Current Electricity • NEET Revision

Example 3.36

Real Voltmeter Error

Question

A voltmeter of resistance 994 Ω is connected across a cell of emf 1 V and internal resistance 6 Ω. Find the potential difference across the voltmeter, that across the terminals of the cell, and the percentage error in the reading of the voltmeter.

1 Vr = 6 ΩVRₖ = 994 Ω

Given

Asked

(i) PD across voltmeter, (ii) PD across cell terminals, (iii) % error in voltmeter reading.

Formula

\(I = \dfrac{E}{R_V + r}\)
\(V = I R_V = E - Ir\)
\(\%\ \text{error} = \dfrac{E - V}{E} \times 100\)

Solution

  1. Current drawn by the voltmeter: \(I = \dfrac{1}{994 + 6} = \dfrac{1}{1000} = 10^{-3}\ \text{A}\).
  2. PD across voltmeter: \(V = I R_V = 10^{-3} \times 994 = 0.994\ \text{V}\).
  3. PD across cell terminals: \(V = E - Ir = 1 - 10^{-3}\times 6 = 0.994\ \text{V}\). (Same value — the voltmeter is connected directly across the terminals, so both PDs are across the same two points.)
  4. An ideal voltmeter should read the emf \(1\ \text{V}\). Error \(= 1 - 0.994 = 0.006\ \text{V}\).
  5. \(\%\ \text{error} = \dfrac{0.006}{1}\times 100 = 0.6\%\).
\(V_{\text{voltmeter}} = V_{\text{terminals}} = 0.994\ \text{V};\;\) error \(= 0.6\%\)

Short Cut

Fraction of emf lost inside the cell \(= \dfrac{r}{R_V + r}\). So \(\%\ \text{error} = \dfrac{r}{R_V+r}\times 100 = \dfrac{6}{1000}\times 100 = 0.6\%\) — no need to compute \(I\) or \(V\) at all.