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A moving coil galvanometer of resistance $100 \Omega$ is used as an ammeter using a resistance $0.1 \Omega$. The maximum deflection current in the galvanometer is $100 \mu \mathrm{A}$. Find the minimum current in the circuit, so that ammeter shows maximum deflection?
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Verified Answer
The correct answer is:
100.1 mA
We are given with following ammeter
Now,
$\begin{aligned}
& V_{A B}=I_g G=\left(I-I_g\right) S \\
& \quad\left(100 \times 10^{-6}\right) 100=\left(I-100 \times 10^{-6}\right) 0.1 \\
& \quad I-100 \times 10^{-6}=10^{-1} \\
& I=0.1+100 \times 10^{-6} \\
& \quad=0.1+0.0001=0.1001 \mathrm{~A}=100.1 \mathrm{~mA}
\end{aligned}$
So, circuit current is $100.1 \mathrm{~mA}$.
Now,
$\begin{aligned}
& V_{A B}=I_g G=\left(I-I_g\right) S \\
& \quad\left(100 \times 10^{-6}\right) 100=\left(I-100 \times 10^{-6}\right) 0.1 \\
& \quad I-100 \times 10^{-6}=10^{-1} \\
& I=0.1+100 \times 10^{-6} \\
& \quad=0.1+0.0001=0.1001 \mathrm{~A}=100.1 \mathrm{~mA}
\end{aligned}$
So, circuit current is $100.1 \mathrm{~mA}$.
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