The potential difference across a resistor ‘r’ carrying current ‘I’ is Ir. 1. 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 2. Find the percentage error in measuring the potential difference by a voltmeter. 3. At what value of RV, does the voltmeter measures the true potential difference?

1. V = lr (without voltmeter R_V) V'= lrR_V r+R_V = Ir 1+ r R_V V'<V 2. Percentage error, ( V-V' V ) 100= ( r r+R_V ) 100 3. R_V ,V'=lr=V

The potential difference across a resistor ‘r’ carrying current ‘…

Question

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?

✓

$\text{V}'=\frac{\text{lrR}_\text{V}}{\text{r}+\text{R}_\text{V}}=\frac{\text{Ir}}{1+\frac{\text{r}}{\text{R}_\text{V}}}$

$\text{V}'<\text{V}$

$\Big(\frac{\text{V}-\text{V}'}{\text{V}}\Big)\times100=\Big(\frac{\text{r}}{\text{r}+\text{R}_\text{V}}\Big)\times100$

$\text{R}_\text{V}\xrightarrow{\ \ \ \ \ \ }\infty,\text{V}'=\text{lr}=\text{V}$

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