Wattmeter MCQ || Wattmeter Questions and Answers

Answer.2. 3.33, 1.67

Explanation:

In two wattmeter method, the power factor is given by

$cos\phi = \cos \left[ {{{\tan }^{ – 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} – {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)} \right]$

Given that power factor = 0.866

$\phi = {\tan ^{ – 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} – {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)$

Cosφ = 0.866 ⇒ φ = 30°

$\begin{array}{*{20}{l}} {}\\ {{{30}^ \circ } = {{\tan }^{ – 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} – {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)}\\ {\frac{\begin{array}{l} \\ \sqrt 3 \left( {{W_1} – {W_2}} \right) \end{array}}{{{W_1} + {W_2}}} = \frac{1}{{\sqrt 3 }}} \end{array}$

Given the power input is 5 kW, i.e. (W₁ + W₂) = 5

${\frac{{\sqrt 3 \left( {{W_1} – {W_2}} \right)}}{5} = \frac{1}{{\sqrt 3 }}}$

(W₁ − W₂) = 5/3

W₁ = 10/3 kW

W₁ = 5/3 kW

Answer.2. Φ = 60°

Explanation:

The reading of two wattmeters can be expressed as

W₁ =  V_LI_Lcos(30 + φ)
W₂ =  V_LI_Lcos(30 − φ)

(i) When PF is unity ( φ = 0°)

W₁ =  V_LI_Lcos30°
W₂ =  V_LI_Lcos30°

Both wattmeters read equal and positive reading i.e upscale reading

(ii) When PF is 0.5 (φ = 60°)

W₁ =  V_LI_Lcos90° = 0
W₂ =  V_LI_Lcos30°

Hence total power is measured by wattmeter W₂ alone

(iii) When PF is less than 0.5 but greater than 0 i.e ( 90° > φ > 60°)

W₁ = Negative
W₂ = positive (since cos(−φ) = cosφ)

The wattmeter W₂ reads positive (i.e.upscale) because for the given conditions (i.e. ( 90° > φ > 60°), the phase angle between voltage and current will be less than 90°. However, in wattmeter W₁, the phase angle between voltage and current shall be more than 90° and hence the wattmeter gives a negative (i.e. downscale) reading.

Wattmeter cannot show negative reading as it has only a positive scale. An indication of negative reading is that the pointer tries to deflect in a negative direction i.e. to the left of zero.

In such a case, reading can be converted to positive by interchanging either pressure coil connections or by interchanging current coil connections. Remember that interchanging connections of both the coils will have no effect on wattmeter reading.

 (iv) When P.F reads zero (φ = 90°)

Such a case occurs when the load consists of pure inductance or pure capacitance

W₁ =  V_LI_Lcos(30 + 90°) = – V_LI_Lsin30°
W₂ =  V_LI_Lcos(30 − 90°) = V_LI_Lsin30°

In this condition, the two wattmeter reads equal and opposite i.e W₁ + W₂ = 0

Answer.2. 0.2 & 6 √3 kVAR

Explanation:

Given that,

Reversing the connections

W1 = 4 kW, W2 = – 2 kW

Total reactive power (Q) = √3[4 – (- 2)] = 6 √3 kVAR

$\begin{array}{l} \phi = {\rm{ta}}{{\rm{n}}^{ – 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} – {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\\ \\ \phi = {\rm{ta}}{{\rm{n}}^{ – 1}}\left( {\frac{{\sqrt 3 \left( {4 – ( – 2)} \right)}}{{4 – 2}}} \right)\\ \\ \phi = {\rm{ta}}{{\rm{n}}^{ – 1}}\left( {\frac{{6\sqrt 3 }}{2}} \right) \end{array}$

ϕ = 79.10^o

cosϕ = 0.2 (approx)

Answer.2. Two wattmeter method

Explanation:

1. The instrument most commonly used for active power measurement is the wattmeter.
2. It is essentially an inherent combination of an ammeter and a voltmeter and, therefore, consists of two coils known as the current coil and pressure coil.
3. The current coil is inserted in series with the line carrying the current to be measured and the pressure coil in series with a high non-inductive resistance connected across the load or supply terminals.

The wattmeter method is not used for resistance Measurement.

Answer.4. 120

Explanation:

Dynamometer type 3-phase power factor meter:

• The power factor meter circuit includes two fixed coils namely the pressure coil and a current coil.
• The pressure coil is connected across the circuit while the current coil is connected such that it can carry a circuit current or a definite fraction of current.
• The power factor meter circuit includes two moving coils. These are rigidly connected to each other so that their axes are at 120° to each other.
• These coils move together and carry the pointer which indicates the power factor of the circuit.
• Unlike, the single-phase dynamometer type power factor meter, it has both voltage coils in series with pure resistances. Instead of fixing the voltage coils 90° apart, we fix those coils 120° apart.

Answer.1. Both balanced and unbalanced circuits

Explanation:

A 3-wattmeter method is used to measure power in 3 phases using a 4 wire system which is based on Blondel’s theorem.

The three-wattmeter method of power measurement will work regardless of whether the load is balanced or unbalanced, wye-or delta-connected.

The three-wattmeter method is well suited for power measurement in a three-phase system where the power factor is constantly changing. The total average power is the algebraic sum of the three wattmeter readings.

Answer.1. 640 W

Explanation:

Given,

Vpc = V = 200 ∠0° V

Z = Impedance = 4 +j3 = 5∠36.87°

I = Current from supply = V/Z = 200∠0°/5∠36.87° = 40∠-36.87°A

Icc = 40 × (5/50) = 4 A

ϕ = 36.87°

Average power = 200 ×  4 ×  cos 36.87° = 640 W 

Answer.1. In series with the fixed coil

Explanation:

When a dynamometer instrument is used as a wattmeter, the fixed coils are connected in series with the load and carry the load current while the moving coil is connected across the load through a series multiplier R and carries a current proportional to the load voltage.

The fixed coil (or coils) is called the current coil and the movable coil is known as the potential coil.

Answer.3. 1

Explanation:

Given that, W₁ = 500 W

W₂ = 500 W

W₁ – W₂ = 0

W₁ + W₂ = 1000 W

$\phi = {\rm{ta}}{{\rm{n}}^{ – 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} – {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)$

tanφ = √3 × 0/1000

⇒ ϕ = 0°

⇒ Power factor = cos ϕ = 1

Answer.1. Average value of active power

Explanation:

The power measured by the dynamometer wattmeter at any instant is the instantaneous power absorbed by the three loads connected in three phases.  In fact, this power is the average power drawn by the load since the Wattmeter reads the average power because of the inertia of their moving system.

The power measured by each wattmeter varies from instant to instant. Due to the inertia of the moving system, the wattmeter measures the average power. The sum of the two wattmeter readings gives the total 3-phase power.