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  • To measure a potential difference across a resistor of resistance a vol

To measure a potential difference across a resistor of resistance \mathrm{R \, \Omega} a voltmeter of resistance \mathrm{R_v} is used. To measure the potential with a minimum accuracy of \mathrm{95 \%} then

Option: 1

\mathrm{R_v=5 R}


Option: 2

\mathrm{R_v=15 R}


Option: 3

\mathrm{R_v=10 R}


Option: 4

\mathrm{R_v \geq 19 R}


Answers (1)

best_answer

Initailly Potential difference across R

\mathrm{V_1}=\mathrm{iR}
Now  \mathrm{i_v=\frac{i R}{R+R v}}


The potential difference as measured by the voltmeter
\mathrm{v_2=R_v \times i_v=\frac{i R R_v}{R+R_v} \\ }
\mathrm{\frac{v_2}{v_1} \geq 0.95 }
\mathrm{ \frac{R_v}{R+R_v} \geq 0.95 \Rightarrow R_v \geq 0.95 R v+0.95 R}
\mathrm{ 0.05 R v \geq 0.95 R \Rightarrow R v \geq 19 R }

 
 

Posted by

seema garhwal

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