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- Thread starter Jahnavi
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berkeman

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Start with R = 1kOhm. Then yes, calculate the delta-R caused by the finite impedances of the Voltmeter and Ammeter in each circuit. Then maybe verify it with R = 10kOhm to be sure.

Also, one of the circuits can probably be eliminated right away by inspection. Can you say which one and why?

- #3

cnh1995

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It means you need to find the arrangement that gives most accuate value for R.I am not able to understand how percentage

error in R is calculated using the given circuits.

As berkeman said, one of the three options can be eliminated by inspection.

- #4

Jahnavi

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calculate the delta-R caused by the finite impedances of the Voltmeter and Ammeter in each circuit.

What exactly is ΔR ? Is it the difference between the resistance R and the net resistance across the battery ?

- #5

Jahnavi

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It means you need to find the arrangement that gives most accuate value for R.

How is R measured in these circuits ?

- #6

berkeman

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I don't think the battery's internal resistance is part of this question. It was not mentioned in the problem statement.What exactly is ΔR ? Is it the difference between the resistance R and the net resistance across the battery ?

Delta-R is the change in the measured resistance from what R really is. So for example, you already know that you could get a measured R = 1kOhm + 10^-3 Ohms if you measure both the resistor R and the Ammeter in series...

- #7

berkeman

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With the Voltmeter and Ammeter as shown in the circuits...How is R measured in these circuits ?

- #8

Jahnavi

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With the Voltmeter and Ammeter as shown in the circuits...

I might not be explaining myself well . A voltmeter measures voltage and an ammeter measures current .

How is R being measured ? What is causing a change in the actual value of R ?

- #9

cnh1995

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Ideally, the voltage across R should be equal to the battery voltage. So the ratio V/I (measured using the meters) should give you the value of R.How is R being measured ? What is causing a change in the actual value of R ?

However, real meters introduce their own resistances in the circuit and you get a

- #10

Jahnavi

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Ideally, the voltage across R should be equal to the battery voltage. So the ratio V/I (measured using the meters) should give you the value of R.

However, real meters introduce their own resistances in the circuit and you get aslightlydifferent value of R.

OK.

I am still not able to make a start . What exactly should I calculate ? Please give me a brief strategy to analyze these circuits .

- #11

berkeman

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I did that in post #2...OK.

I am still not able to make a start . What exactly should I calculate ? Please give me a brief strategy to analyze these circuits .

- #12

cnh1995

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As berkeman stated in #2, use actual values for battery voltage and R, and calculate the ratio voltmeter reading/ammeter reading for each circuit. You can work with symbols too, but it will be time consuming.OK.

I am still not able to make a start . What exactly should I calculate ? Please give me a brief strategy to analyze these circuits .

But before that, eliminate the incorrect option from the three.

- #13

Jahnavi

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But before that, eliminate the incorrect option from the three.

Without actually using any given data , Is it option C) ? Because voltmeter reading V will be nearly equal to the battery voltage but Ammeter reading will be very less as compared to what it should be . The ratio V/I will give a very high value of resistance . Error would be quite high .Is it correct ?

So it is between options a) and b) ?

- #14

cnh1995

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Yes.Without actually using any given data , Is it option C) ? Because voltmeter reading V will be nearly equal to the battery voltage but Ammeter reading will be very less as compared to what it should be . The ratio V/I will give a very high value of resistance . Error would be quite high .

- #15

Jahnavi

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In b) , the voltmeter reading would be less than the battery voltage , but Ammeter reading would be nearly equal to what it should be . V/I ratio would be less than what it should be . Calculated value of R is less than its actual value.

May be now we have to use the values given in the problem .

Right ?

- #16

cnh1995

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Right.

In b) , the voltmeter reading would be less than the battery voltage , but Ammeter reading would be nearly equal to what it should be . V/I ratio would be less than what it should be . Calculated value of R is less than its actual value.

May be now we have to use the values given in the problem .

Right ?

- #17

Jahnavi

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Do I need to write KVL and KCL equations for the two circuits ?

- #18

cnh1995

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You can, or you can also directly use the voltage and current divider equations (which are derived from KCL and KVL).Do I need to write KVL and KCL equations for the two circuits ?

- #19

Jahnavi

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In a) , the net resistance across the battery would be almost equal to R i.e about 10

This is equal to the current which would flow if only R were put across the battery .

But by this approach ,in b) circuit the voltage nearly comes equal to the battery voltage as almost all the battery potential drops across R .

How should I compare ?

- #20

cnh1995

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For option a, you can mentally calculate the value of meausred resistance.

For b, you need to use series-parallel combimations and voltage divider. Where are you having difficulty?

- #21

Jahnavi

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I will do calculations for circuit b) by myself .

- #22

cnh1995

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I think you should work with symbols from the beginning and substitute the actual values in the final step. That would be easier.

I will do calculations for circuit b) by myself .

Let's call the battery voltage V, voltmeter resistance R

What is the ammter current in option a? What is the ratio voltmeter reading/ammter reading in the first circuit?

In option B, you have R in parallel with R

Using voltage divider, what is the voltmeter reading? Using Ohm's law, what is the ammeter reading?

- #23

Jahnavi

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I wonder if we have to do calculations to such precision , then what purpose does this question have in an MCQ ?

Is there an alternative way to choose between a) and b) without doing much calculations ?

- #24

cnh1995

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Actually, you need only one calculation step and I think this is the easiest way.Is there an alternative way to choose between a) and b) without doing much calculations ?

You have to be very precise while calibrating the meters. Perhaps this question is all about using basic concepts like voltage division, equivalent resistance etc and knowing how the meters can affect the circuit parameters.I wonder if we have to do calculations to such precision , then what purpose does this question have in an MCQ ?

Interestingly, it turns out that the answer to the question actually depends on R. Depending on the value of R, either of a and b can be more accurate than the other.

- #25

CWatters

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You should also consider answer d. I think that would be my answer.

- #26

Jahnavi

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Interestingly, it turns out that the answer to the question actually depends on R. Depending on the value of R, either of a and b can be more accurate than the other.

I agree

In a) ∆R = R

In b) ∆R = R

In this problem with the given values , ∆R of circuit a) is less.

But since ∆R of b) depends on R , if value of R is small like 10 Ohms then circuit b) would give more accurate value .

- #27

cnh1995

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Right.I agree

In a) ∆R = R_{a}

In b) ∆R = R^{2}/(R+R_{v})

In this problem with the given values , ∆R of circuit a) is less.

But since ∆R of b) depends on R , if value of R is small like 10 Ohms then circuit b) would give more accurate value .

So you can see the circuit option a) is preferable practically. The error does not depend on voltmeter resistance and ammeter resistance is very small.

If the value of R is very small, b) is more accurate, but for measuring such small resistances, other more accurate methods (like Kelvin bridge) are used.

- #28

Jahnavi

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Thanks !

- #29

CWatters

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Edit: Ah forget this. Makes no difference.

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