An ammeter having internal resistance of \( 0.81\,\Omega \) can read upto 1 A. To convert it into an ammeter reading 0 - 10 A we have to connect:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → a \( 0.09\,\Omega \) resistor in parallel
To extend the range of an ammeter a low resistance, the shunt S, is connected in parallel with it. The shunt takes the extra current, and the ammeter and the shunt have the same potential difference across them.
\( I_g = 1\,\mathrm{A},\quad G = 0.81\,\Omega,\quad I = 10\,\mathrm{A} \)
\( I_g G = (I - I_g)\,S \)
\( \displaystyle S = \frac{I_g G}{I - I_g} = \frac{1\times0.81}{10-1} = 0.09\,\Omega \)