A galvanometer of 60Ω resistance shows full scale deflection for a current of 8 mA. The shunt required to convert the galvanometer into an ammeter of range 0 to 6 A will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $8 × 10^{-2}Ω$
$R_g$, Resistance of galvanometer = 60 Ω
$I_g$, full scale deflection current = 8 mA
$I_{max}$, Desire Ammeter range = 6A
∴ Current in Shunt Resistance, $I_{shunt}=6A-0.008A=5.992A$
∴ Voltage across the galvanometer, $V_g=I_g×R_g$ [By ohm's law]
$=0.008×60$
$=0.48V$
$V_{shunt}=V_g=0.48V$
$R_{shunt}=\frac{V_{shunt}}{I_{shunt}}=\frac{0.48}{5.992}=0.08Ω$