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(A) The resistance of the ammeter is $G = 20 \Omega$ and the full-scale deflection current is $i_g = 1 	ext{ mA} = 10^{-3} 	ext{ A}$.Four resistors

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$6$

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(A) The resistance of the ammeter is $G = 20 \Omega$ and the full-scale deflection current is $i_g = 1 	ext{ mA} = 10^{-3} 	ext{ A}$.Four resistors of $16 \Omega$ each are connected in parallel. The equivalent shunt resistance $S$ is given by $\frac{1}{S} = \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} = \frac{4}{16} = \frac{1}{4} \Omega^{-1}$.Thus,$S = 4 \Omega$.For an ammeter,the shunt resistance $S$ is connected in parallel to the galvanometer $G$. The current $i$ to be measured is divided into $i_g$ through the galvanometer and $i_s$ through the shunt.The voltage across the parallel combination is $i_s S = i_g G$.$(i - i_g) S = i_g G$.$(i - 10^{-3}) 	imes 4 = 10^{-3} 	imes 20$.$i - 10^{-3} = \frac{20 	imes 10^{-3}}{4} = 5 	imes 10^{-3}$.$i = 5 	imes 10^{-3} + 1 	imes 10^{-3} = 6 	imes 10^{-3} 	ext{ A} = 6 	ext{ mA}$.

An ammeter of resistance $20 \Omega$ gives full scale deflection when $1 \text{ mA}$ current flows through it. What is the maximum current that can be measured by connecting $4$ resistors each of $16 \Omega$ in parallel with the meter (in $\text{ mA}$)?

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