You are given two resistors `X` and `Y` whose resistances are to be determined using an ammeter of resistance `0.5 Omega` and a voltmeter of reistance `20 k Omega`. It is known that `x` is in the range of a few ohms. While `Y` is in the range of several thousand ohms. In each case, while of the following two connections (Fig. `6.35`) would you choose for resistance measurement? Justify your answer quantiatively. [Hint : For each connection, determine the error in resistnce measurement. The connection that corresponds to a smaller error (for a given range of resistance) is to be perferred.]

Answer

circuit (a) should be used to measure `Y` circuit (b) should be used to measure `Y`

Explanation

Let us measure a resistance `R` using circuit (a) .
True value`= R`,
measured value `= (V)/(I) = R + R_(A)`
percentage error `= (R + R_(A) - R)/(R ) xx 100`
`= (R_(A) xx 100)/(R ) = (0.5 xx 100)/(R ) = (50)/(R )`

So the more the value of `R`, the lesser the percentage error. Hence, circuit (A) should be used to measure `Y`. Now, let us measure `R` using circuit (b).
True value `= R`
Measured value `= (V)/(I) = (V)/(I_(R) + I_(V)) = (V)/((V)/(R) + (V)/(R_(V)))= (R R_(V))/(R + R_(V))`
Percentage error `= (R + (R R_(V))/(R + R_(V)))/(R ) xx 100 = (1)/(1 + R_(V)//R) xx 100`
The lesser the value of `R`, the lesser the percentage error. Hence,circuit (b) should be used to measure `X`.