A millivoltmeter with a range of $10 \mathrm{~mV}$ is to be fundamentally converted into an ammeter

Question

A millivoltmeter with a range of $10 \mathrm{~mV}$ is to be fundamentally converted into an ammeter capable of measuring up to $20 \mathrm{~A}$. The value (in ohms) of the necessary shunt required will be approximately

Professor PC Thomas Classes · Answer

(d) : The maximum voltage the meter can handle is $V_g = 10 \mathrm{~mV} = 10 	imes 10^{-3} \mathrm{~V}$. When converted to an ammeter of range $I = 20 \mathrm{~A}$, the shunt $S$ is connected in parallel. Since the shunt resistance is extremely small compared to the galvanometer resistance, almost all the current $I$ passes through the shunt. Therefore, the voltage drop across the parallel combination is roughly the voltage drop across the shunt: $V_g \approx I 	imes S$. Thus, $S = \frac{V_g}{I} = \frac{10 	imes 10^{-3}}{20} = 0.5 	imes 10^{-3} = 0.0005 \Omega$.