# Voltmeter Ammeter and percentage error in R

• Jahnavi
In summary,The student is not able to understand how percentage error in R is calculated using the given circuits.The student should 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.
Jahnavi

## The Attempt at a Solution

I am not able to understand how percentage
error in R is calculated using the given circuits.

Should I assign currents in the branches and write KVL equations ?

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I think I can see the answer by inspection, but I'd still do a couple of calculations to verify my assumptions.

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?

cnh1995
Jahnavi said:
I am not able to understand how percentage
error in R is calculated using the given circuits.
It means you need to find the arrangement that gives most accuate value for R.
As berkeman said, one of the three options can be eliminated by inspection.

berkeman said:
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 ?

cnh1995 said:
It means you need to find the arrangement that gives most accuate value for R.

How is R measured in these circuits ?

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

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...

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

berkeman said:
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 ?

Jahnavi said:
How is R being measured ? What is causing a change in the actual value of R ?
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 a slightly different value of R.

Jahnavi and berkeman
cnh1995 said:
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 a slightly different 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 .

Jahnavi said:
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 .
I did that in post #2...

Jahnavi said:
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 .
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.
But before that, eliminate the incorrect option from the three.

Jahnavi
cnh1995 said:
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) ?

Jahnavi said:
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 .
Yes.

Jahnavi
In a) voltmeter reading would be equal to the battery voltage but current would be little less .Ammeter reading would be less than what it should be . Calculated value of R is more than the actual value .

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 ?

Jahnavi said:
In a) voltmeter reading would be equal to the battery voltage but current would be little less .Ammeter reading would be less than what it should be . Calculated value of R is more than the actual value .

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 ?
Right.

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

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

Consider a 10 V battery .

In a) , the net resistance across the battery would be almost equal to R i.e about 103 .The current flowing through the ammeter would be {106/(106+10-3)} x (10/103) = 10-2 A .

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 ?

You need to calculate the exact values of V/I in both the cases.

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?

Can you show me how you would calculate exact V/I in a) circuit ?

I will do calculations for circuit b) by myself .

Jahnavi said:
Can you show me how you would calculate exact V/I in a) circuit ?

I will do calculations for circuit b) by myself .
I think you should work with symbols from the beginning and substitute the actual values in the final step. That would be easier.

Let's call the battery voltage V, voltmeter resistance Rv and ammeter resistance Ra.

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 Rv. You can call their equivalent resistance as R1.
Using voltage divider, what is the voltmeter reading? Using Ohm's law, what is the ammeter reading?

Jahnavi
I can do all what you have suggested .

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 ?

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

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

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.

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

cnh1995 said:
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 = Ra

In b) ∆R = R2/(R+Rv)

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 .

cnh1995
Jahnavi said:
I agree

In a) ∆R = Ra

In b) ∆R = R2/(R+Rv)

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 .
Right.
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.

Jahnavi
Thanks !

It doesn't say you have to use both meters at the same time.

Edit: Ah forget this. Makes no difference.

## What is a voltmeter and ammeter?

A voltmeter is a device used to measure the voltage or potential difference between two points in an electrical circuit. An ammeter is a device used to measure the electric current flowing through a circuit.

## How are a voltmeter and ammeter used together?

A voltmeter and ammeter are used together to measure the voltage and current in a circuit. The voltmeter is connected in parallel to the component being measured, while the ammeter is connected in series. This allows for the measurement of both voltage and current simultaneously.

## What is percentage error in R?

Percentage error in R refers to the difference between the actual resistance of a component in a circuit and the measured value, expressed as a percentage of the actual value. It is calculated using the formula: (|Ractual - Rmeasured| / Ractual) x 100%

## Why is percentage error important in electrical measurements?

Percentage error is important because it indicates the accuracy of the measurements being taken. A high percentage error can indicate a faulty component or errors in measurement technique, while a low percentage error suggests a more accurate measurement.

## How can percentage error in R be reduced?

Percentage error in R can be reduced by using more precise measuring instruments, ensuring proper connections in the circuit, and taking multiple measurements to obtain an average value. It is also important to use components with known and accurate values for comparison.

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