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: |
$10^{-2}$ Ohm in series. $10^{-2}$ Ohm in parallel. $10^{-1}$ Ohm in parallel. $10^{-1}$ Ohm in series. |
$10^{-2}$ Ohm in parallel. |
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, $V=I_s×R_s$ $⇒R_s=\frac{V}{I_s}=\frac{0.2}{19.998}≃0.01Ω$ |