Question

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 resistance 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, which of the following two connections would you choose for resistance measurement? Justify you answer quantitatively

Answer 1

Let a resistance have true value R measured value $\frac{V}{I}=R+R^{\prime }$

Percentage error $=\frac{△R}{R}\times 100$

Percentage error $=\frac{R+R^{\prime }-R}{R}\times 100$

$=\frac{R^{\prime }\times 100}{R}$

$=\frac{0.5\times 100}{R}$

$=\frac{50}{R}$

So the more value of R less will be percentage errors and vice versa so this type of circuit should be used for y because it has more resistance .

Now for circuit (b)

True value = R

Measured value $\frac{V}{I}=\frac{V}{I_R+I_V}$

$=\frac{V}{\frac{V}{R}+\frac{V}{R_V}}=\frac{R{R}_V}{R+R_V}$

Percentage error $=\frac{R-\frac{R{R}_V}{R+R_V}}{R}\times 100\frac{1}{1+\frac{R_V}{R}}\times 100$

So lesser the value of R ,lesser the percentage errors