Question

The emf of a cell of internal resistance 2 Ω is measured using a voltmeter of resistance 998 Ω. The error in the emf measured is:

• A. \( 0.4\% \)
• B. \( 4\% \)
• C. \( 2\% \)
• D. \( 0.2\% \)

Hint:

A voltmeter with high resistance minimizes measurement errors in emf. The error is proportional to the ratio of internal resistance to total circuit resistance.

Answer 1

The Correct Option is D

Step 1: Define the Voltage Measurement Formula The measured voltage across a voltmeter is: \[ V = E \times \frac{R_v}{R_v + r} \] where: 
- \( E \) is the actual emf, 
- \( R_v = 998 \) Ω (voltmeter resistance), 
- \( r = 2 \) Ω (internal resistance). 
Step 2: Compute the Voltage Drop \[ V = E \times \frac{998}{998+2} = E \times \frac{998}{1000} = 0.998E \] 
Step 3: Compute the Percentage Error \[ \text{Error} = \left( 1 - \frac{V}{E} \right) \times 100 = (1 - 0.998) \times 100 = 0.2\% \]