Find the value of resistance in the circuit

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

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I don't understand how can current in ammeter be zero .Why is 2V battery not sending current in the right loop ?

 

[cnh1995]

I don't understand how can current in ammeter be zero .Why is 2V battery not sending current in the right loop ?

What is the voltage across R?

A specific value of R will make the ammeter read zero. What should be that value?

 

[cnh1995]

A specific value of R will make the ammeter read zero.

In other words, what should be the current through R so that the ammeter reads zero?

 

[Jahnavi]

In other words, what should be the current through R so that the ammeter reads zero?

I think what you are suggesting is that if the potential drop across R is 2V , then there would be no current in the ammeter .

Right ?

 

[cnh1995]

I think what you are suggesting is that if the potential drop across R is 2V , then there would be no current in the ammeter .

Right ?

No.
Potential drop across R is 2V no matter what the value of R is. Do you see why?

How would you apply Kirchhoff's current law at the juntion of R , the 5000 ohm resistor and the left terminal of the ammeter?
What should be the current through R if you know the current through the 5000 ohm resistor?

 

[Jahnavi]

No.
Potential drop across R is 2V no matter what the value of R is. Do you see why?

Yes.

How would you apply Kirchhoff's current law at the juntion of R , the 5000 ohm resistor and the left terminal of the ammeter?
What should be the current through R if you know the current through the 5000 ohm resistor?

Will the current across 5000 ohm be 12/(5000+R) irrespective of E2 ? Doesn't E2 affect this current ?

 

[cnh1995]

Yes.
Will the current across *through 5000 ohm be 12/(5000+R) irrespective of E2 ? Doesn't E2 affect this current ?

What is the voltage drop across the 5000 ohm resistance?
What is the current through the 5000 ohm resistance?

 

[Jahnavi]

Please explain your point . A question for a question is not helping much .

I get the correct answer if I equate [12/(5000+R)]R = 2

 

[cnh1995]

I get the correct answer if I equate [12/(5000+R)]R = 2

That's the correct equation.
Voltage across R is 2V.

Please explain your point . A question for a question is not helping much

I am afraid if I started explaining, I might just give out the answer (and get a warning from some mod). I am not good at the art of explaining something precisely.

How about some reverse engineering?
You know the correct value for R. What can you say about the relation between the currents through R and 5000 ohm resistance? What would it be if R had some other value?

 

[cnh1995]

Please explain your point . A question for a question is not helping much

The 12V battery is driving a current through the 5000 ohm resistor. Say that current is I₁. The current through R is I₂. How are I₁ and I₂ related if the ammeter reads zero?
Have you studied Kirchhoff's current law?

Should the current I₁ split at the junction of R, ammeter and 5000 ohm resistance?

 

[Jahnavi]

Potential drop across R is 2V no matter what the value of R is. Do you see why?

It is because 2V battery is placed across R .Right ?

Suppose there is no 12V battery and no 5000 ohm resistor i.e only the right loop is present .

What is the difference between this case and the setup given in the problem ? Why does current flow in the former but not in the latter despite potential difference across R being 2V in both the cases ?

 

[Jahnavi]

The 12V battery is driving a current through the 5000 ohm resistor. Say that current is I₁. The current through R is I₂. How are I₁ and I₂ related if the ammeter reads zero?

Have you studied Kirchhoff's current law?

Should the current I₁ split at the junction of R, ammeter and 5000 ohm resistance?

No . It should not split .I1 =I2 . I know KCL

 

[cnh1995]

It is because 2V battery is placed across R .Right ?

Suppose there is no 12V battery and no 5000 ohm resistor i.e only the left loop is present .

What is the difference between this case and the setup given in the problem ? Why does current flow in the former but not in the latter despite potential difference across R being 2V in both the cases ?

In the former case, the current is driven by the 2V source.
In the latter case, the current is supplied by the 12V source and the 2V source only maintains a voltage of 2V across R. It does not supply any power. If it did, it would violate KCL and KVL.

 

[cnh1995]

No . It should not split .I1 =I2 . I know KCL

Yes.
Now given that the voltage across R is 2V, what is the voltage across 5000 ohm resistance? Do you know KVL?

 

[Jahnavi]

In the former case, the current is driven by the 2V source.
In the latter case, the current is supplied by the 12V source and the 2V source only maintains a voltage of 2V across R. It does not supply any power. If it did, it would violate KCL and KVL.

OK. This is something new and interesting . Thanks .

But why is only 12V battery driving the current and not 2V battery ? Why doesn't it supply any power ? What if there was a current in the ammeter , wouldn't 2V battery supply current in that case ?

 

[cnh1995]

What if there was a current in the ammeter , wouldn't 2V battery supply current in that case ?

Yes. But for all values of R except the one which is the answer to this problem.
This specific value of R prevents any current from flowing through the ammeter.
Have you studied Wheatstone's bridge? This is exactly like a balanced Wheatstone's bridge.

 

[Jahnavi]

Yes.
Now given that the voltage across R is 2V, what is the voltage across 5000 ohm resistance? Do you know KVL?

10 V . Current through 5000 ohm will be 10/5000 ohm .This same current will flow through R i.e (1/500)R = 2 gives R .

 

[cnh1995]

10 V . Current through 5000 ohm will be 10/5000 ohm .This same current will flow through R i.e (1/500)R = 2 gives R .

That's correct.

 

[Jahnavi]

So role of 2V battery is to maintain a potential difference of 2V across R irrespective of whether any current flows through Ammeter or not .

Is that correct ?

 

[cnh1995]

So role of 2V battery is to maintain a potential difference of 2V across R irrespective of whether any current flows through Ammeter or not .

Is that correct ?

Yes.

You can see for any other value of R, there will be a current through the ammeter and the 2V source will be supplying or absorbing power depending on the direction of the current through the ammeter.

 

[Jahnavi]

Yes. But for all values of R except the one which is the answer to this problem.
This specific value of R prevents any current from flowing through the ammeter.
Have you studied Wheatstone's bridge? This is exactly like a balanced Wheatstone's bridge.

I am not sure about Wheatstone bridge , but I think this same mechanism happens in potentiometer . Right ?

 

[cnh1995]

I am not sure about Wheatstone bridge , but I think this same mechanism happens in potentiometer . Right ?

Yes.

In this problem, suppose you removed the 2V source. Now because of the 12V source only, there will be a current in the leftmost loop. The voltage across R is still 2V.
This means the 12V source establishes a 2V potential difference across R. Now when you connect the 2V source back in the circuit, you are connecting a 2V source between two points where the potential difference is already 2V.
Hence, adding the 2V source doesn't make any difference.

 

[Jahnavi]

Thanks a lot

 

[cnh1995]

Thanks a lot

No probs!
Good night!

 

[Jahnavi]

Everything else remaining same in the circuit , if I remove the 5000 ohm resistor and try to write KVL in the two loops I end up with an absurd result 10 =0 .

Is it because now both the batteries are trying to maintain their respective potential differences across resistor R ?

Now if I put a resistor somewhere in the left or right loop then again things fall in place .

Why does this happen ?

 

[haruspex]

Is it because now both the batteries are trying to maintain their respective potential differences across resistor R ?

Yes. You have created a short, much like simply connecting the terminals of the battery to each other. In the real world, there would be some resistance in the wires and internal resistance in the batteries. In particular, the lower voltage battery may have considerable resistance to being recharged.

 

[cnh1995]

Everything else remaining same in the circuit , if I remove the 5000 ohm resistor and try to write KVL in the two loops I end up with an absurd result 10 =0 .

If you just remove it, there won't be any problem. But if you replace it with a wire (short it), you are creating a contradiction i.e. an invalid condition in circuit theory.

You can't connect two ideal voltage sources in parallel unless they have same voltage and you can't connect two ideal current sources in series unless they have same current.
Also, you can't short circuit an ideal voltage source and open circuit an ideal current source.

 

[Jahnavi]

Yes. You have created a short, much like simply connecting the terminals of the battery to each other. In the real world, there would be some resistance in the wires and internal resistance in the batteries. In particular, the lower voltage battery may have considerable resistance to being recharged.

Fine .

One thing is still confusing me .Potential difference across R is 2V if some current flows in/out of 2V battery .Potential difference across R is 2V if no current flows in/out of 2V battery .

So , why is it that for a particular value of R , no current flows through 2V battery considering the fact that potential difference across R always remain 2V ?

 

[cnh1995]

So , why is it that for a particular value of R , no current flows through 2V battery considering the fact that potential difference across R remains 2V throughout ?

For this particular value of R, there would be 2V across it with or without the 2V battery.
The KVL and KCL take care of the mathematical explanation.

How about some hand waving?

In this problem, suppose you removed the 2V source. Now because of the 12V source only, there will be a current in the leftmost loop. The voltage across R is still 2V.
This means the 12V source establishes a 2V potential difference across R. Now when you connect the 2V source back in the circuit, you are connecting a 2V source between two points where the potential difference is already 2V.

The 2V source doesn't supply or absorb any power is because it doesn't have to do it. It is directly connected across R, hence, as per circuit theory, its job is to maintain a voltage of 2V across R. But it is already taken care of by the 12V source, hence, to maintain the 2V p.d. across R, the 2V source doesn't need to supply/absorb current.

If R had some other value, the potential difference across R established by the 12V sourve would be more/less than 2V. Hence, when you connect the 2V source back in the circuit, it is forcing the voltage across R to be 2V, which will alter the voltage across the 5000 resistor. This in turn changes the current through the 5000 resistance (I₁) while the current through R (I2) is unchanged. The difference between I1 and I2 flows through the ammeter.
Here, the 2V battery has to absorb/supply current to maintain 2V across R. In the previous case, there was already a 2V p.d. across R.

 

[Jahnavi]

For this particular value of R, there would be 2V across it with or without the 2V battery.
The KVL and KCL take care of the mathematical explanation.

How about some hand waving?

The 2V source doesn't supply or absorb any power is because it doesn't have to do it. It is directly connected across R, hence, as per circuit theory, its job is to maintain a voltage of 2V across R. But it is already taken care of by the 12V source, hence, to maintain the 2V p.d. across R, the 2V source doesn't need to supply/absorb current.

If R had some other value, the potential difference across R established by the 12V sourve would be more/less than 2V. Hence, when you connect the 2V source back in the circuit, it is forcing the voltage across R to be 2V, which will alter the voltage across the 5000 resistor. This in turn changes the current through the 5000 resistance (I₁) while the current through R (I2) is unchanged. The difference between I1 and I2 flows through the ammeter.
Here, the 2V battery has to absorb/supply current to maintain 2V across R. In the previous case, there was already a 2V p.d. across R.

You have not only removed my confusion but have also very nicely explained the underlying concept of a potentiometer .

Thank you so much