Question

$A100\mu A$ galvanometer with internal resistance $1000\Omega$ is to be converted into an ammeter of 10 A. Find the shunt to be connected

• A. $0.1\Omega$
• B. $1\Omega$
• C. $0.01\Omega$
• D. $0.001\Omega$

Answer 1

Answer: (C)

$0.01\Omega$

To convert galvanometer into an ammeter a small resistance(shunt) is to be connected in parallel with it. Consider the circuit shown in figure.
Since, galvanometer and shunt are parallel to each other voltage across them is same.
Hence, $(I-I_g)S=I_{g}R$
$\Rightarrow S=\frac{I_{g}R}{I-I_g}$
$S=\frac{100\times {10}^{-6}\times 1000}{10-100\times {10}^{-6}}$

$S=\frac{{10}^{-1}}{10-0.0001}$

$S=\frac{0.1}{9.9999}$

$S=0.01\Omega$