Let II be the actual magnitude of current passing through the resistor rr when the voltmeter is not connected across it.
So actual voltage across the resistor will be
V_"actual"=I*rVactual=I⋅r
When a voltmeter of resistance RR (which is > > rr) 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'=(R*I)/(r+R)
So the changed potential across the resistor will be
V_"changed"=r*I'=(r*R*I)/(r+R)
So change in the reading of voltmeter will be
DeltaV=V_"actual"-V_"changed"
=I*r-(r*R*I)/(r+R)
=(I(r^2+r*R-r*R))/(r+R)
=(I*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%
