A galvanometer of resistance $100 \Omega$ is converted to a voltmeter of range $10 \mathrm{~V}$ by…

A galvanometer of resistance $100 \Omega$ is converted to a voltmeter of range $10 \mathrm{~V}$ by connecting a resistance of $10 \mathrm{k} \Omega$. The resistance required to convert the same galvanometer to an ammeter of range $1 \mathrm{~A}$ is
  1. $0.4 \Omega$
  2. $0.3 \Omega$
  3. $1.2 \Omega$
  4. $0.1 \Omega$

Solution

Here, $V=10 \mathrm{~V}$, $\begin{aligned} & R=10 \mathrm{k} \Omega, \\ & G=100 \Omega=0.1 \mathrm{k} \Omega \\ & V=I_g(G+R) ; I_g \approx 1 \mathrm{~mA}\end{aligned}$ $\begin{aligned} & \left(I-I_g\right) S=I_g G \\ & \text { or } \quad S=\frac{I_g G}{I-I_g}\end{aligned}$ $=\frac{1 \times 10^{-3} \times 100}{1-1 \times 10^{-3}}=0.1 \Omega$

Asked in: NEET 2020 (Phase 1)

Practice more Current Electricity questions on Aicharya