Question:medium
A galvanometer of resistance 100 $\Omega$ gives full scale deflection for a current of 1 mA. It is converted into an ammeter of range 0 – 10 A. The shunt required is: ____.
A galvanometer of resistance 100 $\Omega$ gives full scale deflection for a current of 1 mA. It is converted into an ammeter of range 0 – 10 A. The shunt required is: ____.
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The shunt resistance is always much smaller than the galvanometer resistance. If your calculated $S$ is larger than $G$, you have likely swapped your current values!
Updated On: May 4, 2026
- 0.10 $\Omega$
- 0.001 $\Omega$
- 1.0 $\Omega$
- 0.01 $\Omega$
Show Solution
The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
To convert a galvanometer into an ammeter, a very low resistance called a "shunt" ($S$) is connected in parallel with the galvanometer. This allows most of the current to bypass the delicate galvanometer coil.
Step 2: Key Formula or Approach:
\[ S = \frac{I_g \cdot G}{I - I_g} \] Where: - $G$ = Galvanometer resistance - $I_g$ = Full scale deflection current - $I$ = Desired ammeter range
Step 3: Detailed Explanation:
Given: $G = 100\,\Omega$, $I_g = 1\,\text{mA} = 0.001\,\text{A}$, $I = 10\,\text{A}$. 1. Since $I_g$ is very small compared to $I$, we can approximate $I - I_g \approx I$: \[ S = \frac{0.001 \times 100}{10 - 0.001} \approx \frac{0.1}{10} \] 2. Calculate the shunt resistance: \[ S = 0.01\,\Omega \]
Step 4: Final Answer:
The required shunt resistance is 0.01 $\Omega$.
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