Question

The potential difference across a resistor r carrying current I is Ir.
(i) Now if the potential difference across r is measured using a voltmeter of resistance R, show that the reading of voltmeter is less than the true value.
(ii) Find out the % error in measuring the potential difference by a voltmeter.
(iii)At what value of R does the voltmeter measures the true potential difference ?

Answer 1

Let a cell of emf ξ and internal resistance r is connected with external voltmeter of resistance is R as shown in figure.

Let i₁ current flow through the resistance R₁.
Now, applying Ohm's law between resistance and voltmeter,
i₁R = (i - i₁)R₁
⇒i₁R = iR₁ - i₁R₁
⇒(R + R₁)i₁ = iR₁
⇒ i₁ = iR₁/(R + R₁) ---------(1)

Now, current through resistor R₁ = (i - i₁) = i - iR₁/(R + R₁) = iR/(R + R₁)

Now, reading of voltmeter , V = (i - i₁)R₁
= iRR₁/(R + R₁) or, iR₁ - iR₁²/(R + R₁) < iR₁
Hence, here it is clear that reading of voltmeter is less than true value.

(ii) % error = reading of voltmeter/true value × 100
= (iRR₁)/(R + R₁)/iR₁ × 100
= R/(R + R₁) × 100

(iii) we can see that, R$\tends\to\infty$
Then, iR₁²/(R + R₁) $\tends\to\,0$
Hence, if resistance of voltmeter is infinite or very very large then voltmeter will measure true value .