Question

In torque-slip curve of a 1-\(\phi\) induction motor, the backward torque is

• A. \( \text{negative from slip s=0 to s=1 and positive from slip s=1 to s=2} \)
• B. \( \text{negative from slip s=0 to s=2} \)
• C. \( \text{negative from slip s=0.5 to s=1.5} \)
• D. \( \text{negative from slip s=1 to s=2 and positive from slip s=0 to s=1} \)

Hint:

In double revolving field theory, the backward torque opposes the forward direction of rotation across all normal speeds. This means its sign remains consistently negative throughout the entire slip range from $s = 0$ to $s = 2$.

Answer 1

The Correct Option is B

Concept: According to the Double Revolving Field Theory, any single-phase alternating pulsating magnetic field can be resolved into two counter-rotating magnetic fields of equal magnitude:
• A forward rotating field moving at synchronous speed ($N_s$).
• A backward rotating field moving at the same speed in the opposite direction ($-N_s$). Each field produces its own torque-slip characteristic curve. If the rotor slip with respect to the forward field is $s$, then its slip with respect to the backward field is $(2 - s)$. The net torque ($T_{\text{net}}$) developed by the single-phase motor is the algebraic difference between the forward torque ($T_f$) and the backward torque ($T_b$): $$T_{\text{net}} = T_f - T_b$$

Step 1: Understand the direction of torque vectors.
The forward torque attempts to rotate the motor in the positive reference direction. Therefore, $T_f$ is considered positive. The backward torque opposes this movement by attempting to rotate the rotor in the opposite direction. Consequently, relative to the forward direction of motion, the backward field torque acts as a retarding braking force across the entire motoring range.

Step 2: Analyze the range of slip values.
For standard forward motor operation:
• At standalone standstill (stationary rotor, $N = 0$), the slip is $s = 1$.
• At ideal forward synchronous speed ($N = N_s$), the slip is $s = 0$.
• If the rotor is driven in the opposite direction up to backward synchronous speed ($N = -N_s$), the forward slip becomes $s = 2$. Across this entire functional operational range of forward slip from $s = 0$ to $s = 2$, the backward torque opposes the forward coordinate system.

Step 3: Determine the sign of the backward torque.
Because the backward torque acts in opposition to the forward motion across the entire range from $s = 0$ to $s = 2$, it is represented mathematically as a negative torque value on the standard torque-slip grid. Therefore, option (2) is correct.