Question

A voltmeter was used to measure the voltage of a cell with an internal resistance of 1.00 k$\Omega$. What must the internal resistance of the meter be if the relative error in the measurement is to be less than (a) -1.0%,

(b) -0.10%?

Short answer

The internal resistance of the voltmeter is 99$\Omega$ and 999$\Omega$ respectively.

Step-by-step solution

Given information

Cell with an internal resistance of $1k\Omega$and a voltmeter connected to it. We have to calculate the internal resistance of the voltmeter.

Part (a)

The relative error is -1.0%.

The expression for relative error in voltage is given by:

$E=-\frac{R}{R+R_v}\times 100$

here R_v is the internal resistance of the voltmeter.

$$\begin{gathered} -1.0=-\frac{1}{1+R_v}\times 100 \\ 1+R_v=100\Omega \\ {R}_v=99\Omega \end{gathered}$$

Part (b)

The relative error is -0.10%.

The expression for relative error in voltage is given by:

$E=-\frac{R}{R+R_v}\times 100$

here R_v is the internal resistance of the voltmeter.

$$\begin{gathered} -0.10=-\frac{1}{1+R_v}\times 100 \\ 1+R_v=1000\Omega \\ {R}_v=999\Omega \end{gathered}$$