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_"actual"=I*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'=(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%#