The potential difference across the `100 Omega` resistance in the following circuit is measured by a voltmeter of `900 Omega` resistance. The percentage error made in reading the potential difference is

Options

A. `10//9`
B. 0.1
C. 1
D. 10

Answer

3

Explanation

Error is caused because voltmeter draws a current and reduces the voltage to be measured between the two points across a resistance, Consider the case when voltmeter is not connected.
EMF of cell=E volt.
Resistance of circuit `=10+100=110Omega`
`therefore` Current in the circuit (I) `=(E)/(110)` amp
`therefore` Actual potential difference across `100Omega`
`=(E)/(110)xx100`

`therefore V=(10E)/(11)` volt
when voltmeter is connected across 100 `Omega`
Resistance of the parallel combination `=(100xx900)/(100+900)`
`R_(P)=(100xx900)/(1000)=90Omega`
Total resistance of circuit `=(90+10)=100Omega`
`therefore` c urrent in the circuit `I_(1)=(E)/(100)`
`therefore` Potential difference across `R_(P)`
`=I_(1)x xR_(P)=(E)/(100)xx90`
`therefore V_(1)=(9E)/(10)` volt
This is read by the voltmeter

`therefore` Error`=(v-V_(1))` volt.
`therefore` percentage error `=((V-V_(1))/(V))x x100`
`=(1-(V_(1))/(V))xx100=(1-(9E)/(10)xx(11)/(10E))xx100`
`=(1-(99)/(100))100=(1)/(100)xx100=1.0`.