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The operator, 'j', is used to indicate a phasor quantity that has been rotated, counterclockwise, through an angle of 90 degrees.

So, if (for example) the operator is applied to a voltage U, then it is written as jU, which indicates means that the voltage lies along the vertical positive axis. A further operation by j, results in jjU, or j2 U, which means that the voltage lies along the horizontal negative axis -so, j2 is equivalent to -1U (or j is equivalent to the square-root of -1) or, simply, -U.

A further operation by j, results in the voltage lying along the negative vertical axis: that is: jjjU = jj2U=-jU.

But to answer your question, for inductive reactance (XL), we express impedance (Z) a follows: Z = R+jXL and, for capacitive reactance (XC), we express impedance as Z = R - jXC (the L and C should be subscripts).

Strictly-speaking the operator j doesn't actually apply to impedance, because impedance, resistance, and reactance or not phasor (vector) quantities. So if we wanted to be strictly accurate, the above equations should be written as:

(E/I) = (UR/I) + j (UL/I) and (E/I) = (UR/I)- j (UC/I)

...but this is being rather pedantic.

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Reactance is -1/2 pi F C so a 25 uF capacitor at 400 Hz would have a reactance of about -15.9 ohms. The negative sign indicates that capacitive reactance is leading, with current leading voltage.AnswerI would take issue with the previous answer that capacitive reactance is expressed as a negative value, or that it is 'leading'. Reactance is not a vector quantity, so it neither leads nor lags anything. In a (theoretically) purely capacitive circuit, it is the load current that leads the supply voltage. However, when using complex notation, capacitive reactance is expressed as -j 15.9 ohms, where 'j' is called an 'operator' -but even this does not mean that the reactance is 'leading', as it defines reactance in terms of a current phasor -in other words, the '-j' refers to the relative position of current to voltage, not reactance to impedance.


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What is the reactance of a 25 uf capacitor when the applied frequency is 400 hz?

Reactance is -1/2 pi F C so a 25 uF capacitor at 400 Hz would have a reactance of about -15.9 ohms. The negative sign indicates that capacitive reactance is leading, with current leading voltage.AnswerI would take issue with the previous answer that capacitive reactance is expressed as a negative value, or that it is 'leading'. Reactance is not a vector quantity, so it neither leads nor lags anything. In a (theoretically) purely capacitive circuit, it is the load current that leads the supply voltage. However, when using complex notation, capacitive reactance is expressed as -j 15.9 ohms, where 'j' is called an 'operator' -but even this does not mean that the reactance is 'leading', as it defines reactance in terms of a current phasor -in other words, the '-j' refers to the relative position of current to voltage, not reactance to impedance.


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