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Current Electricity — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsCurrent Electricity3 Marks
Question
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?
✓

Answer

  1. V = lr (without voltmeter RV)

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

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

  1. Percentage error,

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

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

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