A moving iron ammeter with a range of 0 to 1 amps has an internal resistance of 50 mΩ and an inductance of 0.1 MH. To increase the range to 0-10 Ampere for all operational frequencies, a shunt coil is connected. The shunt coils resistance in mΩ and time constant in milliseconds are each given as:

Correct option is D

​$$\text{Given:} \\[4pt]\text{Internal resistance of ammeter, } R = 50\,m\Omega \\[4pt]\text{Range of ammeter } = 0\!-\!1\,\text{ampere} \\[4pt]\text{Inductance of ammeter } L = 0.1\,\text{mH} \\[4pt]\text{Desired range: } 0\!-\!10\,\text{ampere} \\[8pt]\textbf{Formula Used:} \\[4pt]\text{The value of the shunt resistance is given by:} \\[4pt]R_{sh} = \dfrac{R\,m}{m - 1} \\[6pt]\text{where } m \text{ is the multiplying factor (range extension factor).} \\[8pt]\text{The time constant } \tau_{sh} \text{ of the shunt is given by:} \\[4pt]\tau_{sh} = \tau_m \\[4pt]\text{Where } \tau_m \text{ is the time constant for the ammeter.} \\[10pt]\textbf{Solution:} \\[4pt]\textbf{Shunt Resistance Calculation:} \\[4pt]\text{Given that the range extension is 10 times, we have:} \\[4pt]m = \dfrac{10}{1} = 10 \\[6pt]\text{The shunt resistance is calculated as:} \\[4pt]R_{sh} = \dfrac{50\,m\Omega}{(10 - 1)} = 5.55\,m\Omega \\[10pt]\textbf{Time Constant Calculation:} \\[4pt]\text{The time constant for the ammeter is:} \\[4pt]\tau_m = \dfrac{L}{R} = \dfrac{0.1\,\text{mH}}{50\,m\Omega} = 2\,\text{seconds} \\[8pt]\text{Thus, the correct values for the shunt resistance and time constant are } R_{sh} = 5.55\,m\Omega \text{ and } \tau_{sh} = 2\,\text{seconds.}$$​​