Question

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$^{−2}$ Ω in series
• B. 10$^{−2}$ Ω in parallel
• C. 10$^{−1}$ Ω in parallel
• D. 10$^{−1}$ Ω in series

Answer 1

The Correct Option is B

Given:

• Galvanometer resistance (G) = 100 Ω
• Full-scale deflection current (I_g) = 2 mA = 0.002 A
• Desired ammeter range (I) = 20 A

To convert a galvanometer into an ammeter, we connect a shunt resistance (S) in parallel.

The shunt resistance is calculated by:

I_gG = (I - I_g)S

S = (I_gG)/(I - I_g)

Substituting the values:

S = (0.002 × 100)/(20 - 0.002) ≈ 0.2/19.998 ≈ 0.01 Ω

The required shunt resistance is approximately 10^-2 Ω connected in parallel.

Answer: 10^-2 Ω in parallel

Answer 2

\(\text{ To convert a galvanometer into an ammeter, a shunt resistance is connected in parallel with the galvanometer.}\)

\(\text{The shunt resistance } R_s \text{ is calculated using the formula:}\)

\(R_s = \frac{I_g R_g}{I - I_g}\) 

\(\text{Where: } I_g = 2 \, \text{mA} = 0.002 \, \text{A}, \, R_g = 100 \, \Omega, \, I = 20 \, \text{A}. \text{ Substituting the values into the formula:}\) 

\(R_s = \frac{0.002 \times 100}{20 - 0.002} = \frac{0.2}{19.998} \approx 0.01 \, \Omega\) 

\(\text{Thus, the required shunt resistance is approximately } 0.01 \, \Omega, \text{ which corresponds to option } 10^{-2} \, \Omega \text{ in parallel.}\)