Step 1: Understanding the Concept:
A galvanometer is an instrument used to detect very small currents. Its internal coil has a maximum current limit called the full-scale deflection current (\(I_g\)).
To convert it into an ammeter that can measure much larger currents (\(I\)), we must bypass most of the current away from the delicate coil.
This is achieved by connecting a very low resistance called a "shunt" (\(S\)) in parallel with the galvanometer.
In this parallel configuration, the potential difference across the galvanometer and the shunt must be the same.
Step 2: Key Formula or Approach:
The relationship for the shunt resistance based on parallel potential equality is:
\[ V_g = V_s \implies I_g G = (I - I_g) S \]
Rearranging to solve for the unknown shunt resistance \(S\):
\[ S = \frac{I_g G}{I - I_g} \]
Where:
\(G = 50\Omega\) (Galvanometer resistance).
\(I_g = 2\text{mA} = 0.002\text{A}\) (Full-scale current).
\(I = 2\text{A}\) (Maximum desired current).
Step 3: Detailed Explanation:
First, ensure all given values are in standard SI units:
- \(G = 50\Omega\)
- \(I_g = 2 \times 10^{-3}\text{A} = 0.002\text{A}\)
- \(I = 2\text{A}\)
Substitute these into the isolated shunt formula:
\[ S = \frac{0.002 \times 50}{2 - 0.002} \]
Numerator calculation: \(0.002 \times 50 = 0.1\text{V}\).
Denominator calculation: \(2 - 0.002 = 1.998\text{A}\).
Now, perform the division:
\[ S = \frac{0.1}{1.998} \approx 0.05005\Omega \]
In competitive exams, since \(I_g\) is extremely small compared to \(I\) (\(0.002\text{A} \ll 2\text{A}\)), we can use the approximation \(I - I_g \approx I\):
\[ S \approx \frac{0.1}{2} = 0.05\Omega \]
The calculated exact value confirms this approximation rounds precisely to \(0.05\Omega\).
Step 4: Final Answer:
The value of the shunt resistance required is \(0.05\Omega\).