Why locus of the current did not start from the origin (where X and Y-axis start) in the case of circle diagra

Question

Why locus of the current did not start from the origin (where X and Y-axis start) in the case of circle diagram of a three-phase induction motor?

Examoogle · Accepted Answer

Option C: even at no load. IM draw the No-Load Current ($I_0$) due to Mechanical Losses And Iron Losses

Explanation & Concepts:
In the circle diagram of a three-phase induction motor, the current locus does not start from the origin because even under no-load conditions, the motor draws a finite no-load current ($I_0$). This no-load current is required to establish the rotating magnetic flux (magnetizing component, $I_m$) and to supply core (iron) and mechanical friction and windage losses (working component, $I_w$).

Working Principle: When a 3-phase supply is connected to the stator, a rotating magnetic field is established. To sustain this field, the motor draws a magnetizing current ($I_m$) that lags the supply voltage by $90^\circ$. Simultaneously, the iron losses in the stator core and mechanical losses due to rotation require an active active current component ($I_w$) in phase with the voltage. The phasor sum of these two components gives the no-load current $I_0 = \sqrt{I_m^2 + I_w^2}$, shifting the starting point of the current locus vector away from the origin.

Key Formulas: $I_0 = \sqrt{I_m^2 + I_w^2}$ — No-load current magnitude; $I_m = I_0 \sin\phi_0$ — Magnetizing component responsible for main field flux; $I_w = I_0 \cos\phi_0$ — Core loss and mechanical loss component; 	ext{Power Factor at No Load} = \cos\phi_0 = \frac{I_w}{I_0} \approx 0.1 	ext{ to } 0.2$ — Typical low no-load power factor

Key Points: The no-load current $I_0$ of a three-phase induction motor is relatively high, usually $30\%$ to $50\%$ of full-load current, due to the presence of an air gap.; The vertical offset from the X-axis represents the active power loss at no load (stator core loss + mechanical friction and windage loss).; The horizontal offset represents the reactive magnetizing component required to establish the air-gap flux.

Advantages: Circle diagrams allow predetermination of motor performance (efficiency, power factor, slip, torque) without direct loading tests.; Provides a simple visual tool to analyze motor operation across all slip ranges.

Disadvantages: Assumes constant parameters (stator/rotor resistance and reactance), which actually vary with saturation and temperature.; Less accurate for high-capacity machines due to non-linear magnetizing characteristics.

Applications: Used in electrical machine testing alongside No-Load and Blocked-Rotor tests.; Performance analysis and design verification of 3-phase induction motors.

Additional Info: Option A is incorrect because stator and rotor copper losses depend primarily on load current. Stator $I^2R$ loss at no load is small, and rotor copper loss at no load is negligible ($s \approx 0$).. Option B is incorrect because being a rotating device explains why mechanical loss exists, but does not fully define the total no-load current which also includes iron/core losses and magnetizing reactive power.. The no-load power factor $\cos\phi_0$ is very low ($0.1$ to $0.2$) due to the predominant magnetizing component $I_m$ required to overcome air gap reluctance.

Answer: C is correct — The current locus starts at the tip of the no-load current phasor $I_0$, which accounts for iron losses, mechanical losses, and magnetizing flux even when no mechanical load is applied.

Exam Tip: Remember that transformer no-load current is only $2\%-5\%$ of rated current because of a continuous iron core, while induction motor no-load current is $30\%-50\%$ of rated current due to the air gap.

Examoogle · Answer

Option A: due to stator and rotor copper loss

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Option B: because its rotating device

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Option D: None of these

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