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 28, 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 Topic:
This problem relates to "Magnetic Effects of Current" and "Current Electricity," specifically the conversion of measuring instruments. A galvanometer is a sensitive device that can only handle tiny currents. To measure much larger currents (converting it to an ammeter), we must provide a bypass path (shunt) so that only a tiny fraction of the total current passes through the galvanometer coil.
Step 2: Key Formulas and Approach:
In a converted ammeter, the galvanometer ($G$) and the shunt ($S$) are in parallel. They share the same voltage.
$I_g \cdot G = (I - I_g) \cdot S$.
Shunt Resistance formula: $S = \frac{I_g \cdot G}{I - I_g}$.
Step 3: Detailed Explanation:
Identify given values: Galvanometer resistance $G = 100 \Omega$. Full scale current $I_g = 1 \text{ mA} = 10^{-3} \text{ A}$. Desired total current range $I = 10 \text{ A}$.
Substitute into the formula: \[ S = \frac{10^{-3} \times 100}{10 - 10^{-3}} \]
Simplify: The numerator is $10^{-1} = 0.1$.
The denominator is $10 - 0.001 = 9.999 \text{ A}$.
Calculate: Since $I_g$ is extremely small compared to $I$, we can approximate $I - I_g \approx I$ for the final division. \[ S \approx \frac{0.1}{10} = 0.01 \Omega \]
This very low resistance shunt ensures that for every 10 Amps entering the ammeter, 9.999 Amps go through the shunt and only 0.001 Amps go through the galvanometer.
Step 4: Final Answer:
The required shunt resistance is 0.01 $\Omega$.
This problem relates to "Magnetic Effects of Current" and "Current Electricity," specifically the conversion of measuring instruments. A galvanometer is a sensitive device that can only handle tiny currents. To measure much larger currents (converting it to an ammeter), we must provide a bypass path (shunt) so that only a tiny fraction of the total current passes through the galvanometer coil.
Step 2: Key Formulas and Approach:
In a converted ammeter, the galvanometer ($G$) and the shunt ($S$) are in parallel. They share the same voltage.
$I_g \cdot G = (I - I_g) \cdot S$.
Shunt Resistance formula: $S = \frac{I_g \cdot G}{I - I_g}$.
Step 3: Detailed Explanation:
Identify given values: Galvanometer resistance $G = 100 \Omega$. Full scale current $I_g = 1 \text{ mA} = 10^{-3} \text{ A}$. Desired total current range $I = 10 \text{ A}$.
Substitute into the formula: \[ S = \frac{10^{-3} \times 100}{10 - 10^{-3}} \]
Simplify: The numerator is $10^{-1} = 0.1$.
The denominator is $10 - 0.001 = 9.999 \text{ A}$.
Calculate: Since $I_g$ is extremely small compared to $I$, we can approximate $I - I_g \approx I$ for the final division. \[ S \approx \frac{0.1}{10} = 0.01 \Omega \]
This very low resistance shunt ensures that for every 10 Amps entering the ammeter, 9.999 Amps go through the shunt and only 0.001 Amps go through the galvanometer.
Step 4: Final Answer:
The required shunt resistance is 0.01 $\Omega$.
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