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VECTOR SUBTRACTION OF TWO PHASORS━━━━━━━━━━━━━━━━━━                                       ⚑ This time, instead of adding...
26/04/2026

VECTOR SUBTRACTION OF TWO PHASORS
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⚑ This time, instead of adding both the horizontal and vertical components, we subtract them.

πŸ“˜ If:

A=x+jy
B=w+jz
πŸ“Œ Then phasor subtraction becomes:
Aβˆ’B=(xβˆ’w)+j(yβˆ’z)

βœ… This gives the new resultant phasor after subtraction.

⚑ THE 3-PHASE PHASOR DIAGRAMS

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πŸ“˜ Previously we looked at single-phase AC waveforms, where one rotating coil generates one sinusoidal voltage.

⚑ But if three identical coils are placed at an electrical angle of 120° to each other on the same rotor shaft, a three-phase voltage supply is generated.
120∘
⚑ BALANCED THREE-PHASE SUPPLY

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πŸ“Œ A balanced three-phase voltage supply consists of:

βœ”οΈ Three sinusoidal voltages
βœ”οΈ Equal magnitude
βœ”οΈ Same frequency
βœ”οΈ 120Β° phase difference between each phase

⚑ STANDARD PHASE COLORS

━━━━━━━━━━━━━━━━━━

πŸ”΄ Red
🟑 Yellow
πŸ”΅ Blue

πŸ“˜ Normal phase sequence:

R→Y→B

⚑ THREE-PHASE PHASOR ROTATION

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⚑ Just like single-phase phasors, three-phase phasors rotate anti-clockwise around a central point at angular velocity:

Ο‰ rad/s

πŸ“Œ All phase voltages are equal in magnitude, only their phase angles are different.

⚑ THREE-PHASE VOLTAGE EQUATIONS

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πŸ“˜ If Red phase is taken as the reference:

VRN​=V∠0∘
VYN​=Vβˆ βˆ’120∘
VBN​=V∠+120∘

πŸ“Œ Yellow phase lags Red by 120Β°
πŸ“Œ Blue phase leads Red by 120Β°

⚑ BALANCED SYSTEM RULE

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πŸ“˜ In a balanced three-phase system, the phasor sum is always zero:

Va​+Vb​+Vc​=0

βœ… This is one of the most important three-phase rules.

⚑ PHASOR DIAGRAMS SUMMARY

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βœ”οΈ Phasor diagrams are graphical representations of AC voltages and currents.
βœ”οΈ They are drawn as rotating vectors.
βœ”οΈ Reference phasor is drawn on the horizontal x-axis.
βœ”οΈ Only sinusoidal AC quantities can be represented.
βœ”οΈ All phasors must have the same frequency.
βœ”οΈ Leading phasors are ahead of reference.
βœ”οΈ Lagging phasors are behind reference.
βœ”οΈ Phasor length usually represents RMS value.
βœ”οΈ Different frequencies cannot be shown correctly on same diagram.
βœ”οΈ Two or more phasors can be added/subtracted into one resultant vector.
βœ”οΈ Horizontal side = Real part (x)
βœ”οΈ Vertical side = Imaginary part (y)
βœ”οΈ Hypotenuse = Resultant (r) vector
βœ”οΈ In balanced 3-phase systems each phasor is displaced by 120Β°.

⚑ NEXT LESSON

━━━━━━━━━━━━━━━━━━

πŸ“˜ In the next tutorial about AC Theory, we will study Complex Numbers in:

βœ”οΈ Rectangular Form
βœ”οΈ Polar Form
βœ”οΈ Exponential Form

πŸ”₯ Follow our page for more electronics knowledge and practical lessons.

✍️ Written by Sisira Senevirathna

PHASOR ADDITION USING RECTANGULAR FORM━━━━━━━━━━━━━━━━━━                                       ⚑ In AC circuit analysis,...
26/04/2026

PHASOR ADDITION USING RECTANGULAR FORM
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⚑ In AC circuit analysis, phasors can be added mathematically using the Rectangular Form. This method is faster and more accurate than drawing large phasor diagrams by scale.

⚑ STEP 1 – RESOLVE VOLTAGE Vβ‚‚

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πŸ“˜ Voltage Vβ‚‚ = 30V points in the reference direction along the horizontal zero axis. Therefore, it has:

βœ”οΈ Horizontal component only
βœ”οΈ No vertical component
Horizontal=30cos0∘=30V
Vertical=30sin0∘=0V
πŸ“Œ So the rectangular expression for Vβ‚‚ is:
V2​=30+j0
⚑ STEP 2 – RESOLVE VOLTAGE V₁

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πŸ“˜ Voltage V₁ = 20V leads Vβ‚‚ by 60Β°. Therefore, it has both horizontal and vertical components.

Horizontal=20cos60∘=20Γ—0.5=10V

Vertical=20sin60∘=20Γ—0.866=17.32V
πŸ“Œ So the rectangular expression for V₁ is:
V1​=10+j17.32

⚑ STEP 3 – ADD COMPONENTS

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πŸ“˜ Now add the horizontal and vertical parts separately.

Real Parts (Horizontal)
VH​=30+10=40V

Imaginary Parts (Vertical)

VV​=0+17.32=17.32V

πŸ“Œ Therefore resultant voltage becomes:

VT​=40+j17.32

⚑ STEP 4 – FIND MAGNITUDE OF Vβ‚œ

━━━━━━━━━━━━━━━━━━

πŸ“˜ Use Pythagoras Theorem to find total voltage.

VT​=/402+17.322
VT​=43.6V

βœ… Resultant magnitude = 43.6V

⚑ STEP 5 – FIND PHASE ANGLE

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πŸ“˜ Use trigonometric ratio

ΞΈ=tanβˆ’1(4017.32​)
θ=23.4∘

VT​=43.6∠23.4∘V

⚑ PHASOR SUBTRACTION

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πŸ“Œ Phasor subtraction is very similar to addition using rectangular form.

πŸ”Ή The only difference is that the vector difference becomes the other diagonal of the parallelogram between voltages V₁ and Vβ‚‚.

πŸ“˜ Simply subtract real parts and imaginary parts separately.

⚑ SUMMARY

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βœ… Convert phasors into horizontal & vertical components
βœ… Add real parts together
βœ… Add imaginary parts together
βœ… Use Pythagoras for magnitude
βœ… Use tangent for phase angle

πŸ”₯ Follow our page for more electronics knowledge and practical lessons.

✍️ Written by Sisira Senevirathna

CONSTRUCTING A GRAPHICAL REPRESENTATION OF A VECTOR━━━━━━━━━━━━━━━━━━                                       ⚑ As the sin...
26/04/2026

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 360Β° or 2Ο€, representing one complete cycle of the waveform.

360
∘
=2Ο€

πŸ“˜ 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.

πŸ“Œ Each position along the horizontal axis indicates the time elapsed since:
t=0
⚑ IMPORTANT VECTOR POSITIONS

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πŸ”Ή When the rotating vector is horizontal, the tip represents angles:

0∘, 180∘, 360∘

πŸ”Ή When the vector is vertical, it represents:

βœ”οΈ Positive peak value at 90Β° or Ο€/2
βœ”οΈ Negative peak value at 270Β° or 3Ο€/2

+Amax​ at 90∘,βˆ’Amax​ at 270∘

πŸ“Œ 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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πŸ”Ή A phasor represents a scaled voltage or current value of a rotating vector that is β€œfrozen” at some instant in time.

πŸ“˜ Example: frozen at angle 30Β°.

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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πŸ“˜ General mathematical expressions:

v(t)=Vmax​sin(Ο‰t)
i(t)=Imax​sin(Ο‰tβˆ’Ξ¦)

πŸ“Œ Here current i lags voltage v by angle Ξ¦.

πŸ”Ή If Ξ¦ = 30Β°, then current lags voltage by 30Β°.

⚑ LEADING AND LAGGING PHASORS

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βœ”οΈ If current comes later β†’ Current lags Voltage
βœ”οΈ If voltage comes earlier β†’ Voltage leads Current

πŸ“Œ 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.

πŸ“Œ Example in RLC Series Circuits where voltages are not in-phase.

βœ… If In-Phase (0Β° Shift)

Two voltages:

βœ”οΈ 50V + 25V = 75V

50+25=75

⚑ If Out-of-Phase

If phase angle exists, normal addition cannot be used.

πŸ“Œ We must use Phasor Diagrams and the Parallelogram Law.

⚑ PHASOR DIAGRAM WORKED EXAMPLE

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Consider:

βœ”οΈ V₁ = 20V
βœ”οΈ Vβ‚‚ = 30V
βœ”οΈ V₁ leads Vβ‚‚ by 60Β°

V1​=20V,V2​=30V,Ξ¦=60∘

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

VT=43.6V∠23.4∘

βœ… Therefore total voltage = 43.6V at angle 23.4Β°

⚑ USING A BIT OF MATHS

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

βœ”οΈ Horizontal = Cosine part
βœ”οΈ Vertical = Sine part

This analytical method is called Rectangular Form.

⚑ RECTANGULAR FORM

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A phasor is divided into:

βœ”οΈ Real part (x)
βœ”οΈ Imaginary part (y)

Z=xΒ±jy

πŸ“Œ This gives a complete mathematical description of magnitude and phase angle.

⚑ SUMMARY

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

πŸ”₯ Follow our page for more electronics knowledge and practical lessons.

✍️ Written by Sisira Senevirathna

PHASOR DIAGRAMS AND PHASOR ALGEBRA━━━━━━━━━━━━━━━━━━                                       ⚑ Phasor Diagrams are a visua...
26/04/2026

PHASOR DIAGRAMS AND PHASOR ALGEBRA
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⚑ Phasor Diagrams are a visual way of representing the magnitude and directional relationship between two or more alternating quantities.

⚑ WHAT ARE PHASOR DIAGRAMS OF A WAVEFORM?

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πŸ“˜ Phasor Diagrams present a graphical representation, plotted on a coordinate system, of the phase relationship between the voltages and currents within passive components or a whole circuit.

πŸ“Œ Generally, phasors are defined relative to a reference phasor, which always points to the right along the horizontal x-axis.

πŸ”Ή Sinusoidal waveforms of the same frequency can have a phase difference between themselves, which represents the angular difference of the two sinusoidal waveforms.

πŸ”Ή Also, the terms lead, lag, in-phase, and out-of-phase are commonly used to indicate the relationship of one sinusoidal waveform to another.

πŸ“˜ The generalised sinusoidal expression is:
A(t)=Am​sin(Ο‰tΒ±Ξ¦)
πŸ“Œ This represents the sinusoid in the time-domain form.

⚑ But when presented mathematically in this way, it can sometimes be difficult to visualize the angular or phasor difference between two or more sinusoidal waveforms.

βœ… One way to overcome this problem is to represent the sinusoids graphically in the phasor-domain form using Phasor Diagrams.

⚑ THE ROTATION OF VECTORS

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πŸ”Ή Basically a rotating vector, also regarded as a Phase Vector, is a scaled line whose length represents an AC quantity that has both:

βœ”οΈ Magnitude (Peak Value)
βœ”οΈ Direction (Phase Angle)

πŸ“Œ It is considered β€œfrozen” at some point in time.

πŸ”Ή A vector has an arrow head at one end which signifies:

➑️ Maximum value of the quantity (Vmax or Imax)
➑️ Direction of rotation

πŸ”Ή Generally, vectors pivot at one fixed point known as the point of origin or initial point.

πŸ“ Usually this is where the coordinate axes intersect:

(0,0)

πŸ”Ή One end of the vector is anchored there, while the arrowed end rotates freely.

⚑ DIRECTION OF ROTATION

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πŸ”„ Anti-clockwise rotation = Positive rotation
πŸ”„ Clockwise rotation = Negative rotation

πŸ“˜ The angular velocity is:

⚑ DIFFERENCE BETWEEN VECTOR AND PHASOR

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πŸ”Ή Although both terms are used for rotating quantities:

βœ”οΈ Vector magnitude = Peak value of the sinusoid
βœ”οΈ Phasor magnitude = RMS value of the sinusoid

πŸ“Œ In both cases, the phase angle, direction, and angular velocity remain the same.

⚑ WHAT DOES A PHASOR DIAGRAM LOOK LIKE?

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πŸ“˜ The phase of an alternating quantity at any instant in time can be represented by phasor diagrams.

πŸ“Œ Thus, phasor diagrams can be thought of as representing functions of time.

πŸ”Ή A complete sine wave can be constructed by a single vector rotating anti-clockwise at angular velocity Ο‰.

πŸ”Ή Then a phasor is a quantity that has both:

βœ”οΈ Magnitude
βœ”οΈ Direction

⚑ PHASOR ALGEBRA

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πŸ“Œ Vectors obey the parallelogram law of addition and subtraction, so they can be added graphically to produce a vector sum.

πŸ“˜ Phasors can also be represented mathematically in:

βœ”οΈ Rectangular Form
a+jb

βœ”οΈ Polar Form
βœ”οΈ Exponential Form

πŸ“Œ Phasor notation defines the effective RMS voltage and RMS current magnitudes.

⚑ SUMMARY

━━━━━━━━━━━━━━━━━━

βœ… Phasor diagrams simplify AC waveform analysis.
βœ… Show phase angle clearly.
βœ… Useful for voltage/current comparison.
βœ… Important for AC circuit calculations.

πŸ”₯ Follow our page for more electronics knowledge and practical lessons.

✍️ Written by Sisira Senevirathna

DIFFERENCE BETWEEN A SINE WAVE AND A COSINE WAVE━━━━━━━━━━━━━━━━━━                                       ⚑ Alternatively...
26/04/2026

DIFFERENCE BETWEEN A SINE WAVE AND A COSINE WAVE
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⚑ Alternatively, we can also say that a sine wave is a cosine wave that has been shifted in the other direction by --90°.

πŸ“Œ Either way, when dealing with sine waves or cosine waves with an angle, the following rules will always apply.

⚑ SINE AND COSINE WAVE RELATIONSHIPS

━━━━━━━━━━━━━━━━━━
cos(Ο‰t+ΞΈ)=sin(Ο‰t+Ξ¦+90∘)

sin(Ο‰t+ΞΈ)=cos(Ο‰t+Ξ¦βˆ’90∘)
πŸ“ˆ When comparing two sinusoidal waveforms, it is more common to express their relationship as either a sine or a cosine with positive-going amplitudes. This is achieved using the following mathematical identities.

⚑ SINE AND COSINE IDENTITIES
sin(AΒ±B)=sinAcosBΒ±cosAsinB
cos(AΒ±B)=cosAcosBβˆ“sinAsinB

⚑ THEREFORE

━━━━━━━━━━━━━━━━━━
βˆ’sin(Ο‰t)=sin(Ο‰tΒ±180∘)
βˆ’cos(Ο‰t)=cos(Ο‰tΒ±180∘)
Β±sin(Ο‰t)=cos(Ο‰tβˆ“90∘)
Β±cos(Ο‰t)=sin(Ο‰tΒ±90∘)
πŸ“Œ By using these relationships above, we can convert any sinusoidal waveform with or without an angular or phase difference from either a sine wave into a cosine wave or vice versa.

⚑ WHY THIS IS IMPORTANT

━━━━━━━━━━━━━━━━━━

πŸ”Ή Helps simplify AC circuit calculations.
πŸ”Ή Makes waveform comparison easier.
πŸ”Ή Useful in phase angle analysis.
πŸ”Ή Important in signal processing and electronics.

⚑ NEXT LESSON – PHASORS

━━━━━━━━━━━━━━━━━━
πŸ“˜ In the next tutorial about Phasors, we will use a graphical method of representing or comparing the phase difference between two sinusoids.

⚑ We can do this by looking at the phasor representation of a single-phase AC quantity along with some phasor algebra relating to the mathematical addition of two or more phasors.

πŸ”₯ Follow our page for more electronics knowledge and practical lessons.

✍️ Written by Sisira Senevirathna

TWO SINUSOIDAL WAVEFORMS – IN-PHASE / OUT-OF-PHASE━━━━━━━━━━━━━━━━━━                                       ⚑ Now, let us...
26/04/2026

TWO SINUSOIDAL WAVEFORMS – IN-PHASE / OUT-OF-PHASE
━━━━━━━━━━━━━━━━━━

⚑ Now, let us consider that the voltage v and the current i have a phase difference of 30°. Therefore:
Ξ¦=30∘=6π​
πŸ“Œ 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 30Β° between the two waveforms is represented by Ξ¦ (phi).

πŸ”Ή The voltage waveform starts at zero (0Β°) 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 30Β° later.

πŸ“ˆ 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 30Β° after the voltage waveform.

⚑ WHAT ARE LEADING AND LAGGING WAVEFORMS?

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πŸ”Ή 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 Ξ¦. In this example: 30Β°.

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

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

πŸ“˜ Expression for lagging current:
v(t)=Vmax​sin(Ο‰t) i(t)=Imax​sin(Ο‰tβˆ’Ξ¦)
πŸ“Œ Where current i lags voltage v by phase angle Ξ¦.

πŸ”Ή 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.

πŸ“˜ Expression for leading current:
v(t)=Vmax​sin(Ο‰t) / i(t)=Imax​sin(Ο‰t+Ξ¦)
πŸ“Œ Where current i leads voltage v by phase angle Ξ¦.

⚑ LEADING VS LAGGING SUMMARY

━━━━━━━━━━━━━━━━━━

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

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

➑️ i lags v by 30°
➑️ 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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πŸ“˜ If a sinusoidal waveform is shifted right or left of 0Β° compared to another sine wave, the expression becomes:

A(t)=Amax​sin(Ο‰tΒ±Ξ¦)=Amax​sin(2Ο€ftΒ±Ξ¦)

πŸ”Ή But if the waveform crosses the horizontal zero axis with a positive-going slope at 90Β° or Ο€/2 radians before the reference waveform, the waveform is called a Cosine Waveform.

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

πŸ”₯ Follow our page for more electronics knowledge and practical lessons.

✍️ Written by Sisira Senevirathna

INSTANTANEOUS WAVEFORM EQUATION━━━━━━━━━━━━━━━━━━                                       ⚑ Instantaneous Waveform Equatio...
26/04/2026

INSTANTANEOUS WAVEFORM EQUATION
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⚑ Instantaneous Waveform Equation

πŸ“˜ Where:

πŸ”Ή Amax – is the maximum amplitude of the waveform.

πŸ”Ή Ο‰t – is the angular frequency of the waveform in radian/sec (could also be written as 2Ο€Ζ’t).

πŸ”Ή Ξ¦ (phi) – is the phase angle in degrees or radians that the waveform has shifted either left or right from the reference point.

πŸ”Ή 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 Ξ¦ > 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.

πŸ”Ή 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 Ξ¦ < 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.

πŸ”Ή Phase Relationship of a Sinusoidal Waveform

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

⚑ As the frequency of the two quantities is the same, their angular velocity Ο‰ 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.

πŸ“ˆ 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.

πŸ“˜ That is: Ξ¦ = 0

Ξ¦=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).

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

πŸ”₯ Follow our page for more electronics knowledge and practical lessons.

✍️ Written by Sisira Senevirathna

PHASE DIFFERENCE AND PHASE SHIFT━━━━━━━━━━━━━━━━━━                                       ⚑ Phase Difference is used to d...
26/04/2026

PHASE DIFFERENCE AND PHASE SHIFT
━━━━━━━━━━━━━━━━━━

⚑ Phase Difference is used to describe the difference in degrees or radians between two or more alternating quantities when they reach their maximum or zero values.

πŸ”Ή What is the Phase Difference Between Two Waveforms?
Phase Difference, also known as Phase Shift or Phase Delay, defines the difference in time between two sinusoidal waveforms of the same frequency. The phase difference of two waveforms indicates how much one leads or lags behind the other.

πŸ“Œ Phasors are an effective way of analyzing the behavior of elements within an AC circuit when the circuit frequencies are the same. The result of adding together two phasors depends on their relative phase, whether they are in-phase or out-of-phase due to some phase difference.

πŸ”Ή Characteristics of a Sinusoidal Waveform
A Sinusoidal Waveform is an alternating quantity that can be presented graphically in the time domain along a horizontal axis using the trigonometric functions of sine or cosine.

πŸ“˜ Expressed as:
A(t) = Amax Γ— sin(Ο‰t)

πŸ“ˆ As a time-varying quantity, sinusoidal waveforms have a positive maximum value at time Ο€/2 (90Β°) and a negative maximum value at time 3Ο€/2 (270Β°), with zero values occurring along the horizontal baseline at:

➑️ 0, Ο€ and 2Ο€ points

πŸ”Ή Horizontal Shifting of an AC Waveform
However, not all sinusoidal waveforms of the same frequency will pass exactly through the zero axis point at the same time. For example, when comparing a voltage waveform to that of a current waveform.

⚑ Thus, compared to one reference waveform, some waveforms may be shifted to the right of 0Β° by some value represented by Ζ’(Ο‰t – tβ‚€), while others may be shifted to the left of 0Β° by some value represented by Ζ’(Ο‰t + tβ‚€). That is, the waveform moves along the zero axis without changing its shape.

πŸ”Ή This difference produces an angular shifting of the sinusoidal waveforms creating what is known as a Phase Difference between them. Any sine wave that does not pass through zero at t = 0 will generally have a phase shift in degrees or radians of some amount.

πŸ”Ή How to Measure Phase Difference in Waveforms?
The difference or phase shift of a Sine Wave is the angle, in degrees or radians, that a waveform has shifted left or right from a certain reference point along the horizontal zero axis compared to another.

πŸ“Œ In other words, it is the lateral difference between two or more waveforms along a common axis of the same frequency.

πŸ”Ή The primary symbol for electrical phase difference is represented by the Greek letter Ξ¦ (phi) or Ο† (phi). Both symbols represent the same angle and therefore phase shift.

πŸ“ Then the difference between phases (Ξ¦) of an alternating waveform can vary from 0Β° to 360Β° or 0 to 2Ο€ radians depending on the angular units used.

⏱️ Phase difference can also be expressed as a time shift of Ο„ (tau) in seconds representing a fraction of the time period T. Example: +10mS or –50uS.

⚑ But generally it is more common to express the difference between two sinusoidal waveforms as an angular measurement.

πŸ“˜ So the equation for the instantaneous value of a sinusoidal voltage or current waveform developed previously must be modified to take account of the phase angle of the waveform. This new general expression becomes.

πŸ”₯ More electronics lessons coming soon! Follow our page and stay updated.

✍️ Written by Sisira Senevirathna

25/04/2026
INSTANTANEOUS WAVEFORM EQUATIONβ–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬                                       ⚑ Instantaneous Waveform Equat...
06/04/2026

INSTANTANEOUS WAVEFORM EQUATION
β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬β–¬

⚑ Instantaneous Waveform Equation

πŸ“Œ Where:
πŸ”Ή Amax – is the maximum amplitude of the waveform.
πŸ”Ή Ο‰t – is the angular frequency of the waveform in radian/sec (could also be: 2Ο€Ζ’t)
πŸ”Ή Ξ¦ (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 β€œbefore” t = 0 then the waveform has shifted to the left. So Ξ¦ > 0. Thus the phase angle will be positive in nature, +Ξ¦ giving a β€œleading phase angle”. In other words it appears earlier in time than 0Β° producing an anticlockwise rotation of the vector.

➑️ Likewise, if the positive slope of the sinusoidal waveform passes through the horizontal x-axis some time β€œafter” t = 0 then the waveform has shifted to the right. So Ξ¦ < 0. Thus the phase angle will be negative in nature -Ξ¦ producing a β€œlagging phase angle” as it appears later in time after 0Β° producing a clockwise rotation of the vector. Both cases are shown below.

πŸ”„ Phase Relationship of a Sinusoidal Waveform

πŸ“Œ Firstly, let’s consider that two instantaneous quantities such as a voltage, v and a current, i have the same frequency Ζ’ in Hertz and both start at t = 0. As the frequency of the two quantities is the same their angular velocity, Ο‰ 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 the current, i.

πŸ“Š 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: Ξ¦ = 0.

βœ… 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 β€œin-phase”.

πŸ“‰ Two Sinusoidal Waveforms – β€œin-phase”

⚑ Now let us consider that the voltage, v and the current, i have a phase difference between themselves of 30Β°. Therefore, (Ξ¦ = 30Β° or Ο€/6 radians).

πŸ“Œ 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 30Β° between the two waveforms is represented by phi, Ξ¦ as shown below.

πŸ“ 30Β° Phase Difference of a Sinusoidal Waveform

πŸ“Š The voltage waveform above starts at zero (0Β°) 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 30Β° later.

πŸ“ 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 30Β° after the voltage waveform.

❓ What are Leading and Lagging Waveforms?

⚠️ Since the two sinusoidal waveforms are no longer β€œin-phase”, they must therefore be β€œout-of-phase” with each other by an amount determined by phi, Ξ¦ and in our simple example this is: 30Β°. So we can say that the two waveforms are now 30Β° β€œout-of-phase” with each other.

⏱️ Then the current waveform can also be said to be β€œlagging” behind the voltage waveform by the phase angle, Ξ¦ of 30Β°. Thus the two waveforms have a Lagging Phase Shift and the expression for both the voltage and current above is given as:

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πŸ“’ Follow us for more electronics insights!

✍️ Sisira Senevirathna

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