Question

In a two wattmeter method, if one wattmeter reads negative then

• A. Power factor $>$ 0.5
• B. Power factor $=$ 1
• C. Power factor $<$ 0.5
• D. Zero power factor

Hint:

The relationship between the total three-phase power factor and the two wattmeter readings can be computed directly using this formula: \[ \tan(\phi) = \sqrt{3} \left( \frac{W_1 - W_2}{W_1 + W_2} \right) \] - If $W_2 = 0 \implies \tan(\phi) = \sqrt{3} \implies \phi = 60^\circ \implies \text{PF} = 0.5$. - If $W_2 < 0 \implies$ the denominator decreases while the numerator increases, meaning $\phi > 60^\circ \implies \text{PF} < 0.5$.

Answer 1

The Correct Option is C

Concept: The two-wattmeter method is a standard technique used to measure the total active power delivered to a balanced or unbalanced three-phase, three-wire load. Let the individual power readings of the two properly connected wattmeters be denoted as $W_1$ and $W_2$. For a balanced three-phase system with a total load phase angle $\phi$ (where $\phi$ is the phase angle between the phase voltage and phase current), the individual power equations recorded by the two instruments are given by: \[ W_1 = V_L I_L \cos(30^\circ - \phi) \] \[ W_2 = V_L I_L \cos(30^\circ + \phi) \] Where $V_L$ represents the line-to-line voltage and $I_L$ represents the line current. Let us analyze how the total load phase angle $\phi$ mathematically affects these two wattmeter readings:

Step 1: Understanding when a wattmeter reads negative. A wattmeter pointer deflections depend directly on the cosine terms in the equations. For standard forward power measurements, the cosine term must be positive. A negative reading occurs when the phase angle inside the cosine function exceeds $90^\circ$. Let us examine $W_2 = V_L I_L \cos(30^\circ + \phi)$: The term $\cos(30^\circ + \phi)$ becomes zero when its argument equals $90^\circ$: \[ 30^\circ + \phi = 90^\circ \quad \Rightarrow \quad \phi = 60^\circ \] If the load phase angle increases beyond $60^\circ$ ($\phi > 60^\circ$), the argument $(30^\circ + \phi)$ enters the second quadrant (between $90^\circ$ and $120^\circ$), causing $\cos(30^\circ + \phi)$ to become negative. Consequently, $W_2$ drops below zero and gives a negative reading.

Step 2: Relate the phase angle constraint to the Power Factor (PF). The power factor of a balanced three-phase system is defined as $\text{PF} = \cos(\phi)$. Let us evaluate the power factor at the boundary condition where $\phi = 60^\circ$: \[ \text{PF} = \cos(60^\circ) = 0.5 \] Since the cosine function decreases monotonically as the angle increases from $0^\circ$ to $90^\circ$: \[ \text{If } \phi > 60^\circ \quad \Rightarrow \quad \cos(\phi) < \cos(60^\circ) \quad \Rightarrow \quad \text{PF} < 0.5 \] Therefore, when one of the wattmeters registers a negative value, it indicates that the system's operating power factor has dropped below 0.5, which corresponds to Option (3).