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In the, Ohm's law experiment to find res...

In the, Ohm's law experiment to find resistance of unknown resistor `R`, following two arrangement (`a`) and (`b`) are possible.

The resistance measured is given by
`R_(measured)=(V)/(i)`
`V`= voltage reading of voltmeter, `i=` current Reading of ammeter.
But unfortunately the ammeters and voltmeter used are not ideal, but having resistance `R_(A)` and `R_(V)` respectively.
You are given two resistor `X` and `Y`. Whose resistance is to be determined , using an ammeter of `R_(A)=0.5Omega` and a voltmeter of `R_(V)=20 KOmega`. It is known that `X` is in range of a few Ohm and `Y` is in range of several kilo ohm. Which of the circuit is preferable to measure `X` and `Y`-Resistor Circute

A

`x to (a)`, `y to (b)`

B

`x to (b)`, `y to (a)`

C

`x to (a)`, `y to (a)`

D

`x to (b)`, `y to (b)`

Text Solution

Verified by Experts

The correct Answer is:
B
Let total current is `I_(0)`. Then in circuit (`a`)

`i=(R_(V))/(R+R_(A)+R_(V))xxI_(0)`
So `(V)/(I)=R+R_(A)=R_(measured)`
in circuit (`b`)

`V=I'xxR_(V)=(RR_(V))/(R+R_(V))xxi`
So `R_(measured)=(RR_(V))/(R+R_(V))`
To measure `x`, circuit (`b`) should be used, as
`R_(measured)=(R R_(V))/(R+R_(V)) gt gt R` as `R gt gt R_(A)`
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