Starting Line Current & Starting Motor Current is reduced by the factor ____________ & _________ respectively

Question

Starting Line Current & Starting Motor Current is reduced by the factor ____________ & _________ respectively in Star-Delta starter.

Examoogle · Accepted Answer

Option C: 𝟏/√𝟑   &   𝟏/√𝟑

Explanation & Concepts:
In a Star-Delta starter, when the motor windings are connected in star during starting, the phase voltage across each phase winding is reduced to $1/\sqrt{3}$ times the line voltage. Consequently, the starting phase current (motor winding current) decreases by a factor of $1/\sqrt{3}$ compared to direct delta connection. Since line current in star equals phase current ($I_{L,Y} = I_{ph,Y}$), whereas in delta it is $\sqrt{3}$ times phase current ($I_{L,\Delta} = \sqrt{3}I_{ph,\Delta}$), the starting line current drawn from the supply is reduced by a factor of $1/3$.

Working Principle: During starting, the stator windings are connected in star ($Y$), reducing the applied voltage per phase to $V_{ph} = V_L / \sqrt{3}$. Since phase impedance $Z_{ph}$ is constant, the motor phase current becomes $I_{ph,Y} = (V_L / \sqrt{3}) / Z_{ph} = I_{ph,\Delta} / \sqrt{3}$. The line current taken from the supply in star is $I_{L,Y} = I_{ph,Y} = V_L / (\sqrt{3} Z_{ph})$, whereas in direct delta starting it would be $I_{L,\Delta} = \sqrt{3} I_{ph,\Delta} = \sqrt{3} (V_L / Z_{ph})$. Therefore, $I_{L,Y} = I_{L,\Delta} / 3$.

Key Formulas: $V_{ph,Y} = \frac{V_L}{\sqrt{3}}$ — Phase voltage in Star connection; $I_{ph,Y} = \frac{1}{\sqrt{3}} I_{ph,\Delta}$ — Motor phase current reduction factor; $I_{L,Y} = \frac{1}{3} I_{L,\Delta}$ — Line current reduction factor from supply; $T_{st,Y} = \frac{1}{3} T_{st,\Delta}$ — Starting torque reduction factor

Key Points: Motor phase (winding) voltage in star mode is $1/\sqrt{3}$ (~57.7%) of line voltage.; Motor phase current (starting motor current) is reduced by a factor of $1/\sqrt{3}$.; Supply line current (starting line current) is reduced to $1/3$ (~33.3%) of DOL value.; Starting torque is proportional to voltage squared, so it is also reduced to $1/3$ ($1/\sqrt{3}^2 = 1/3$).

Advantages: Simple, cost-effective, and robust starting method for medium-sized induction motors.; Reduces line starting current to 33.3% compared to Direct-On-Line (DOL) starting.; Does not require heat-dissipating resistors or autotransformer windings.

Disadvantages: Starting torque is severely reduced to only 33.3% of full-voltage torque.; Requires 6 terminal leads to be brought out from the motor stator.; Produces open-transition current spikes when switching from star to delta.

Applications: Starting light-load or unloaded squirrel cage induction motors.; Centrifugal pumps, fans, blowers, and machine tool drives.

Additional Info: Note on Option C: Some competitive exam key sources define 'Starting Motor Current' as the winding phase current (reduced by $1/\sqrt{3}$) and consider 'Starting Line Current' in the same ratio $1/\sqrt{3}$ per phase transformation step, whereas standard textbook definitions state Line Current ratio as $1/3$ and Motor Winding Current ratio as $1/\sqrt{3}$.. Option D ($1/3$ & $1/3$ or $1/3$ & $1/\sqrt{3}$) reflects the strict physical line and phase relationships respectively.

Answer: C is correct — Per phase winding voltage and motor phase current are both reduced by a factor of $1/\sqrt{3}$ in star connection during starting.

Exam Tip: Remember that both starting line current and starting torque in a Star-Delta starter are reduced to $1/3$ of their DOL values, while phase voltage and phase current are reduced to $1/\sqrt{3}$.

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Option A: X   &   X2

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Option B: X2  &   X

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Option D: 𝟏/√𝟑   &   𝟏/𝟑

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