Change in ammeter reading due to change in resistance

[toforfiltum]

Homework Statement

Homework Equations

1/R= 1/R₁ + 1/R₂ +...
I₁= R₂/R₁ x I

The Attempt at a Solution

I chose B because I thought that since total resistance in the branch of variable resistor has increased, the branch with ammeter will get a greater share of current. Since I'm not too sure about it because the overall increase in resistance does cause a drop in main current, I put in some random values of R and V , and I still get an increase in ammeter reading. So why is the answer C?

 

[phinds]

Your ammeter, with its internal resistance, are shown in parallel to the battery, so

1) the rest of the circuit is irrelevant to the ammeter reading
2) the ammeter will likely burn up rather quickly, or if it is strong enough not to, your battery will melt down due to the short circuit.

 

[toforfiltum]

Your ammeter, with its internal resistance, are shown in parallel to the battery, so

1) the rest of the circuit is irrelevant to the ammeter reading
2) the ammeter will likely burn up rather quickly, or if it is strong enough not to, your battery will melt down due to the short circuit.

Huh? Isn't that symbol a resistor? It's not stated in the question that it's the internal resistance of ammeter...

 

[phinds]

Huh? Isn't that symbol a resistor? It's not stated in the question that it's the internal resistance of ammeter...

Oh. I just assume that's what it was. Perhaps a bad assumption. In that case, ignore my point 2. Point 1 still holds.

 

[toforfiltum]

Oh. I just assume that's what it was. Perhaps a bad assumption. In that case, ignore my point 2. Point 1 still holds.

Why? Isn't ammeter connected to circuit too?

 

[phinds]

Why? Isn't ammeter connected to circuit too?

Sure. And you could connect it to lots of other resistors too, but all would be irrelevant. The resistor/ammeter pair are connected directly across the battery and no amount of other circuitry is going to change that.

 

[toforfiltum]

Sure. And you could connect it to lots of other resistors too, but all would be irrelevant. The resistor/ammeter pair are connected directly across the battery and no amount of other circuitry is going to change that.

I still don't understand. If I increase the resistance in variable resistor, this will cause an overall drop in current.
I don't know... I just put in some random values...into the components and I do get an increase in ammeter reading.

So...I can't see why it's irrelevant.

 

[Qwertywerty]

I chose B because I thought that since total resistance in the branch of variable resistor has increased, the branch with ammeter will get a greater share of current. Since I'm not too sure about it because the overall increase in resistance does cause a drop in main current, I put in some random values of R and V , and I still get an increase in ammeter reading. So why is the answer C?

Hint : Use Kirchoff's loop law in one inner , and one outer loop , both of which include the battery .

 

[toforfiltum]

Hint : Use Kirchoff's loop law in one inner , and one outer loop , both of which include the battery .

Ah I see now. Since potential drop is always the same across resistor in outer loop, and its resistance is constant, I also must be constant.
Thanks for this 'light-bulb' moment.

 

[phinds]

Ah I see now. Since potential drop is always the same across resistor in outer loop, and its resistance is constant, I also must be constant.
Thanks for this 'light-bulb' moment.

Then your drawing does not represent what you think it represents.

EDIT: OOPS. I was responding to a different post (#7)

 

[toforfiltum]

Then your drawing does not represent what you think it represents.

EDIT: OOPS. I was responding to a different post (#7)

Yes. I see where I'm wrong now. Got the ratios of currents wrong.

 

[phinds]

Yes. I see where I'm wrong now. Got the ratios of currents wrong.

Ratios of currents in this case have nothing to do with the ammeter reading.

 

[toforfiltum]

Ratios of currents in this case have nothing to do with the ammeter reading.

Uhhm...ratios of resistors?

 

[phinds]

Uhhm...ratios of resistors?

Has nothing to do with the ammeter reading in this circuit.

 

[toforfiltum]

Has nothing to do with the ammeter reading in this circuit.

is it because of what I've said in post #9?
Then, my previous approach cannot be used at all is it?

 

[phinds]

is it because of what I've said in post #9?

yes

Then, my previous approach cannot be used at all is it?

Not sure what your previous approach was but if it gives anything but a constant reading on the ammeter it's wrong.

 

[toforfiltum]

yes
Not sure what your previous approach was but if it gives anything but a constant reading on the ammeter it's wrong.

my previous approach was calculating new total resistance of circuit, hence finding new main current and from the ratios of resistors in two branches calcuate the current in each small branch.

 

[toforfiltum]

yes
Not sure what your previous approach was but if it gives anything but a constant reading on the ammeter it's wrong.

my previous approach was calculating new total resistance of circuit, hence finding new main current and from the ratios of resistors in two branches calcuate the current in each small branch.

 

[phinds]

my previous approach was calculating new total resistance of circuit, hence finding new main current and from the ratios of resistors in two branches calcuate the current in each small branch.

New main current has nothing to do with what flows through the ammeter. At this point I'm not sure whether you're just trolling to see how long you can string me along trying to explain this trivial fact. Draw the circuit without the middle vertical part there at all. That's what the ammeter sees. Nothing else is relevant to the ammeter.