Correct Answer - Option 1 : 5 hp

__Direct-on-line starting (DOL):__

- As the name suggests, direct-on-line starting means that the motor is started by connecting it directly to the supply at the rated voltage.
**Direct-on-line starting, (DOL), is suitable for stable supplies and small induction motors up to 5HP**

__Drawbacks of DOL:__

- Small motors that do not start and stop frequently need only very simple starting equipment, often in the form of a hand-operated motor protection circuit breaker.
- Full voltage is switched directly onto the motor terminals.
- For small motors, the starting torque will be 150% to 300% of the full-load value, while the starting current will be 300% to 800% of the full-load current or even higher.

The ratio of starting torque to full load torque of DOL:

\(\frac{{{T_{st}}}}{{{T_{f1}}}} = {s_{f1}}{\left( {\frac{{{I_{SC}}}}{{{I_{f1}}}}} \right)^2}\)

Where

Tst is the starting torque

Tf1 full load torque

Sf1 is the slip at full load torque

ISC is the current at starting

If1 is the current at full load

__Star-delta starting:__

- The objective of this starting method, which is used with three-phase induction motors, is to reduce the starting current.
- In the starting position, the current supply to the stator windings is connected in star (Y) for starting.
- In the running position, the current supply is reconnected to the windings in delta (∆) once the motor has gained speed.

Delta starting

Ist(phase) = V1 / Z01

√3 Ist(phase) = Ist(line) = √3 V1 / Z01

Star connected starting

Ist(line) = Ist(phase)

Ist(phase) = V1 / √3 Z01

The ratio of currents of star connection and delta connection:

\(\frac{{{I_{st\left( {star} \right)}}}}{{{I_{st\left( {delta} \right)}}}} = \frac{{\frac{{{V_1}}}{{\sqrt 3 {Z_{01}}}}}}{{\frac{{\sqrt 3 {V_1}}}{{{Z_{01}}}}}} = \frac{1}{3}\)

The starting current with star starting is just 1/3rd when compared to direct delta starting. Therefore starting current is reduced.

The ratio of starting torque to full load torque of star-delta starter:

\(\frac{{{T_{st}}}}{{{T_{f1}}}} = {s_{f1}}\frac{1}{3}{\left( {\frac{{{I_{SC}}}}{{{I_{f1}}}}} \right)^2}\)

Starting torque also reduces to 1/3rd when compared to direct online starting