Question

A galvanometer with resistance $100\Omega$ is converted to ammeter with a resistance of $0.1\Omega .$ The galvanometer shows full scale deflection with a current of $100\mu A$. Then the minimum current in the circuit for full scale deflection of galvanometer will be

• A. $100.1mA$
• B. $10.01mA$
• C. $1.001mA$
• D. $0.1001mA$

Answer 1

The correct option is A $100.1mA$

Given : $I_g=100\mu$ A $={10}^{-4}$ A $R_g=100\Omega$ $S=0.1\Omega$

Let the current in the circuit be $I$.

Using $(I-I_g)S=I_g{R}_g$ $⟹$ $I=\frac{(R_g+S)I_g}{S}$

$\therefore$ $⟹$ $I=\frac{(100+0.1)\times {10}^{-4}}{0.1}=100.1$ mA