The potential difference across a resistor $‘r\ ’$ carrying current $‘I\ ’$ is $Ir.$
- Now if the potential difference across $‘r\ ’$ is measured using a voltmeter of resistance $‘RV\ ’,$ show that the reading of voltmeter is less than the true value
- Find the percentage error in measuring the potential difference by a voltmeter.
- At what value of $RV,$ does the voltmeter measures the true potential difference?
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- $V = lr ($without voltmeter $R_V)$
- $\Big(\frac{\text{V}-\text{V}'}{\text{V}}\Big)\times100=\Big(\frac{\text{r}}{\text{r}+\text{R}_\text{V}}\Big)\times100$
- $\text{R}_\text{V}\xrightarrow{\ \ \ \ \ \ }\infty,\text{V}'=\text{lr}=\text{V}$
- $\text{V}'=\frac{\text{lrR}_\text{V}}{\text{r}+\text{R}_\text{V}}=\frac{\text{Ir}}{1+\frac{\text{r}}{\text{R}_\text{V}}}$
$\text{V}'<\text{V}$ - Percentage error,
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