Question

An ammeter has a current range of 0–5 A, and its internal resistance is 0.2 Ω. In order to change the range to 0–25 A, we need to add a resistance of ________.

• A. 0.8 Ω in series with the meter
• B. 0.05 Ω in parallel with the meter
• C. 1.0 Ω in series with the meter
• D. 0.04 Ω in parallel with the meter

Hint:

To extend ammeter range → use low-value shunt in parallel. Use current division to find value!

Answer 1

The Correct Option is B

Step 1: Concept of Shunt Resistance 
To extend the range of an ammeter, a low-value resistor (called a shunt) is connected in parallel so that only part of the current flows through the meter.
Step 2: Given Data 
- Meter current range \( I_m = 5\,\text{A} \)
- Total required range \( I = 25\,\text{A} \)
- Internal resistance of meter \( R_m = 0.2\,\Omega \) 
Step 3: Apply Current Division Principle 
Let the shunt resistance be \( R_s \). The current through shunt = \( I - I_m = 20\,\text{A} \) The voltage across the meter and the shunt must be equal:
\[ I_m R_m = (I - I_m) R_s \Rightarrow 5 \times 0.2 = 20 \times R_s \Rightarrow 1 = 20 R_s \Rightarrow R_s = 0.05\,\Omega \] Conclusion: 
We must connect a \( 0.05\,\Omega \) resistor in parallel to the ammeter. So, Option (2) is correct.