# Question ab54a

##### 1 Answer
Jul 24, 2017

Let $I$ be the actual magnitude of current passing through the resistor $r$ when the voltmeter is not connected across it.

So actual voltage across the resistor will be
${V}_{\text{actual}} = I \cdot r$

When a voltmeter of resistance $R$ (which is > > $r$) is attached across it in parallel combination,main current is divided in two paths and the current through the resistor drops.

Then the changed current through the resistor is
$I ' = \frac{R \cdot I}{r + R}$

So the changed potential across the resistor will be

${V}_{\text{changed}} = r \cdot I ' = \frac{r \cdot R \cdot I}{r + R}$

So change in the reading of voltmeter will be

$\Delta V = {V}_{\text{actual"-V_"changed}}$

$= I \cdot r - \frac{r \cdot R \cdot I}{r + R}$

$= \frac{I \left({r}^{2} + r \cdot R - r \cdot R\right)}{r + R}$

$= \frac{I \cdot {r}^{2}}{r + R}$

Hence percentage of error in measuring the voltage across $r$ will be

=(DeltaV)/V_"actual"xx100%

= ((I*r^2)/(r+R))/(I*r)xx100%

= (r/(r+R))xx100%#