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What is the voltage across R?Jahnavi said:I don't understand how can current in ammeter be zero .Why is 2V battery not sending current in the right loop ?
In other words, what should be the current through R so that the ammeter reads zero?cnh1995 said:A specific value of R will make the ammeter read zero.
cnh1995 said:In other words, what should be the current through R so that the ammeter reads zero?
No.Jahnavi said: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 said: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?
What is the voltage drop across the 5000 ohm resistance?Jahnavi said:Yes.
Will the current across *through 5000 ohm be 12/(5000+R) irrespective of E2 ? Doesn't E2 affect this current ?
That's the correct equation.Jahnavi said:I get the correct answer if I equate [12/(5000+R)]R = 2
I am afraid if I started explaining, I might just give out the answer (and get a warning from some modJahnavi said: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?Jahnavi said:Please explain your point . A question for a question is not helping much
cnh1995 said:Potential drop across R is 2V no matter what the value of R is. Do you see why?
cnh1995 said: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?
In the former case, the current is driven by the 2V source.Jahnavi said: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 ?
Yes.Jahnavi said:No . It should not split .I1 =I2 . I know KCL![]()
cnh1995 said: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.
Yes. But for all values of R except the one which is the answer to this problem.Jahnavi said:What if there was a current in the ammeter , wouldn't 2V battery supply current in that case ?
cnh1995 said:Yes.
Now given that the voltage across R is 2V, what is the voltage across 5000 ohm resistance? Do you know KVL?
That's correct.Jahnavi said: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 .
Yes.Jahnavi said: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 said: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.
Yes.Jahnavi said:I am not sure about Wheatstone bridge , but I think this same mechanism happens in potentiometer . Right ?
No probs!Jahnavi said:Thanks a lot![]()
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.Jahnavi said:Is it because now both the batteries are trying to maintain their respective potential differences across resistor R ?
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.Jahnavi said: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 .
haruspex said: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.
For this particular value of R, there would be 2V across it with or without the 2V battery.Jahnavi said: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 ?
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.cnh1995 said: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.
cnh1995 said: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.
Resistance is a measure of how much a material or component impedes the flow of electricity in a circuit. It is measured in ohms (Ω) and is denoted by the symbol R.
Knowing the value of resistance in a circuit is important because it helps us understand how the circuit will behave and how much current will flow through it. It also allows us to calculate the power dissipated in the circuit, which is important for determining the efficiency of the circuit.
The value of resistance in a circuit can be calculated using Ohm's Law, which states that resistance is equal to the voltage (V) divided by the current (I). So, R = V/I. It can also be calculated using the equation R = ρL/A, where ρ is the resistivity of the material, L is the length of the conductor, and A is the cross-sectional area of the conductor.
The value of resistance in a circuit is affected by the type of material used in the conductor, the length and cross-sectional area of the conductor, and the temperature of the material. It also depends on the type of circuit, as different circuit configurations have different values of resistance.
The value of resistance in a circuit can be measured using a multimeter, which is a device that measures voltage, current, and resistance. It can also be measured using an ohmmeter, which is a specialized type of multimeter that is used specifically for measuring resistance.