QJEEM24Ph5758: JEE Main 2024 Physics - Class 12 JEEM24Ph5758 Solution: A galvanometer having coil resistance shows a full scale…

To determine the value of the shunt resistor required to convert the galvanometer into an ammeter capable of measuring a current of $8 \mathrm{~A}$ , we need to use the concept of shunting where the additional resistor (shunt) is placed in parallel with the galvanometer.

The formula for calculating the value of the shunt resistor $R_s$ is given by:

$R_s = \frac{R_g \cdot I_g}{I - I_g}$

Where:

$R_g$ = resistance of the galvanometer = $10 \Omega$

$I_g$ = full-scale deflection current of the galvanometer = $3 \mathrm{~mA}$ = $0.003 \mathrm{~A}$

$I$ = total current to be measured = $8 \mathrm{~A}$

Substituting these values into the formula:

$R_s = \frac{10 \Omega \cdot 0.003 \mathrm{~A}}{8 \mathrm{~A} - 0.003 \mathrm{~A}}$

Perform the calculations:

$R_s = \frac{0.03 \Omega \cdot \mathrm{~A}}{7.997 \mathrm{~A}}$

$R_s \approx 3.75 \times 10^{-3} \Omega$

Thus, the value of the shunt resistor should be:

Option A: $3.75 \times 10^{-3} \Omega$