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Step-by-Step Solution
Step 1: Identify the Known Quantities
The galvanometer has a resistance of $50\,\Omega$ and can safely carry a maximum current of $0.002\,\text{A}$ (2 mA). We want to convert this galvanometer into an ammeter that can measure up to $0.5\,\text{A}$.
Step 2: Understand the Need for a Shunt Resistor
When converting a galvanometer into an ammeter, we place a small resistance (called a "shunt") in parallel with the galvanometer. Most of the current flows through this shunt resistor, protecting the galvanometer from excess current.
Step 3: Apply the Formula for the Shunt Resistor
Let $R_g$ be the galvanometer resistance, $I_g$ be its maximum current, and $R_s$ be the shunt resistance. If the total desired current range is $I$, the current through the shunt $I_s$ will be $I - I_g$. The voltage across the galvanometer and the shunt must be the same since they are in parallel, hence:
$$
I_g R_g = I_s R_s.
$$
Since $I_s = I - I_g$, we can write:
$$
R_s = \frac{I_g \, R_g}{I - I_g}.
$$
Step 4: Substitute the Given Values
Given:
$R_g = 50\,\Omega$,
$I_g = 0.002\,\text{A}$,
$I = 0.5\,\text{A}.$
Substitute these into the formula:
$$
R_s = \frac{0.002 \times 50}{0.5 - 0.002} \, \Omega = \frac{0.1}{0.498} \, \Omega.
$$
Step 5: Calculate the Numerical Value
$$
R_s \approx 0.2008\,\Omega \approx 0.2\,\Omega.
$$
This is the required shunt resistance that must be placed in parallel with the galvanometer.
Step 6: Final Answer
Therefore, a resistor of approximately $0.2\,\Omega$ should be connected in parallel with the galvanometer to convert it into an ammeter of range $0\,\text{A}$ to $0.5\,\text{A}$.
