Test: Phasor Diagrams - Electrical Engineering (EE) MCQ

Given that

V(t) = 100√2 cos (100πt),

i(t) = 10√2 sin(100π + π/4).

To consider voltage as a reference phasor, The First step would be to make Cosine as Sine by adding 90º to its phase.

V(t) = 100√2 cos (100πt)

= 100√2 sin (100πt + π/2)

i(t) = 10√2 sin(100π + π/4)

Phasor diagram for above two equations can drown as below.

As you can see by considering voltage as reference current leg behind by 45º.

So RMS value of current in phasor form is

= 

Kirchhoff’s Current Law (KCL): It states that the algebraic sum of currents entering a node or closed boundary is zero.

Mathematically, KCL implies that

Where N is the number of branches connected to the node

And i_n is the n^th current entering or leaving the node.

By this law current entering a node may be regarded as positive and current leaving a node may be regarded as negative or vice-versa.

Hence, the sum of currents entering a node is equal to the sum of the currents leaving the node.

Calculation:

Given that, three currents i₁, i₂, and i3 meet at a node,

From the above concept,

i₁,+ i₂, + i₃ = 0

or,  i₃ = - (i₁ + i₂)

Since the current is given in phasor form, hence the addition of current i₁ and i₂ can be done by using the parallelogram method,

We have,

i₁ = 10 sin (400t + 60°) A .... (1)

i₂ = 10sin (400t - 60°) A .... (2)

The phasor diagram can be drawn as,

By using the parallelogram method,

Where, it is the resultant current

We have, θ = (60° + 60°) = 120°

From equation (1), (2) & (3),

= 10

Since i₁ & i₂ has the same phase angle as well as the magnitude,

∴ i_r = 10sin 400t A

Since the sum of all current is zero, hence,

i₃ = - i_r = - 10sin 400t A

Types of loads and Phasor Diagrams:

Pure Inductor load:

• The load which contains only inductance (L) and not any other quantities like resistance and capacitance in the circuit is called a Pure inductive load.
• In this type of load, the current lags behind the voltage by an angle of 90 degrees.

Pure capacitor:

• The load which contains only capacitance (C) and not any other quantities like resistance and inductance in the circuit is called a Pure capacitive load.
• In this type of load, the current leads the voltage by an angle of 90 degrees.

RL load:

• The load which contains only Resistance (R) and Inductance (L) and not any other quantities like capacitance in the circuit is called a RL load.
• In this type of load, the current lags behind the voltage by an angle between 0 to 90 degrees.

RLC Load:

• When a pure resistance of R ohms, a pure inductance of L Henry, and a pure capacitance of C farads are connected together in series combination with each other then RLC Seriesload is formed.
• As all the three elements are connected in series so, the current flowing through each element of the load will be the same as the total current I flowing in the load.

Case I

RLC load with inductive reactance more than the capacitive reactance

Where

V_L = Inductor voltage

V_C = Capacitor voltage

I = Current

V = Supply voltage

V_R = Resistor Voltage

Case II

RLC load with inductive reactance lesser than the capacitive reactance

So If we observe the phasor diagram between V and I in all cases, RL load or RLC with inductive reactance more than the capacitive reactance (Case II) matches the given phasor.

Concept:

A phasor is a vector representation of a sinusoidal function that has both a magnitude and a phase angle.

Representations:

Rectangular form: z = x + jy

Polar form: z = r ∠ ϕ

Exponential form: z = z e^jϕ 

Comparison:

Time Domain: V(t) = V_m cos (ωt + ϕ)

It represents the voltage as a function of time as shown:

Phasor Domain: V = V_m ∠ ϕ 

It gives the magnitude and phase and can be represented as:

Where, 

Vm is the magnitude of the voltage function

ϕ is the phase angle of the voltage function

Note:

Power factor = cos ϕ 

• The phase angle is negative for lagging power factor
• The phase angle is positive for leading power factor

In the phasor representation of an alternating quantity, the sinusoidally varying alternating quantity can be represented graphically by a straight line with an arrow.

Given-

I_m = 10 A,

Power factor = 0.866 lagging (∵ inductive circuit)

∴ cosϕ = cos^-1(0.866)

∴ ϕ = -30° (∵ lagging power factor)

So that phasor representation of current is given by

I = 10 ∠-30° A

Resistance:

• If the pure resistive element is present in the circuit, then the current is in phase with the voltage. So angle is 0°
• The power factor in this case is unity.

Inductance:

• If the pure inductive element is present in the circuit, then the current lags the voltage by 90 degrees.
• The power factor in this case is lagging.

Capacitance:

• If the pure capacitive element is present in the circuit, then the current leads the voltage by 90 degrees.
• The power factor, in this case, is leading.

Hence the correct option for angle between the current and voltage in pure resistive circuit is zero or 0°.

Concept:

Coils induce EMF if moved with constant velocity in the magnetic field in accordance with Faraday's law of electromagnetic induction, and this EMF is also called Motional EMF. This is a basic law of electromagnetism.

Sum of 3 equal phasors at 0°,120°, -120° apart is zero.

Calculations :

Consider, E₁ = 100∠0 V,  E₂ = 100∠120 V,  E₃ = 100∠-120 V

Resultant EMF, E_R = E₁ + E₂ + E₃

E_R = 100∠0 + 100∠120 + 100∠-120

E_R = 100e^j0 + 100e^j120+100e^−j120

=100[e^j0+e^j120+e^−j120]

=100[1+2cos(120)]

= 0

Phasor diagrams are a graphical way of representing the magnitude and directional relationship between two or more alternating quantities.

Basically, a rotating vector called a Phasor is a scaled line whose length represents an AC quantity that has both magnitude and direction which is frozen at some point in time.

Characteristics:

• It is a vector quantity
• It is a complex quantity
• It is a time dependent quantity

When the Alternating emf is running in the circuit is

e = e₀ sin ωt

Current in the inductive circuit is –

• From above it is clear that current leads the voltage by π/2.

Hence

the Phase difference between current and voltage in pure capacitive circuit is 90° or π/2.

If the magnetization is non linear in nature then it will cause a saturation in the core and harmonics will be introduced to cause humming sounds.

According to the primary and secondary equivalent circuits of a transformer equation stated in option 1 correctly suits with the kirchoff’s voltage law for primary side of a transformer, similarly equation for secondary side can also be written down.

When phasors are drawn representing the maximum values instead of the rms value, the shape of the diagram remains unaltered and hence the phase angle remains the same.

Ammeters and voltmeters are calibrated to read the rms value, hence the phasors are drawn representing the rms values.

We know that the rms value is 1/√2 times the maximum value, hence the rms value is 0.707 times the maximum value.

Graph of E_f vs F_f is the open circuit characteristic of the machine.