A galvanometer with resistance 100 Ohm gives full scale deflection with a current of 2 mA. The resistance required to convert galvanometer into ammeter of range 0 to 20 A is nearly:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $10^{-2}$ Ohm in parallel.
To convert a galvanometer into an ammeter, we need to connect shunt resistance $(R_s)$ in parallel with galvanometer $(R_g)$
Now,
V = Voltage across galvanometer
$=I_g×R_g=0.002×100=0.2V$
Using Ohm's law,
$V=I_s×R_s$
$⇒R_s=\frac{V}{I_s}=\frac{0.2}{19.998}≃0.01Ω$