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Physics Question on Current Electricity

A galvanometer with resistance 100 Ω gives full-scale deflection with a current of 2 mA. The resistance required to convert the galvanometer into an ammeter of range 0 to 20 A is nearly:

A

10−2^{−2} Ω in series

B

10−2^{−2} Ω in parallel

C

10−1^{−1} Ω in parallel

D

10−1^{−1} Ω in series

Answer

10−2^{−2} Ω in parallel

Explanation

Solution

\text{ To convert a galvanometer into an ammeter, a shunt resistance is connected in parallel with the galvanometer. The shunt resistance } R_s \text{ is calculated using the formula:}$$R_s = \frac{I_g R_g}{I - I_g}
Where: Ig=2 mA=0.002 A, Rg=100 Ω, I=20 A. Substituting the values into the formula:\text{Where: } I_g = 2 \, \text{mA} = 0.002 \, \text{A}, \, R_g = 100 \, \Omega, \, I = 20 \, \text{A}. \text{ Substituting the values into the formula:}
Rs=0.002×10020−0.002=0.219.998≈0.01 ΩR_s = \frac{0.002 \times 100}{20 - 0.002} = \frac{0.2}{19.998} \approx 0.01 \, \Omega
Thus, the required shunt resistance is approximately 0.01 Ω, which corresponds to option 10−2 Ω in parallel.\text{Thus, the required shunt resistance is approximately } 0.01 \, \Omega, \text{ which corresponds to option } 10^{-2} \, \Omega \text{ in parallel.}