Question

The potential difference across the 100 ohm resistance in the following circuit is measured by a voltmeter of 900ohm resistance.
The percent error in reading the potential difference is

Answer 1

Answer:

The percent error in reading the potential difference is 1 % .

Explanation:

Let , source voltage be V .

Therefore , voltage across $100\ \Omega$ before connecting voltmeter is :

$V_1=\dfrac{100}{100+10}V\\\\V_1=\dfrac{100}{110}V$ ( By voltage divider rule )

Now , voltage across $100\ \Omega$ after connecting voltmeter is :

$V_2=\dfrac{R_{eq}}{R_{eq}+10}V$       ........ ( 1 )

Here ,

$\dfrac{1}{R_{eq}}=\dfrac{1}{100}+\dfrac{1}{900}\\\\R_{eq}=\dfrac{900\times 100}{1000}\ \Omega\\\\R_{eq}=90\ \Omega$

Putting value of $R_{eq}$ in equation 1 .

We get :

$V_2=\dfrac{90}{100}V\\\\V_2=0.9V$

Therefore , percentage error is :

$Percentage\ Error=\dfrac{V_1-V_2}{V_1}\times 100\\\\\dfrac{\dfrac{10}{11}V-0.9V}{\dfrac{10}{11}V}\times 100\\\\\dfrac{\dfrac{10}{11}-0.9}{\dfrac{10}{11}}\times 100=1\ \%$

Therefore , the percent error in reading the potential difference is 1 % .