Current through an Ammeter in an Electric Circuit

[VSayantan]

Homework Statement

[/B]
It is required to find the current through the ammeter, marked A, in the above figure.

Homework Equations

$V=iR$

The Attempt at a Solution

I've redrawn the circuit as follows -

But don't know how to obtain the value $i_3$

 

[Abhishek kumar]

Homework Statement

View attachment 215842[/B]
It is required to find the current through the ammeter, marked A, in the above figure.

Homework Equations

$V=iR$

The Attempt at a Solution

I've redrawn the circuit as follows -
View attachment 215844
But don't know how to obtain the value $i_3$

Apply loop rule

 

[phinds]

You are making this a lot harder than it is. What is the relationship between the two 100ohm resistors to the left of the ammeter?

 

[cnh1995]

Have you studied Thevenin's theorem?

 

[VSayantan]

You are making this a lot harder than it is. What is the relationship between the two 100ohm resistors to the left of the ammeter?

They are connected in parallel.

 

[VSayantan]

Have you studied Thevenin's theorem?

Yes.
But it needs two terminals, right?

 

[phinds]

They are connected in parallel.

So, what do you DO with that information ... ?

 

[VSayantan]

So, what do you DO with that information ... ?

The equivalent resistance is ${\frac {100} {2}}\Omega$
But most of the current will flow through the left-most path, since it has no resistance.
I don't know what to do with the $50 \Omega$ resistance.

 

[phinds]

The equivalent resistance is ${\frac {100} {2}}\Omega$
But most of the current will flow through the left-most path, since it has no resistance.
I don't know what to do with the $50 \Omega$ resistance.

You REALLY want this to be hard, it seems. Try redrawing the circuit, collapsing as many resistances as possible into single resistances.

 

[phinds]

By the way, here's a hint. In terms of solving this circuit for the current through the ammeter, there are only 3 resistors in the reduced drawing. When you get to where this becomes immediately apparent, you will be well on the way to finding it easy to solve resistive circuits.

 

[cnh1995]

Yes.
But it needs two terminals, right?

Well, there will be two open terminals if you open circuit the ammeter to find V_th.

But I believe you should follow phinds' suggestion first (he and I posted almost at the same moment). It is a lot easier.

 

[VSayantan]

You REALLY want this to be hard, it seems. Try redrawing the circuit, collapsing as many resistances as possible into single resistances.

Is this the reduced circuit?

 

[cnh1995]

View attachment 215852
Is this the reduced circuit?

No, the second circuit is the reduced circuit. You can find the ammeter current using simple current division.
What is the resistance of an ammeter, if it is not mentioned in the problem?

 

[phinds]

View attachment 215852
Is this the reduced circuit?

No, you are still missing a valid step (your last step was, as cnh pointed out, invalid). Think about how you could reduce it from 5 resistors to 3 without changing the flow through the ammeter.

 

[VSayantan]

No, the second circuit is the reduced circuit. You can find the ammeter current using simple current division.
What is the resistance of an ammeter, if it is not mentioned in the problem?

The resistance of the ammeter is not given. So, I assume it to have zero resistance.

The equivalent resistance of the last circuit is $60\Omega$?

Then the current through the $5\Omega$ resistance is $100 mA$

 

[phinds]

Now you're just floundering but you got part way there. This is the circuit you need to validly simplify:

What is the effect on the current through the ammeter if you move the 5 ohm resistor together and the 50 ohm resistors together?

 

[VSayantan]

Now you're just floundering but you got part way there. This is the circuit you need to validly simplify:
View attachment 215854

What is the effect on the current through the ammeter if you move the 5 ohm resistor together and the 50 ohm resistors together?

Like this?
Then the potential is same because the resistors are all connected in parallel. Right?

 

[cnh1995]

View attachment 215857
Like this?
Then the potential is same because the resistors are all connected in parallel. Right?

Correct, but not ALL resistances are in parallel. Now what is the current through the ammeter? You already know the current through the 10 ohm resistance. How will it split between the two parallel branches?

 

[.Scott]

Correct, but not ALL resistances are in parallel. Now what is the current through the ammeter? You already know the current through the 10 ohm resistance. How will it split between the two parallel branches?

I don't think he knows the current through the 10 ohm yet. I believe he will need to compute the total resistance in the circuit to compute that current.

 

[VSayantan]

Correct. Now what is the current through the ammeter? You already know the current through the 10 ohm resistance. How will it split between the two parallel branches?

The current through the $10~ohm$ resistance is $100~mA$.
So, the potential across the $10~ohm$ resistance is $1.0~V$.
Therefore the potential across the parallel branches must be $5.0~V$.
Since the $100~ohm$ resistance, connected with the ammeter, is made of two equal resistances the potential is equally divided, but same current flows through each component. Giving a value of $\frac {5.0~V} {100~\Omega}$, i.e., $50.0~mA$

 

[cnh1995]

I don't think he knows the current through the 10 ohm yet.

He does. See #15.

The current through the $10~ohm$ resistance is $100~mA$.
So, the potential across the $10~ohm$ resistance is $1.0~V$.
Therefore the potential across the parallel branches must be $5.0~V$.
Since the $100~ohm$ resistance, connected with the ammeter, is made of two equal resistances the potential is equally divided, but same current flows through each component. Giving a value of $\frac {5.0~V} {100~\Omega}$, i.e., $50.0~mA$

Looks good.

 

[VSayantan]

Looks good.

Thanks a lot @cnh1995 and @phinds for guiding me through the problem.

I would love to know how do I solve the problem using Thevenin's theorem.

Do I need to disconnect the ammeter and obtain the equivalent resistance?

 

[cnh1995]

Do I need to disconnect the ammeter and obtain the equivalent resistance?

Suppose the ammeter terminals are a and b. First, you need to disconnect the ammeter and find the open circuit voltage between the two terminals ( where the ammeter was placed earlier) i.e. Vab.
Then you have to short the voltage source and find the equivalent resistance R between a and b.
Then Vab/R will give you the ammeter current.

 

[phinds]

View attachment 215857
Like this?
Then the potential is same because the resistors are all connected in parallel. Right?

No, forget the potential. You are looking for current. The current through the 5ohm restistors HAS to be the same because they are in series and the same for the 50 ohm resistors. Now that you have the simplified circuit, it should be very ease to solve for the current through the ammeter. When you have done this sort of thing for a while, you should be able to go from the original circuit to the final reduced circuit all in one obvious step.

 

[VSayantan]

No, forget the potential. You are looking for current. The current through the 5ohm restistors HAS to be the same because they are in series and the same for the 50 ohm resistors. Now that you have the simplified circuit, it should be very ease to solve for the current through the ammeter. When you have done this sort of thing for a while, you should be able to go from the original circuit to the final reduced circuit all in one obvious step.

You explained so easily!

 

[VSayantan]

Suppose the ammeter terminals are a and b. First, you need to disconnect the ammeter and find the open circuit voltage between the two terminals ( where the ammeter was placed earlier) i.e. Vab.
Then you have to short the voltage source and find the equivalent resistance R between a and b.
Then Vab/R will give you the ammeter current.

Is it ok?
So, the equivalent Thevenin potential is $$V_{Th} = \frac {6\times 100}{100+10}~V$$
That is $$V_{Th}=5.45~V$$

The equivalent Thevenin resistance is $$R_{Th} = [{\frac {1}{{\frac {1}{10}}+{\frac {1}{100}}}}+50+50]\Omega$$
That is $$R_{Th}=100.909~\Omega$$

Then the current through the ammeter is $V_{Th}/R_{Th}$?

 

[cnh1995]

Your calculation of Rth is incorrect. You have (10Ω||100Ω) in series with 100Ω.

Keep Vth and Rth in fractional form (p/q). It will be easier to simplify.

 

[VSayantan]

Your calculation of Rth is incorrect. You have (10Ω||100Ω) in series with 100Ω.

Keep Vth and Rth in fractional form (p/q). It will be easier to simplify.

Thanks for pointing out @cnh1995
I typed it wrong.
The current comes out to be $$i=54.009~mA$$
Which is pretty close to $50~mA$, as obtained previously.

 

[cnh1995]

The current comes out to be

i=54.009 mAi=54.009 mA​

i=54.009~mA
Which is pretty close to 50 mA50 mA50~mA, as obtained previously.

No, it must be the same as the one obtained previously.
You have Vth=6*100/110 V.
Rth= 100+ (1000/110) ohm.
Simplify and take the ratio.

 

[VSayantan]

No, it must be the same as the one obtained previously.
You have Vth=6*100/110 V.
Rth= 100+ (1000/110) ohm.
Simplify and take the ratio.

Oh!
Yes, you are right.

The current is $$i=\frac {600}{110}~V \times \frac {110}{12000}{\Omega}^{-1}$$
i.e., $$i=50.0~mA$$

 

[cnh1995]

Oh!
Yes, you are right.

The current is $$i=\frac {6000}{110}~V \times \frac {110}{12000}{\Omega}^{-1}$$
i.e., $$i=50.0~mA$$

There's again one typo in your equation, but the final result is correct.

 

[phinds]

You're still not getting the obvious simplified circuit. Draw the circuit WITH the voltage source AND the ammeter AND 3 resistors, one in series w/ the voltage source, one in series w/ the ammeter and one in the bottom line.

You want to get to the point where when your eyes see this:

Your brain sees this, pretty much automatically

That IS of course based on the specific question being answered. This reduction might not be as helpful if you were asked to find the voltage across the left-most resistor (although even for that, I would still do this reduction, get the current in that line and then the voltage)