Electrical and Electronic Measurements Test 3

Concept:

In a two-wattmeter method,

Total power (P) = P₁ + P₂

The power factor of the system = cos ϕ

\(\tan \phi = \sqrt 3 \frac{{[{P_1} - {P_2}]}}{{\left[ {{P_1} + {P_2}} \right]}}\)

Where P₁ & P₂ are power of wattmeters

Calculation:

P₁ = 5 kW, P₂ = 1 kW

Solution:

P_t = P₁ – P₂

= 5 – 1 = 4 kW

p.f = cos ϕ

and \(\phi = {\tan ^{ - 1}}\left( {\sqrt 3 \frac{{\left( {{P_1} - {P_2}} \right)}}{{\left( {{P_1} + {P_2}} \right)}}} \right) = {\tan ^{ - 1}}\left( {\sqrt 3 \left( {\frac{6}{4}} \right)} \right)\)

= 68.94°

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

Therefore the pf is 0.3592 lag

V = √2 × 200 sin (314 πt), I = √2 cos (314 πt)

Now, P = V × I

= √2 × √2 × 200 × sin (314 πt) × cos (314 πt)

(as sin (314 πt) = 0)

⇒ P = 0

Hence there will be no deflection.

V = V₁ sin (ωt + θ₁) + V₂ sin (2ωt + θ₂) + ….

I = I₁ sin (ωt + ϕ₁) + I₂ (2ωt + ϕ₂) + …

\(P = VI = \frac{1}{2}{V_1}{I_1}\cos \left( {{\theta _1} - {\phi _1}} \right) + \frac{1}{2}{V_2}{I_2}cos\left( {{\theta _2} - {\phi _2}} \right) + \ldots\)

Calculation:

V = 80 sin ωt + 60 sin (3ωt – 60°) + 5sin (5ωt + 30°)

I = 12 sin ωt + 8 sin(5 ωt + 180°)

\(P = \frac{1}{2} \times 80 \times 12\cos 0 + \frac{1}{2} \times 60 \times 0 + \frac{1}{2} \times 5 \times 8\cos \left( { - 150^\circ } \right)\)

= 40 × 12 + 0 - 4 × 5 × (0.866)

Concept:

The error due to pressure coil inductive reactance is,

% error = ± (tan ϕ tan β) × 100

‘+’ sign signifies the load is inductive.

‘-’ sign signifies the load is capacitive.

Where β = pressure coil impedance angle

ϕ = power factor angle

Calculation:

Power factor = 0.7, f = 50 Hz, R_Pc = 1000 Ω, L = 0.5 H, cos ϕ = 0.7

X_L = 2π f × L

= 2 × π × 50 × 0.5 = 157.07 Ω

ϕ = cos^-1 (0.5) = 60°

tan ϕ = tan 60° = 1.732

Now, we know that  \(\tan \beta = \frac{{{X_p}}}{{{R_p}}}\)

Where X_p = reactance of pressure coil

R_p = resistance of pressure coil

\(\tan \beta = \frac{{{X_p}}}{{{R_p}}} = \frac{{157.07}}{{1000}} = 0.15707\)

∴ % error = + (tan ϕ tan β) × 100

= (1.732 × 0.15707) × 100

= 27.20%

Given that, Voltage (V) = 300 V

Current (I) = 16 A

Current coil resistance (R_CC) = 0.2 Ω

Pressure coil resistance (R_PC) = 20 kΩ

Power factor = cos ϕ = 0.4

Power (P) = VI cos ϕ = 300 × 16 × 0.4 = 1.92 kW

As the pressure coil is connected across load side, the error is due to pressure coil resistance.

Error \( = \frac{{{V^2}}}{{{R_{PC}}}} = \frac{{{{300}^2}}}{{20 \times {{10}^3}}} = 4.5\;W\)

W₁ = 1000 W, W₂ = 650 W, V = 440 V

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

\(ta{m^{ - 1}}\left( {\sqrt 3 \left( {\frac{{1000 - 650}}{{1000 + 650}}} \right)} \right)\)

= 20.17°

W₁ = V_L I_L cos(30 - ϕ)

= √3 V_ph I_ph cos(30 - ϕ)

⇒ 1000 = √3 (440) (I_ph) cos (30-20.17)

⇒ I_ph = 1.33 A

Impedance \(= \left| z \right| = \left| {\frac{{{V_{ph}}}}{{{I_{ph}}}}} \right| = \frac{{440}}{{1.33}} = 330.83\;{\rm{\Omega }}\)