Electricity: Basic circuit question with a switch

AI Thread Summary
In the discussed circuit, closing the switch results in an increase in the ammeter's reading due to the addition of parallel resistance, which allows more current to flow. The voltage remains constant from the source, but the terminal voltage decreases because of the internal resistance of the voltage source, which is affected by the increased current. The conversation emphasizes the importance of understanding physical concepts alongside calculations to avoid errors in reasoning. Additionally, the discussion touches on the distinction between total power and wasted power, clarifying that wasted power refers to energy lost as heat in the internal resistance of the battery. Overall, the participants engage in a detailed exploration of circuit behavior and the implications of resistance and voltage changes.
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


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
 
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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:
 
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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You do?
Well, don't keep me in suspense... what is it? :confused:
 
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 gohttp://img15.imageshack.us/img15/2825/fixxooros.jpg

So clearly I was wrong.

The current INCREASES when the switch is closed.

As for the physical explanation, let me think about it.
 
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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.
 
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:
 
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.
 
Femme_physics said:
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?


Femme_physics said:
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! :!)


Femme_physics said:
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
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
Femme_physics said:
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
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
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
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
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
Femme_physics said:
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:



Femme_physics said:
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 modeled 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
If I calculate V across all the resistors will it be sufficient proof?PS made a reply earlier you might have missed
 
  • #19
Femme_physics said:
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.


Femme_physics said:
Well, one of the words

מתח

Does mean voltage.

Right.


Femme_physics said:
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.


Femme_physics said:
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
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
Femme_physics said:
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.


Femme_physics said:
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
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

Right? :smile:
 
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  • #23
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
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

When the circuit is closed the power is bigger!

Right? :smile:
 
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  • #25
Femme_physics said:
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:
V_1 = V - I_1 \cdot r = 1 \textrm{ [V]} - 0.5 \textrm{ [A]} \cdot 1 \mathrm{ [\Omega]} = 0.5 \textrm{ [V]}
Femme_physics said:
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:
 
  • #26
Not the signs - the symbols!

I meant symbols! *smackey forehead*

I'll leave the final question for tomorrow. The 2nd term, by the way, is the day after tomorrow^^ So tomorrow is super electronics practice day! I hope to have you with me :smile:
 
  • #27
And here is the final question that made me skip the question

Q: In which of the two cases (open switch or closed switch) will the wasted power of the circuit bigger. By how much?


What do they mean? Didn't we do that in the earlier clause? Comparing powers? What exactly is "wasted power"? If I'm translating it correctךט

ההספק המבוזבז
 
  • #28
Good morning Fp! :smile:

There! We're back to deducing what was intended by a word.

Actually, in your previous question I was already slightly concerned about the word "general power".
Why didn't they simply ask for "total power"?
Or can it be that it should be "useful power" instead of "general power"?

What you have is that the total power generated (which you calculated) consists of 2 parts.
One part that is dissipated in the circuit itself in a useful manner.
And one part that cannot be used, the part that is wasted as heat. That would be the part dissipated in the internal resistor of the battery.
 
  • #29
Ohhhhhhhhhhhhhhhhhhhhh. So that's what they're asking, I think. Yea, you're right.

The previous one meant total power (I think. The word in exact translation from hebrew means "general", or "inclusive")

כללי

But I still think they meant total.

As for this question, then yes, they meant the dissipated power at the internal resistance! Brilliant :biggrin:

Thank you for helping me understand it, ILS!


You think I'm ready for the test now? I've attached all the problems from Term A (Hebrew though-- but I think just by looking at the circuit and the English SYMBOLS (not signs, SYMBOLS :wink:) in the question ye'd be able to deduce it. Besides, we solved question 1 and 3 on the forum. Question 2 and 4 I did successfully on my own! :smile:
 

Attachments

  • #30
I like Serena said:
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.

"It is terminal voltage": voltage measured across the terminals of a "loaded" source, that is a source which is connected into a circuit.

The voltage across the terminals of a source, not connected to anything, that is no current flows through it, is the "electromotive force (abbreviate: emf) usually denoted by E or ε.
If r is the internal resistance of the source

Uterminal=E-rI

"General power" can be "generator power" the power supplied by the source to the circuit = (terminal voltage )* (generator current).
ehild
 
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  • #31
So my answer is still correct it's only a notation mishap?
 
  • #32
Yeah. I think you're ready for the test now. o:)

Btw, what is the answer to the last question?
 
  • #34
You rock! :)
 
  • #35
You rock harder! Thanks! ^^
 

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