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CONSTRUCTING A GRAPHICAL REPRESENTATION OF A VECTOR
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⚡ As the single vector rotates in an anti-clockwise direction, its tip at point A will rotate one complete revolution of 360° or 2π, representing one complete cycle of the waveform.

360
∘
=2π

📘 If the length of its rotating tip is transferred at different angular intervals in time to a graph, a smooth sinusoidal waveform would be drawn starting at the left with zero time.

📌 Each position along the horizontal axis indicates the time elapsed since:
t=0
⚡ IMPORTANT VECTOR POSITIONS

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🔹 When the rotating vector is horizontal, the tip represents angles:

0∘, 180∘, 360∘

🔹 When the vector is vertical, it represents:

✔️ Positive peak value at 90° or π/2
✔️ Negative peak value at 270° or 3π/2

+Amax​ at 90∘,−Amax​ at 270∘

📌 Therefore, the time axis of the waveform represents the angle in degrees or radians through which the phasor has moved.

⚡ PHASOR AT A PARTICULAR TIME

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🔹 A phasor represents a scaled voltage or current value of a rotating vector that is “frozen” at some instant in time.

📘 Example: frozen at angle 30°.

Φ=30∘

⚡ This is very useful when comparing two waveforms such as voltage and current on the same axis.

⚡ PHASE DIFFERENCE BETWEEN TWO WAVEFORMS

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📘 General mathematical expressions:

v(t)=Vmax​sin(ωt)
i(t)=Imax​sin(ωt−Φ)

📌 Here current i lags voltage v by angle Φ.

🔹 If Φ = 30°, then current lags voltage by 30°.

⚡ LEADING AND LAGGING PHASORS

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✔️ If current comes later → Current lags Voltage
✔️ If voltage comes earlier → Voltage leads Current

📌 One phasor is always selected as the reference phasor, and all others are measured relative to it.

⚡ PHASOR ADDITION OF PHASOR DIAGRAMS

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⚡ One major use of phasors is adding sinusoids of the same frequency.

📌 Example in RLC Series Circuits where voltages are not in-phase.

✅ If In-Phase (0° Shift)

Two voltages:

✔️ 50V + 25V = 75V

50+25=75

⚡ If Out-of-Phase

If phase angle exists, normal addition cannot be used.

📌 We must use Phasor Diagrams and the Parallelogram Law.

⚡ PHASOR DIAGRAM WORKED EXAMPLE

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Consider:

✔️ V₁ = 20V
✔️ V₂ = 30V
✔️ V₁ leads V₂ by 60°

V1​=20V,V2​=30V,Φ=60∘

📌 By drawing the two phasors to scale and constructing a parallelogram, the resultant total voltage becomes:

VT=43.6V∠23.4∘

✅ Therefore total voltage = 43.6V at angle 23.4°

⚡ USING A BIT OF MATHS

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📘 Instead of drawing diagrams, we can calculate the result using horizontal and vertical components.

✔️ Horizontal = Cosine part
✔️ Vertical = Sine part

This analytical method is called Rectangular Form.

⚡ RECTANGULAR FORM

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A phasor is divided into:

✔️ Real part (x)
✔️ Imaginary part (y)

Z=x±jy

📌 This gives a complete mathematical description of magnitude and phase angle.

⚡ SUMMARY

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✅ Phasors simplify AC calculations
✅ Easy comparison of voltage/current phase
✅ Useful in RLC circuits
✅ Can be solved graphically or mathematically

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✍️ Written by Sisira Senevirathna

TWO SINUSOIDAL WAVEFORMS – IN-PHASE / OUT-OF-PHASE
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⚡ Now, let us consider that the voltage v and the current i have a phase difference of 30°. Therefore:
Φ=30∘=6π​
📌 As both alternating quantities rotate at the same speed, they will have the same frequency. Therefore, this phase difference will remain constant for all instants in time. Thus, the phase shift of 30° between the two waveforms is represented by Φ (phi).

🔹 The voltage waveform starts at zero (0°) along the horizontal reference axis. However, at that same instant of time, the current waveform is still negative in value and does not cross this reference axis until 30° later.

📈 Then we can see that there exists a difference between the phases of the two waveforms as the current waveform crosses the horizontal reference axis, reaching its maximum peak and zero values 30° after the voltage waveform.

⚡ WHAT ARE LEADING AND LAGGING WAVEFORMS?

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🔹 Since the two sinusoidal waveforms are no longer in-phase, they must therefore be out-of-phase with each other by an amount determined by Φ. In this example: 30°.

✅ So we can say that the two waveforms are now 30° out-of-phase with each other.

🔹 Then the current waveform can also be said to be lagging behind the voltage waveform by the phase angle 30°. Thus the two waveforms have a Lagging Phase Shift.

📘 Expression for lagging current:
v(t)=Vmax​sin(ωt) i(t)=Imax​sin(ωt−Φ)
📌 Where current i lags voltage v by phase angle Φ.

🔹 Likewise, if the current i crosses the reference axis and reaches its peak before the voltage v, then the current waveform is said to be leading the voltage.

📘 Expression for leading current:
v(t)=Vmax​sin(ωt) / i(t)=Imax​sin(ωt+Φ)
📌 Where current i leads voltage v by phase angle Φ.

⚡ LEADING VS LAGGING SUMMARY

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✅ i lags v = current comes later than voltage.
✅ i leads v = current comes earlier than voltage.
✅ Phase angle describes the relationship between two same-frequency waveforms.

📌 In our example above, the two waveforms are out-of-phase by 30°. So we can correctly say:

➡️ i lags v by 30°
➡️ or v leads i by 30°

Both statements are correct depending on which waveform is used as the reference.

⚡ IMPORTANCE IN AC CIRCUITS

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⚡ In AC power circuits, describing the relationship between voltage and current sine waves is very important. This forms the basis of AC circuit analysis and Power in AC Circuits.

⚡ THE COSINE WAVEFORM

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📘 If a sinusoidal waveform is shifted right or left of 0° compared to another sine wave, the expression becomes:

A(t)=Amax​sin(ωt±Φ)=Amax​sin(2πft±Φ)

🔹 But if the waveform crosses the horizontal zero axis with a positive-going slope at 90° or π/2 radians before the reference waveform, the waveform is called a Cosine Waveform.

cos(x)=sin(x+90∘)
📌 The cosine wave has the same shape as the sine wave, but it is shifted by +90° or one-quarter cycle ahead.

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✍️ Written by Sisira Senevirathna

INSTANTANEOUS WAVEFORM EQUATION
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⚡ Instantaneous Waveform Equation

📘 Where:

🔹 Amax – is the maximum amplitude of the waveform.

🔹 ωt – is the angular frequency of the waveform in radian/sec (could also be written as 2πƒt).

🔹 Φ (phi) – is the phase angle in degrees or radians that the waveform has shifted either left or right from the reference point.

🔹 If the positive slope of the sinusoidal waveform passes through the horizontal axis before t = 0, then the waveform has shifted to the left. So Φ > 0.

⚡ Thus the phase angle will be positive in nature, +Φ, giving a leading phase angle. In other words, it appears earlier in time than 0°, producing an anticlockwise rotation of the vector.

🔹 Likewise, if the positive slope of the sinusoidal waveform passes through the horizontal x-axis some time after t = 0, then the waveform has shifted to the right. So Φ < 0.

⚡ Thus the phase angle will be negative in nature, -Φ, producing a lagging phase angle as it appears later in time after 0°, producing a clockwise rotation of the vector. Both cases are shown below.

🔹 Phase Relationship of a Sinusoidal Waveform

📌 Firstly, let’s consider that two instantaneous quantities such as a voltage v and a current i have the same frequency ƒ in Hertz and both start at t = 0.

⚡ As the frequency of the two quantities is the same, their angular velocity ω must also be the same. So at any instant in time we can say that the phase of voltage v will be the same as the phase of current i.

📈 Then the angle of rotation within a particular time period will always be the same. The difference between the two quantities of v and i will therefore be zero.

📘 That is: Φ = 0

Φ=0

⚡ As the frequency of the voltage v and the current i are the same, they must both reach their maximum positive, negative and zero values during one complete cycle at the same time (although their amplitudes may be different).

✅ Then the two alternating quantities, v and i, are said to be IN-PHASE.

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✍️ Written by Sisira Senevirathna