An ammeter having internal resistance of $0.81 \Omega$ can read upto 1. 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
$R_A$, Internal Resistance of ammeter = 0.81 Ω
$I_A$, Maximum current for ammeter = 1 A
$I_{new}$, New desired current range = 10 A
$∴I_{shunt}=I_{new}-I_A$
$=10A-1A=9A$
and,
$V_A=I_A×R_A$ [Ohm's law]
$=(1×0.81)Ω=0.81V$
$∴R_{shunt}=\frac{V_A}{I_{shunt}}=\frac{0.81}{9}=0.09Ω$