Question

The short circuit current of an 3-phase induction motor is 5 times the full load current and full load slip is 0.035. To start the motor using auto transformer starter with 40 percent of full load torque, the suitable percent tapping on auto transformer will be

• A. 67.6%
• B. 45.7%
• C. 48.3%
• D. 42.3%

Answer 1

The correct option is A 67.6%

$\text{We know that,}$
$\frac{\text{Starting torque}}{\text{Full load torque}}=\frac{T_{st}}{T_{fl}}=x^2{\left(\frac{I_{SC}}{I_{fl}}\right)}^2\times S_{fl}$

$0.4=x^2\times 5^2\times 0.035$

$x^2=\frac{0.4}{25\times 0.035}$

$x=\sqrt{\frac{0.4}{25\times 0.035}}=0.676$

$\text{The percentage of tapping is 67.6\%.}$