# A direct on-line starter is used for starting motors up to

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A direct on-line starter is used for starting motors up to
1. 5 hp
2. 10 hp
3. 20 hp
4. 40 hp

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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

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