A moving coil galvanometer has a resistance of 50Ω and gives full scale deflection for 10 mA. How could it be converted into an ammeter with a full scale deflection for 1 A?

Question

A moving coil galvanometer has a resistance of 50Ω and gives full scale deflection for 10 mA. How could it be converted into an ammeter with a full scale deflection for 1 A? (A)  50/99 Ω in series (B)  50/99 Ω in parallel (C)  0.01 Ω in series (D)  0.01 Ω in parallel

Tardigrade Admin · Accepted Answer

S=(ig× G/i-ig)=(10× 10-3× 50/1-10-3× 10) =(50/99)Ω in parallel

Q. A moving coil galvanometer has a resistance of $50Ω$ and gives full scale deflection for 10 mA. How could it be converted into an ammeter with a full scale deflection for 1 A?

$50/99 Ω$ in series

$50/99 Ω$ in parallel

$0.01 Ω$ in series

$0.01 Ω$ in parallel

$S = \frac{i _{g} \times G}{i - i _{g}} = \frac{10 \times 10 ^{- 3} \times 50}{1 - 10 ^{- 3} \times 10}$ $= \frac{50}{99} Ω$ in parallel

Q. A moving coil galvanometer has a resistance of 50Ω and gives full scale deflection for 10 mA. How could it be converted into an ammeter with a full scale deflection for 1 A?

50/99Ω in series

50/99Ω in parallel

0.01Ω in series

0.01Ω in parallel