If the impedance at the load has the form of \$Z = R + jX\$, where \$R\$ and \$X\$ are positive real numbers, then then network is called inductive. If \$Z = R - jX\$, then the network is called capacitive.

Why is this the case and what do we do when there is both capacitance and inductance present at the load (i.e. \$RLC\$ network)?


4 Answers 4


A capacitance \$C\$ has impedance

$$Z_{C} = \frac{1}{j\omega C}$$

But note that

$$\frac{1}{j} = -j$$

so equivalently

$$Z_{C} = -j\frac{1}{\omega C}$$

A capacitor's impedance therefore has a negative imaginary part.

An inductor \$L\$ has impedance

$$Z_{L} = j\omega L$$

and therefore has a positive imaginary part.

So if the load has a positive imaginary part then it behaves more like an inductor, and if it has a negative imaginary part then it behaves more like a capacitor.

If both inductors and capacitors are present then simply find the equivalent impedance of the load network. If the imaginary part of the equivalent impedance is positive then the load is inductive, if it is negative then it is capacitive, and if it is zero then it is resistive. An inductive load network has an overall higher impedance as the frequency increases even if there are capacitor(s) in the network, and a capacitive load network has an overall lower impedance as the frequency increases even if there are inductor(s) in the network.

It can be important to distinguish between the two cases in order to understand the frequency response of a circuit. For example, if a voltage amplifier has an inductive load then the amplifier will perform better at higher frequencies where the load has a higher impedance (the ideal load for a voltage amplifier is infinite in order to maximize the voltage gain). However, if the voltage amplifier's load is capacitive then it will perform better at lower frequencies since the load will then have a higher impedance at lower frequencies.

  • \$\begingroup\$ Hi, could you also elaborate on the physical meaning of having a inductive or capacitive network? Why do people feel the need to delineate between these two in the field? \$\endgroup\$
    – Fraïssé
    Sep 26, 2014 at 15:52
  • \$\begingroup\$ @IllegalImmigrant I added to the end of my answer. Hopefully that will help. \$\endgroup\$
    – Null
    Sep 26, 2014 at 16:09

At any given frequency the network has an impedance that is just a single (complex) number, and if the imaginary part is negative then it's capacitive (at that frequency) and if the imaginary part is positive then it's inductive (at that frequency). If the real part is much bigger than the imaginary part, then it behaves mostly like a resistor.

It's not uncommon for parts to be mostly resistive at one frequency, mostly capacitive at another, and inductive at another frequency. See, for example, ferrite beads.

If you think of the applied voltage waveform as being composed of many different frequencies, the current will be the sum of the responses of each frequency component to the impedance at that frequency (assuming a linear time-invariant system).


Why is this the case...

Impedance is defined as the ratio of the voltage phasor and a current phasor. For more information on phasors, you can check here.

Basically, a phasor is a complex number, and as such, supports several types of representation, including the rectangular (which is what you expressed) and the polar form or module / angle. If we express the current and voltage as complex numbers, the impedance is a complex number, but you can not say it's a phasor.

Phasors are applied to the analysis of steady state of an electrical circuit. In the case of an inductive circuit, the current is delayed in phase with the voltage, while in a capacitive circuit, the current is advanced in phase with the voltage. How does this relate to the impedance? as the impedance is a complex number, the angle of the same, corresponds to the phase shift between voltage and current.
If this angle is positive, the voltage is ahead of the current, or the current is late with respect to the voltage, so this is an inductive circuit. A similar analysis can be raised to a capacitive circuit.
A positive angle corresponds to a complex number whose real and imaginary parts are positive.

...what do we do when there is both capacitance and inductance present at the load (i.e. RLC network)?

In a circuit containing both capacitive and inductive elements prevails one identity, that is, the circuit will eventually inductive or capacitive, according to the value of the total impedance equivalent. Is it possible cancel each inductive and capacitive elements? Yes. In this case, it is said that the circuit is in resonance, and from the point of view of the source, is a pure resistive circuit, although containing reactive elements.

  • \$\begingroup\$ Hi, thank you. What do you mean by "eventually inductive or capacity"? What does it physically mean for a circuit to be one or the other \$\endgroup\$
    – Fraïssé
    Sep 26, 2014 at 15:52
  • \$\begingroup\$ @IllegalImmigrant When calculating the equivalent impedance of a circuit, the reactive part is a positive number (inductive loop) or a negative number (capacitive circuit). Then, while the circuit containing both types of components, the overall effect will be of only one kind or another. Mathematically, the complex number representing the impedance will have a positive imaginary part (inductive) or, negative (capacitive). \$\endgroup\$ Sep 26, 2014 at 15:56
  • \$\begingroup\$ @IllegalImmigrant From the physical standpoint, this means that the current from the source to the load, will be delayed in phase with regard to the voltage (inductive circuit) or, advance in phase with regard to the voltage (capacitive circuit). \$\endgroup\$ Sep 26, 2014 at 15:58

Inductive reactance(\$X_L\$) and Capacitive reactance(\$X_C\$) both depend on the frequency. $$X_L=2\pi f L$$

$$X_C=\frac{1}{2\pi f C}$$

\$X_L\$ would increase as the frequency increases and \$X_C\$ would decrease as the frequency increases. \$j\$ indicates it as a complex term of Impedance \$Z\$.


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