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Medium Resistance Measurement by Ammeter Voltmeter Metheod
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## Procedure:

This method is used for measurement of medium resistances. This method is very popular since the instruments required for this test are usually available in the laboratory. Two types of connections employed for an ammeter-voltmeter method are shown in fig.1 and fig.2. In both the cases, if readings of ammeter and voltmeter are taken, then the measured value of resistance is given by:

$R_m=\frac{Voltmeter readings}{Ammeter Readings}=\frac{V}{I}$

The measured value of resistance Rm, would be equal to the true value, R, if the ammeter resistance is zero and the voltmeter resistance is infinite, so that the conditions in the circuit are not disturbed. But in practice this is not possible and ,hence both the methods give inaccurate results.

## Considering the circuit :

In this circuit ammeter measures the true value of the current through the resistance but the voltmeter does not measure the true voltage across the resistance. The voltmeter indicates the sum of the voltages across the ammeter and the measured resistance.

Let Ra be the resistance of the ammeter.

So the voltage across the ammeter = $V_a=IR_a$

Now, measured value of resistance,

$R_{m1} = \frac{V}{I}=\frac{(V_R+V_a)}{I}=\frac{(IR+IR_a)}{I} = R + R_a$

So the value of resistance,

$R = R_{m1 }-R_a=R_{m1 } (1-\frac{R_a}{R_{m1 }})$

Thus the measured value of resistance is higher than the true value. It is also clear from above that true value is equal to the measured only if the ammeter resistance ,R_a is zero.

Relative error,    $\epsilon_r= \frac{(R_{m1}-R)}{R}= \frac{R_a}{R}$

The relative error would be small if the value of resistance under measurement is large as compared to the internal resistance of the ammeter. Therefore, the  circuit in fig.1 is suitable for measuring high resistance values.

## Considering the circuit:

in this circuit the voltmeter measures the true value of voltage but the ammeter measures the sum of currents through the resistance and the voltmeter.

So, the current through the voltmeter,

$I_v=\frac{V}{I_v}$

Measured value of resistance,

$R_{m2 }= \frac{V}{I} =\frac{V}{(IR+IV)} =\frac{V}{(\frac{V}{R}+\frac{V}{R_v})} = \frac{R}{(1+\frac{R}{R_V})}$

True value of resistance,

$R =\frac{R_{m2} Rv}{R_v-{R_{m2}}} =R_{m2}(\frac{1}{1-\frac{R_{m2}}{R_v}})$

It is clear that the true value of resistance is equal to the measured value only if the resistance of voltmeter, R_v, is infinite. However , if the resistance of voltmeter is very large as compared to the resistance under measurement :

Rv >>Rm2 , and therefore, Rm2/Rv is very small.

Thus, we have,

R = Rm2(1 + Rm2/Rv)

Thus the measured value of resistance is smaller than the true value.

Relative error,

$\epsilon _r=\frac{R_{m2}-R}{R_v}=\frac{R_{m2}^2}{R_vR}$

The value of Rm2 is approximately equal to R.

Relative error,

$\epsilon _r=\frac{-R}{R_v}$

The relative error for the two cases are equal when:  Ra/R = R/Rv

Or, when true value of resistance,

$R=\sqrt{R_aR_v}$

Cite this Simulator:

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