To convert a galvanometer into an ammeter, a shunt resistor is connected in parallel. This shunt allows the majority of the current (20 A) to bypass the galvanometer, enabling it to measure higher currents without damage.
First, the full-scale voltage across the galvanometer, \( V_g \), is calculated using Ohm's Law:
\( V_g = I_g \times R_g \)
With \( I_g = 2 \, \text{mA} = 0.002 \, \text{A} \) and \( R_g = 100 \, \Omega \), the voltage is:
\( V_g = 0.002 \, \text{A} \times 100 \, \Omega = 0.2 \, \text{V} \)
Next, the shunt resistance, \( R_s \), is determined using the formula:
\( R_s = \frac{V_g}{I - I_g} \)
Where \( I = 20 \, \text{A} \) is the total current. The calculation for \( R_s \) is:
\( R_s = \frac{0.2 \, \text{V}}{20 \, \text{A} - 0.002 \, \text{A}} \approx \frac{0.2 \, \text{V}}{19.998 \, \text{A}} \approx 0.01 \, \Omega \)
Therefore, a resistance of approximately \( 10^{-2} \, \Omega \) connected in parallel is required for the desired range. The correct answer is:
10$^{-2}$ Ω in parallel