From the circuit theory it is known that an element whose response is linear is called linear element such as resistor, capacitor. For example, in case of resistor the slope between V and I is a straight line, but I have some confusion what about current source or voltage source? Is it linear element or non-linear?

Also if we see the graph of its V-I we can see that it is also constant: straight vertical or horizontal line.

  • \$\begingroup\$ I've never seen a linear correlation between V and I for a capacitor but it's still a linear component so maybe your premise is incorrect. \$\endgroup\$ – Andy aka Oct 26 '16 at 17:21
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    \$\begingroup\$ @Andyaka, in the phasor domain: \$v = i(1/j\omega{}C)\$. That's a linear relationship. \$\endgroup\$ – The Photon Oct 26 '16 at 17:23
  • \$\begingroup\$ @mtg: This old question applies, in cases of non-ideal sources (non-zero resistance for voltage sources or non-zero conductance for current sources). \$\endgroup\$ – The Photon Oct 26 '16 at 17:25
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    \$\begingroup\$ Linear in a geometric or algebraic sense is not the same as linear in a systems sense. For example, y=mx +c is geometrically linear but, unless c =0, it would not be linear if it related to the input/output characteristic of an amplifier. To illustrate: twice the input would not produce twice the output, to use but one property of a linear system. \$\endgroup\$ – Chu Oct 26 '16 at 19:49
  • \$\begingroup\$ I stand corrected. I found two references that both say the independent voltage and current sources and be viewed as a special nonlinear resistor. One bases its argument on the fact that its graph doesn't go through the origin unless it has the value of zero. The other uses the argument that the source graph moves parallel when the value is changed and not along the line. \$\endgroup\$ – owg60 Oct 27 '16 at 23:49

Only elements that have both inputs and outputs can be considered to be linear. "Linear" is a term that applies to a system with inputs and outputs, since it describes a particular kind of relationship between them. This means that simple voltage and current sources (when seen as ideal circuit elements) are neither linear nor non-linear.

In essence, a system is linear if it follows the superposition principle. Note that in reality, all physical circuit elements are non-linear, but in practice they behave as linear systems for some input range.

See also this answer for the difference between linear and affine, which is important for some circuit elements.

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    \$\begingroup\$ Good answer. It's probably worth adding that a network that contains a source itself won't obey the superposition principle, so we should probably not call it a linear network. \$\endgroup\$ – The Photon Oct 26 '16 at 17:31
  • \$\begingroup\$ @ThePhoton Yes, that's a good point. Alternatively, any circuit that produces a non-zero output for zero input (from \$t=-\infty\$) is (strictly speaking) non-linear. \$\endgroup\$ – MBaz Oct 26 '16 at 17:56
  • \$\begingroup\$ Exactly... Linearity can only be defined within a range of operation. Outside of that, all elements are non linear, including a simple wire. \$\endgroup\$ – soosai steven Oct 26 '16 at 18:41
  • \$\begingroup\$ accept Only elements that have both inputs and outputs can be considered to be linear. ? cant not understand. this statement. \$\endgroup\$ – mtg Oct 29 '16 at 17:08

Ideal sources are linear. On way to think about this is solving simple circuits in the VI plane. Since I=V/R, a resistor which is a linear element is a line through the origin with slope 1/R. If we connect this to a voltage source, we solve it graphically. The graph of the voltage source is a vertical line at the value of the voltage. The intersection of resistor line and the voltage line will give the current. Now to be linear means that add to or multiply the input, the voltage source in this case, the output will be will be added to or multiplied by that same amount. So now consider if you took 2*V. The vertical line will be twice as far from the origin and so will the current. This is true because all the components including the source are linear. If you put a diode equation on this graph doubling the voltage would not double the current. The diode is a nonlinear element. You can see that the ideal current source is also linear because its graph is a horizontal line at the given current.

Another example of this is if we put a voltage divider across an ideal voltage source. Let's say you set it up to divide by 5. If I put 5 volts in, I get 1 volts out. Now if you multiply this voltage by 3 you get 15 volts in and 3 Volts out. The output is three times as much because all the elements including the source are linear.

  • \$\begingroup\$ thank you sir for you responce...so it mean both voltage and current sources are linear..? \$\endgroup\$ – mtg Oct 29 '16 at 17:11
  • \$\begingroup\$ No, I was wrong. See the last comment I made under your original question. \$\endgroup\$ – owg60 Oct 29 '16 at 17:20

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