In mathematics, i=√(-1), meaning "the square root of minus 1".
In physics and engineering calculations j is also used to represent √(-1).
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For even more information about the meaning of "i" as used in mathematics, physics and electrical engineering, see the answer to the Related Question "What's the physical meaning of i aka the square root of -1 Does it have any physical existence and if not then why is it used to describe some real physical quantities like some terms in electricity?". You can reach it using the link shown below!
Additional AnswerWhile, in mathematics, the symbol, 'i', represents an imaginary number, in electrical engineering, we use the letter 'j' -this is because a lower-case 'i' is used to represent an instantaneous-value of current, so 'j' is used to avoid confusion.
You can think of the symbol 'j' as an 'operation' on a phasor (vector). +j indicates that the phasor has been rotated counterclockwise by 90o. -j indicates that the phasor has been rotated clockwise by 90o.
So, suppose a phasor represents a current, I, represented by an arrowed line in the horizontal, positive, position (0o). Then:
So the operator 'j' can be used to indicate whether a phasor is lying at 0o (e.g. I), 90o (e.g. jI), 180o(e.g. -I), or 270o (e.g. -jI).
In our example, we have used current, but the j-operator can be applied to voltage, impedance, etc.
Although this might sound rather complicated, the use of the j-operator allows very complex A.C. circuits to be then solved mathematically, rather than having to resort to what can be exceptionally complicated phasor diagrams.
For example, suppose we have an impedance written as (10 +j20) ohms, this indicates a resistance of 10 ohms, and an inductive reactance of 20 ohms. Suppose we have a second impedance written as (30 - j15) ohms, this indicates a resistance of 30 ohms and an capactive reactance of 15 ohms. We can add the 'real' components algebraically, and we can add the 'imaginary' components algebraically, giving us a total impedance of (10 + 30) + j(20 - 15), or (40 +j5) ohms. If we were to solve the addition of these two impedances using a phasor diagram, it would be more difficult and time consuming than by using j-notation.
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