Question

A galvanometer of resistance 50 Ω is converted into ammeter by connecting a low resistance (shunt) of value 1 Ω in parallel to the galvanometer S. If full scale deflection current of galvanometer is 10 mA, then the maximum current to be measured by the ammeter is –

• A. 500 Ma
• B. 510 mA
• C. 610 mA
• D. 490 mA

Answer 1

Answer: (B)

510 mA

Answer 2

$G\frac{I_{g}}{I–I_{g}}$

l = $50 \times \frac{10}{I–10}$

I – 1 0 = 500

I = 510 mA