Question

When the supply voltage to an induction motor is reduced by 10%, the maximum torque will be decreased approximately by

• A. \( 5% \)
• B. \( 10% \)
• C. \( 20% \)
• D. \( 40% \)

Hint:

For small percentage changes, you can use calculus approximations (binomial theorem): $$T \propto V^2 \quad \Rightarrow \quad \frac{\Delta T}{T} \approx 2 \cdot \frac{\Delta V}{V}$$ Given a $10%$ reduction in voltage ($\frac{\Delta V}{V} = 10%$): $$\frac{\Delta T}{T} \approx 2 \times 10% = 20%$$ This approximation provides the correct answer immediately.

Answer 1

The Correct Option is C

Concept: The maximum torque ($T_{\max}$), also known as the breakdown torque or pull-out torque of a three-phase induction motor, is derived from its equivalent circuit model. The mathematical relationship shows that the maximum torque is directly proportional to the square of the applied stator supply voltage ($V$): $$T_{\max} \propto V^2$$ If the voltage changes from an initial value $V_1$ to a new value $V_2$, the torque scales quadratically. This relationship allows us to determine the percentage change in maximum torque for a given percentage reduction in supply voltage.

Step 1: Express the new voltage in terms of the initial voltage.
The supply voltage is reduced by $10%$. This means the new voltage $V_2$ is $90%$ of the original voltage $V_1$: $$V_2 = V_1 - 0.10 \cdot V_1 = 0.9 \cdot V_1$$

Step 2: Apply the quadratic torque relationship.
Using the proportionality $T_{\max} \propto V^2$, let us write the ratio of the new maximum torque ($T_{\max2}$) to the original maximum torque ($T_{\max1}$): $$\frac{T_{\max2}}{T_{\max1}} = \left(\frac{V_2}{V_1}\right)^2$$ Substitute $V_2 = 0.9 \cdot V_1$ into this ratio: $$\frac{T_{\max2}}{T_{\max1}} = (0.9)^2 = 0.81$$ This shows that the new maximum torque is $81%$ of its original value: $$T_{\max2} = 0.81 \cdot T_{\max1}$$

Step 3: Calculate the percentage decrease in maximum torque.
The percentage reduction is given by the fractional change multiplied by $100$: $$% \text{ Decrease} = \left(\frac{T_{\max1} - T_{\max2}}{T_{\max1}}\right) \cdot 100%$$ $$% \text{ Decrease} = (1 - 0.81) \cdot 100% = 0.19 \cdot 100% = 19%$$ An exact reduction of $19%$ is approximately equal to $20%$. Reviewing the available options:
• (1) $5%$
• (2) $10%$
• (3) $20%$
• (4) $40%$ Option (3) is the correct choice.