Question

We need to measure emf of a cell whose internal resistance is $2\Omega$. The percentage error in reading of voltmeter will be, (resistance of voltmeter is $98\Omega )$

• A. 4%
• B. 2%
• C. 3%
• D. 1%

Answer 1

The correct option is B 2%

Let , $ϵ$ represent emf of cell,
$i=\frac{ϵ}{R+r}$
Reading of voltmeter = iR volt
Percentage error $=\frac{ϵ-iR}{ϵ}\times 100$
$=\frac{ϵ-\frac{ϵ}{R+r}.R_v}{ϵ}\times 100=\frac{r}{R+r}\times 100$
$=\frac{2}{\left(}\times 100=2\%$