Wattmeter Reactive Power Measurement MCQ || Measurement of 3 Phase Reactive Power Using Wattmeter Questions and Answers

Answer.3. Moving iron instrument

Explanation:

• We can use electrostatic instruments to measure voltage and power. These types of wattmeters are used for the measurement of a small amount of power, practically when the voltage is high and the power factor is low. It is also used for measurement of dielectric loss of cables on alternating voltage and for calibration of wattmeter and energy meters.
• Thermocouple instruments are used to measure current, voltage, and power.
• We can use the moving iron instrument to measure current and voltage only.
• Electrodynamic instruments are the most preferred instruments for measurement.

Answer.4. 0.866

Explanation:

Given that,

\({W_2} = \frac{{{W_1}}}{2}\)

We know that,

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

\( = {\tan ^{ – 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} – \frac{{{W_1}}}{2}} \right)}}{{\left( {{W_1} + \frac{{{W_1}}}{2}} \right)}}} \right)\)

\(= {\tan ^{ – 1}}\left( {\frac{{\sqrt 3 \left( {2{W_1} – {W_1}} \right)}}{{2{W_1} + {W_1}}}} \right)\)

\(= {\tan ^{ – 1}}\left( {\frac{{\sqrt 3 {W_1}}}{{3{W_1}}}} \right) = {\tan ^{ – 1}}\left( {\frac{1}{{\sqrt 3 }}} \right) = 30^\circ \)

Power factor, cos ϕ = cos 30° = 0.866

Answer.3. +597 W

Explanation:

Given

V_RY = 415 V

Cosφ = 0.8 Lag

φ = 36.96

V_RY = 415∠30°

θ = 30° − 36.96

θ = −6.86°

i.e

I_R = 415∠30°/100∠36.96°

= 1.45∠−6.86

${W = 4.15 \times \frac{{415}}{{\sqrt 3 }}{\rm{cos}}\left( {120 + 6.86} \right)}$

W = −596.46 watt

Answer.1. 0.2

Explanation:

Power measured by 1- ϕ  wattmeter is given by P = V_pcI_cc cos ϕ

Where

V_pc = Voltage of potential coil

I_cc = Current through the current coil

Here, V_pc = 300 V , I_cc = 5 A , P = 300 W (Full Scale)

300 = V_pc I_cc cos ϕ

300 = 300 × 5 cos ϕ

cos ϕ  = 0.2

Answer.2.

Explanation:

The power measured by the wattmeter is given by

W = I_RV_YBcos(I_RV_YB)

From the phasor diagram, we can observe that the phase angle between IR and VYB is 90 – ϕ.​

⇒ W = V_YBI_R cos (90 – ϕ)

⇒ W = V_LI_L sin ϕ

⇒ W = √3 V_phI_ph sin ϕ 

But we know that three-phase reactive power Q = 3 V_phI_ph sin ϕ

So, so by multiplying the wattmeter reading by √3 we can get three-phase reactive power.

Q (total) = √3 W = 3-ϕ reactive power

•  For the measurement of reactive power, the current coil is connected to one of the phases of a 3-ϕ network.
• And the potential coil is to be connected between the remaining two phases.

Answer.1. Both wattmeters will indicate the same value but with opposite signs

Explanation:

Given that load is pure inductive – Hence it is zero power factor (lag)

Cos ϕ = 0

⇒ ϕ = 90°

The reading of two wattmeters can be expressed as

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

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

W₁ = -W₂

Answer.2. 132.79 μF

Explanation:

Given,

Supply voltage V = 440 V

Supply frequency f = 50 Hz

Wattmeter reading W1 = 2000 W

W2 = – 1000 W

Total active power P = W1 + W2 = 1000 W

Power factor angle ϕ1

${\phi _1} = {\tan ^{ – 1}}\left( {\sqrt 3 \frac{{{W_1} – {W_2}}}{{{W_1} + {W_2}}}} \right)$

${\phi _1} = {\tan ^{ – 1}}\left( {\sqrt 3 \frac{{2000 + 1000}}{{2000 – 1000}}} \right)$

ϕ1 = 79.1∘

Total Real power P = √3 V_L I_L cos ϕ1

Line current I_L = (P / √3 V_L) = (1000 / √3 × 440) = 6.94 A

Z_ph = V_ph/I_ph (In star connection V_ph = V_L/√3 and I_L = I_ph)

Z_ph = 440 / (√3 × 6.94) = 36.61 Ω

Resistance R_ph = Z_ph cos ϕ1 = 36.61 cos 79.1° = 6.922 Ω

Reactance X_ph = Z_ph sin ϕ1 = 36.61 sin 79.1° = 35.95 Ω

After capacitance is added to each phase of the load, also total real power is measured by only one wattmeter then

Power factor angle ϕ₂

${\phi _2} = {\tan ^{ – 1}}\left( {\sqrt 3 \frac{{2000 + 0}}{{2000 – 0}}} \right)$

Reactance after adding capacitance X’_ph

X’_ph = R_ph tan ϕ₂ = 6.922 × tan 60^∘ = 11.98 Ω 

Capacitive reactance X_C = X_ph – X’_ph = 35.95 – 11.98 = 23.97 Ω

Capacitive reactance X_C = 1 / 2πfC 

23.97 = 1 / 2π(50)C

⇒ C = 132.79 μF

Answer.1. 45.82 kVAR

Explanation:

Given that,

Apparent power (S) = 50 kVA

Reading of wattmeter = Active power (P) = 20 kW

Apparent Power

$S = \sqrt {{P^2} + {Q^2}}$

$ \Rightarrow 50 = \sqrt {{{20}^2} + {Q^2}}$

Q = 45.82 kVAR

Answer.2. 4 kW, 0.3592 lag

Explanation:

P₁ = 5 kW, P₂ = 1 kW

P_t = P₁ – P₂

= 5 – 1 = 4 kW

p.f = cos ϕ

${\phi _1} = {\tan ^{ – 1}}\left( {\sqrt 3 \frac{{{W_1} – {W_2}}}{{{W_1} + {W_2}}}} \right)$

${\tan ^{ – 1}}\left( {\sqrt 3 \left( {\frac{6}{4}} \right)} \right)$

= 68.94°

Hence, p.f. = cos ϕ = cos 68.94° = 0.3592

Therefore the pf is 0.3592 lag

Answer.3. 39.69 kW

Explanation:

P = 30 W, cos ϕ = 0.4

If P₁ and P₂ are two individual wattmeter readings then according to the problem,

P = P₁ + P₂ = 30 kW

Power factor angle ϕ = cos^-1 (0.4) = 66.4°

$\phi = {\tan ^{ – 1}}\left( {\sqrt 3 \frac{{\left( {{P_1} – {P_2}} \right)}}{{\left( {{P_1} + {P_2}} \right)}}} \right)$

${66.4^ \circ } = {\tan ^{ – 1}}\left( {\frac{{\sqrt 3 \left( {{P_1} – {P_2}} \right)}}{{30}}} \right)$

(2.288 × 30)/√3 = P₁ − P₂

P₁ – P₂ = 39.69 kW

Answer.1. W₁ = 36 kW, W₂ = 0 kW

Explanation:

Given that, line voltage (VL) = 500 V

Cos ϕ = 0.5

Cos ϕ = 0.5 ⇒ ϕ = 60°

W₁ + W₂ = 36 kW

${\phi _1} = {\tan ^{ – 1}}\left( {\sqrt 3 \frac{{{W_1} – {W_2}}}{{{W_1} + {W_2}}}} \right)$

$\Rightarrow \tan 60 = \frac{{\sqrt 3 \left( {{W_1} – {W_2}} \right)}}{{\left( {{W_1} + {W_2}} \right)}}$

⇒ W₁ – W₂ = 36

⇒ W₁ = 36 kW, W₂ = 0 kW