Solveeit Logo

Question

Physics Question on Current electricity

A galvanometer coil has a resistance of 10Ω10 \, \Omega and the ammeter shows full scale deflection for a current of 1mA1\,mA. The shunt resistance required to convert the galvanometer into an ammeter of range 0100mA0-100\,mA is about

A

10Ω10 \, \Omega

B

1Ω1 \, \Omega

C

0.1Ω0.1 \, \Omega

D

12.48Ω12.48 \, \Omega

Answer

0.1Ω0.1 \, \Omega

Explanation

Solution

Let shunt resistance be SS
Voltage across shunt is
Now, V=(IIg).SV=\left(I-I_{g}\right) .S
and IgRG=(IIg).SI_{g} \cdot R_{G}=\left(I-I_{g}\right) .S
S=IgRG(IIg)\Rightarrow S=\frac{I_{g} R_{G}}{\left(I-I_{g}\right)}
=103×10(100×1031×103)=\frac{10^{-3} \times 10}{\left(100 \times 10^{-3}-1 \times 10^{-3}\right)}
=1099=0.1Ω=\frac{10}{99}=0.1 \,\Omega