Starting Torque of 3-Phase Induction Motor; Torque Equation of 3-Phase Induction Motor

Digital ElectronicsElectronElectronics & Electrical

The torque (τ) developed by the rotor of a 3-phase induction motor is directly proportional to −

  • Rotor current (I2)
  • Rotor EMF (E2)
  • Rotor circuit power factor (cos Ο•2)


$$\mathrm{\tau \propto 𝐸_2𝐼_2 cos \varphi_2}$$

$$\mathrm{⇒ \tau = 𝐾𝐸_2𝐼_2 cos \varphi_2 … (1)}$$

Where, K is the constant of proportionality.

Starting Torque of 3-Phase Induction Motor


  • Rotor resistance/Phase = 𝑅2
  • Rotor reactance/Phase at standstill = 𝑋2
  • Rotor EMF/Phase at standstill = E2

∴ Rotor impedance/Phase at standstill,

$$\mathrm{𝑍_2 = \sqrt{𝑅_2^2 + 𝑋2^2}}$$

Rotor current/Phase at standstill,

$$\mathrm{𝐼_2 =\frac{𝐸_2}{𝑍_2}=\frac{𝐸_2}{\sqrt{𝑅_2^2 + 𝑋2^2}}}$$

And, Rotor power factor at standstill,

$$\mathrm{cos\varphi_2 =\frac{𝑅_2}{𝑍_2}=\frac{𝑅_2}{\sqrt{𝑅_2^2 + 𝑋2^2}}}$$

∴ Starting torque,

$$\mathrm{\tau_𝑠 = 𝐾𝐸_2𝐼_2 cos\varphi_2 = 𝐾𝐸_2 × (\frac{𝐸_2}{\sqrt{𝑅_2^2 + 𝑋2^2}}) × (\frac{𝑅_2}{\sqrt{𝑅_2^2 + 𝑋2^2}})}$$

$$\mathrm{⇒ \tau_𝑠 =\frac{𝐾 𝐸_2^2 𝑅_2}{(𝑅_2^2 + 𝑋_2^2)}… (2)}$$

In general, the stator supply voltage V is constant so that the magnetic flux per pole set up by the stator is also constant. Therefore, the EMF E2 in the rotor will be constant.

$$\mathrm{\tau_𝑠 =\frac{𝐾_1 𝑅_2}{(𝑅_2^2 + 𝑋2^2)}… (3)}$$

Where, K1 = K E22 is another constant.

Condition for Maximum Starting Torque of 3-Phase Induction Motor

As the starting torque is given by,

$$\mathrm{\tau_𝑠 =\frac{𝐾_1 𝑅_2}{(𝑅_2^2 + 𝑋2^2)}=\frac{𝐾_1}{(𝑅_2 +\frac{𝑋_2^2}{𝑅_2})}… (4)}$$

To be the maximum of the stating torque, the denominator of the eqn. (4) should be minimum, i.e.,

$$\mathrm{\frac{𝑑}{𝑑𝑑}(𝑅_2 +\frac{𝑋_2^2}{𝑅_2}) = 0}$$

$$\mathrm{⇒ 1 −\frac{𝑋_2^2}{𝑅_2^2} = 0}$$

$$\mathrm{⇒ 𝑅_2 = 𝑋2 … (5)}$$

Hence, the stating will be maximum when,

$$\mathrm{𝐑𝐨𝐭𝐨𝐫\:π‘πžπ¬π’π¬π­πšπ§πœπž/𝐏𝐑𝐚𝐬𝐞 = 𝐑𝐨𝐭𝐨𝐫\:π‘πžπšπœπ­πšπ§πœπž/𝐏𝐑𝐚𝐬𝐞\:𝐚𝐭\:𝐬𝐭𝐚𝐧𝐝𝐬𝐭𝐒π₯π₯}$$

Under the condition of maximum starting torque, the rotor power factor angle Ο•2 = 45° and the rotor power factor is 0.707 lagging.

Effect of Change of Supply Voltage on Starting Torque of 3-Phase Induction Motor

$$\mathrm{\because \tau_𝑠 =\frac{𝐾 𝐸_2^2 𝑅_2}{(𝑅_2^2 + 𝑋_2^2)}}$$

$$\mathrm{\because 𝐸2 \propto Supply\:voltage (𝑉)}$$

$$\mathrm{\therefore \tau_𝑠 =\frac{𝐾 𝑉_2 𝑅_2}{(𝑅_2^2 + 𝑋_2^2)}}$$

$$\mathrm{⇒ \tau_𝑠 \propto 𝑉^2 … (6)}$$

Hence, the starting torque is directly proportional to the square of the supply voltage. Therefore, the starting torque is very sensitive to changes in the value of the supply voltage.

Important −

  • Starting Torque of Squirrel Cage Motors – For the squirrel cage motors, the starting torque is very low about 1.5 to 2 times of the full-load value.
  • Starting Torque of Wound Rotor Motors – In case of slip ring induction motors, the resistance of the rotor circuit can be increased by inserting external resistance. By adding the proper value of the external resistance (i.e., R2 = X2), maximum starting torque can be obtained.

Numerical Example

A 100 kW, 3 kV, 50 Hz, 8-pole, star connected induction motor has a star connected slip ring rotor with a turn ratio of 2.5 (stator/rotor). The rotor resistance is 0.2 Ω/phase and its per phase leakage inductance is 4 mH. The stator impedance may be neglected. Find the starting torque on rated voltage with short circuited slip rings.


Trasformation ratio,

$$\mathrm{𝐾 =\frac{Rotor\:turns/phase}{Stator\:turns/phase} =\frac{1}{2.5}= 0.4}$$

$$\mathrm{Rotor \:resistance/phase\: referred \:to \:stator, 𝑅′_2 =\frac{𝑅_2}{𝐾_2} =\frac{0.2}{0.4^2}= 1.25 \Omega}$$

The reactance of the rotor circuit is,

$$\mathrm{𝑋_2 = 2πœ‹π‘“πΏ = 2πœ‹ × 50 × (4 × 10^{−3}) = 1.256 \Omega;}$$

Rotor reactance/phase referred to stator,

$$\mathrm{𝑋′_2 =\frac{𝑋_2}{𝐾^2} =\frac{1.256}{0.4^2} = 7.85 \Omega}$$

Now, supply voltage⁄phase,

$$\mathrm{𝐸_1 =\frac{3000}{\sqrt{3}}= 1732 V}$$

Therefore, the starting torque of the motor is,

$$\mathrm{\tau_𝑠 =\frac{𝐾 𝐸_2^2 𝑅_2}{(𝑅_2^2 + 𝑋_2^2)}=\frac{3}{2πœ‹π‘_𝑠}×\frac{𝐸_1^2𝑅'_2}{(𝑅′_2)^2 + (𝑋′2)^2}}$$


$$\mathrm{𝑁_𝑠 =\frac{120𝑓}{𝑃}=\frac{120 × 50}{8}= 750 RPM = 12.5 rps}$$


$$\mathrm{𝐾 =\frac{3}{2πœ‹π‘_𝑠}; and 𝐸2 \propto 𝐸1}$$

$$\mathrm{\therefore \tau_𝑠 =\frac{3}{2πœ‹π‘_𝑠}×\frac{𝐸_1^2𝑅'_2}{(𝑅′_2)^2 + (𝑋′2)^2}}$$

$$\mathrm{= (\frac{3}{2πœ‹ × 12.5}) × (\frac{(1732)^2 × 1.25}{1.252 + 7.85^2})}$$

$$\mathrm{= 2267 Nm}$$

Published on 30-Aug-2021 11:59:38