jpg [Broken]


In the circuit in the drawing switch S is open. Both resistors are idential, both having resistance R each. Given the source is E, and R = r (the resistance of the voltage source). The ammeter resistance is negligible.

Closing the switch

A) Will the ammeter's reading go up, down, or won't change? Base your answer on a physical consideration or prove it by calculation.

((There were several questions but I'll just start with A for now))

The Attempt at a Solution




The ammeter's reading decreases since there is more resistance.

Proof

http://img543.imageshack.us/img543/1949/ianswer.jpg [Broken]
 
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Answers and Replies

  • #2
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Hi again Fp! :smile:

Do you have a physical consideration for your answer?

As for your calculation, I'm afraid you made a simple conceptual mistake.
But before I tell you what it is, I'd like you to think about the physical consideration, and I kinda hope it will *click* then! :smile:
 
  • #3
Femme_physics
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Actually, I just copied it from my test (it's the question I skipped and ended up deleting)
, but I see what you mean now
 
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  • #4
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You do?
Well, don't keep me in suspense... what is it? :confused:
 
  • #5
Femme_physics
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I can't believe I would've been wrong at THIS question if I handed it on the test. Why am I so careless?!? Here we go


http://img15.imageshack.us/img15/2825/fixxooros.jpg [Broken]

So clearly I was wrong.

The current INCREASES when the switch is closed.

As for the physical explaination, let me think about it.
 
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  • #6
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Aaaaaaaaaaaaaaaah! Now I get it! :smile:

As for being careless, that's what we train and exercise for.
The reason I'm asking for the physical consideration, is because it will give you a visual (or rather conceptual) cue about what you're doing.

The point being that when your calculated results don't match your visual cues, you'll try and find out what you did wrong and fix it.
 
  • #7
Femme_physics
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As for being careless, that's what we train and exercise for.

True, I learned that in mechanics, I think I got it in me after passing both the internal and external tests successfully. Sadly, I did the electronics test before them...before I became a super-test-taking-ninja!

The reason I'm asking for the physical consideration, is because it will give you a visual (or rather conceptual) cue about what you're doing.

I agree. Then, for the physical explanation, I'd say the current has more flowing options. Does that work? :smile:
 
  • #8
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Question 2: Does the voltage go up, down, or doesn't change? Base the answer on a physical consideration or prove it ia calculation


Physical Explanation:


It doesn't change. The voltage remains constant regardless of the alteration in the construct of the circuit since V is the source, while R and I are determined by the construct of the circuit.
 
  • #9
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True, I learned that in mechanics, I think I got it in me after passing both the internal and external tests successfully. Sadly, I did the electronics test before them...before I became a super-test-taking-ninja!

Yieaah!
Can you catch a speeding bullet's kinetic energy now, when such a question is fired at you?


I agree. Then, for the physical explanation, I'd say the current has more flowing options. Does that work? :smile:

Yes, that works.
I can't believe how far you've come!!! :!!)


Question 2: Does the voltage go up, down, or doesn't change? Base the answer on a physical consideration or prove it ia calculation

Physical Explanation:

It doesn't change. The voltage remains constant regardless of the alteration in the construct of the circuit since V is the source, while R and I are determined by the construct of the circuit.

Uh oh, I take that back.
The idealized voltage source indeed keeps a constant voltage, but......
 
  • #10
for physical explanations for increase in current when a parallel resistance is added, think of the first wire getting fatter thus allowing more current to pass and since the voltage is same, desired result follows.
 
  • #11
Femme_physics
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The idealized voltage source indeed keeps a constant voltage, but......

What? But what?

for physical explanations for increase in current when a parallel resistance is added, think of the first wire getting fatter thus allowing more current to pass and since the voltage is same, desired result follows.

Yep! The water analogy is also great :smile:
 
  • #12
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What? But what?

If you look at your problem, you'll see that an internal resistance was specified.
This means that the terminal voltage is not constant.
(Hehe, I managed to insert "terminal voltage" into a sentence! :smile:)

Now what would the consequences be that the voltage source is not perfect? :confused:
 
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  • #13
Femme_physics
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I thought that the terminal voltage is constant. It's just

Vf = Vi - I1r

I don't see anything about rates of change or "time" in this formula

Sorry for the mechanics notation. Force of habit! :biggrin:
 
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  • #14
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Mechanics and electronics are interwoven (didn't you study mechatronics?)
So it's sort of fine to use the same symbols. :wink:
Although..... in mechanics you have a velocity v (lower case letter), whereas in electronics you have a voltage V (capital letter).

Nope. It depends on the current I1.
The more current flows the lower the terminal voltage.

Just like in real life.....
If you make a battery do too much, at some point it won't be able to provide voltage anymore to make a current flow. :smile:
 
  • #15
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Mechanics and electronics are interwoven (didn't you study mechatronics?)

Am "studying", not studied! But yes, I do realize they're intrervoven, but having the mathmatical introspection as to how, well-- that I didn't fully grasp yet.

If you make a battery do too much, at some point it won't be able to provide voltage anymore to make a current flow.

Fair enough, but I don't see "time" in ohm's law. Ohm's law doesn't care about time!
 
  • #16
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Come to think about it, there was a word I wasn't sure how to translate



in hebrew it's

מתח ההדקים

I just figured it means voltage according to my dictionaries and translators

Well, one of the words

מתח

Does mean voltage. The other words

הדקים

I'm not so sure about, but I just figured it's a fancy way of writing voltage. Since we're out of school semester, I can't get access to ask my teacher. I tried emailing him but he hasn't emailed me back.

Maybe Quabache can help us here.. :smile:
 
  • #17
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Am "studying", not studied! But yes, I do realize they're intrervoven, but having the mathmatical introspection as to how, well-- that I didn't fully grasp yet.

All in good time. You'll get there. And it will be amazing! :smile:



Fair enough, but I don't see "time" in ohm's law. Ohm's law doesn't care about time!

No, "time" is not in Ohm's law.
However, voltage and current are.

Remember that a voltage source is modelled as an ideal voltage source in series with an internal resistor.
When the current increases, so does the voltage drop over the internal resistor.
In effect, the voltage source delivers less voltage on its "terminals".
(There! I got to insert "terminals" again in a sentence! :smile:)
 
  • #18
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If I calculate V across all the resistors will it be sufficient proof?


PS made a reply earlier you might have missed
 
  • #19
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Come to think about it, there was a word I wasn't sure how to translate



in hebrew it's

מתח ההדקים

I just figured it means voltage according to my dictionaries and translators

According to google translate, it's "power terminals".
That makes sense, since it would indicate the "power" available at the "terminals" of the battery.
A "terminal" would be a point on the outside of a device where you can connect something.


Well, one of the words

מתח

Does mean voltage.

Right.


The other words

הדקים

I'm not so sure about, but I just figured it's a fancy way of writing voltage. Since we're out of school semester, I can't get access to ask my teacher. I tried emailing him but he hasn't emailed me back.

Maybe Quabache can help us here.. :smile:

According to google translate it mean "thin", but I have no clue to what that might mean.


If I calculate V across all the resistors will it be sufficient proof?

Sure. Let's see if it will match with the physical consideration that the voltage should have dropped. :)
 
  • #20
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Hmm....then I think I understand. They're really asking for the voltage of the internal resistance, right?


Power terminals?


But the word

מתח

means voltage.

How did we get to "power"?!?
 
  • #21
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Hmm....then I think I understand. They're really asking for the voltage of the internal resistance, right?

No, they're asking for the voltage on the outside of the voltage source.
That is the original voltage reduced by the voltage across the internal resistance.


Power terminals?

But the word

מתח

means voltage.

How did we get to "power"?!?

Can't help you there! :wink:
But it doesn't matter.
"On the terminals" means on the outside of the voltage source.
 
  • #22
Femme_physics
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No, they're asking for the voltage on the outside of the voltage source.
That is the original voltage reduced by the voltage across the internal resistance.
OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOH!


Reading comprehension, once again!

In that case the voltage GOES DOWN

http://img39.imageshack.us/img39/4374/vdownl.jpg [Broken]

Right? :smile:
 
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  • #23
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Well, you mixed up your symbols a bit in the first 2 lines, writing V when you should be writing I.

But yes! You're right! :smile:
 
  • #24
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Well, you mixed up your symbols a bit in the first 2 lines, writing V when you should be writing I.

Well I meant the voltage for the internal resistance. So I included "V" for voltage and I included the sign for internal resistance, "r". Are you sure I got the signs wrong?

But yes! You're right!

w00t! So now question 3...

At which of the two cases, will be the general power of the circuits bigger? By how much?


http://img41.imageshack.us/img41/2420/powerssss.jpg [Broken]

When the circuit is closed the power is bigger!!!

Right? :smile:
 
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  • #25
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Well I meant the voltage for the internal resistance. So I included "V" for voltage and I included the sign for internal resistance, "r". Are you sure I got the signs wrong?

Not the signs - the symbols!

"r" is not a a sign. It is the internal resistance of the voltage source.
If you multiply it by the current, you'll get the voltage difference across the internal resistance.

So you would have for instance:
[tex]V_1 = V - I_1 \cdot r = 1 \textrm{ [V]} - 0.5 \textrm{ [A]} \cdot 1 \mathrm{ [\Omega]} = 0.5 \textrm{ [V]}[/tex]



w00t! So now question 3...

At which of the two cases, will be the general power of the circuits bigger? By how much?

When the circuit is closed the power is bigger!!!

Right? :smile:

Right! And power to you! :smile:
 

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