Are there any electrical components that 'consume' current? That is, the current coming out of the component is different than the current going into it? For example, the following components consume voltage (outside of a voltage source):

  • Resistor
  • Diode
  • LED
  • etc...

Are there any similar components that do the same with current?

  • 4
    \$\begingroup\$ If it has only 2 connections what goes in, must come out. It's in series. \$\endgroup\$ Jan 31 '20 at 7:06
  • 3
    \$\begingroup\$ I think a black hole may do the job, within limits. The problem is that we're a bit short on these as actual electrical components, just now. But if you wait, they might make a come-back. Another option might arrive if you can provide a source of positrons -- say, a bunch of \$^{15}O\$? -- or if you can soak up electrons by converting \$^{40}K\$ to \$^{40}Ar\$. \$\endgroup\$
    – jonk
    Jan 31 '20 at 7:09
  • 1
    \$\begingroup\$ My usual "there are no umm... uninformed questions" approach failed me big time on this one. I am almost tempted to put a bounty on it. \$\endgroup\$
    – Maple
    Jan 31 '20 at 7:21
  • \$\begingroup\$ The voltage equivalent would be a component that connects in parallel to another component but has a different voltage across it. I wouldn't call it "consume". I'm not sure what you would call that. It's impossible anyway. \$\endgroup\$
    – user253751
    Jan 31 '20 at 11:17

Current is the flow of charge. If current flows in, but doesn't flow out, then charge must be accumulating. An example of such a phenomenon would be a gold-leaf electroscope.

In electronics we usually don't consider that effect or model it with lumped parts like parasitic capacitors to ground, so the current into the electroscope is matched by a hidden ground current.

Since Q = C\$\cdot\$V, for very small (parasitic) capacitances, significant charge means enormous voltages.

  • 1
    \$\begingroup\$ The results of accumulating charge is an increasing/decreasing electric field which means also current (without flow of charge): displacement current \$\endgroup\$
    – Curd
    Jan 31 '20 at 10:21
  • \$\begingroup\$ @Curd Maxwell’s equations.. \$\endgroup\$ Jan 31 '20 at 10:22

Voltage is not in fact consumed by any of those components.

Voltage is electrical potential energy, similar to the way that height is gravitational potential energy: to raise a mass m by delta-height h requires mechanical energy of mgh, and dropping mass m by delta-height h releases mechanical energy of mgh. In that sense, a ladder has a delta-height across it, but the ladder does not "consume" height.

More precisely, voltage is a field with a gradient, like a hillslope. A resistor is a conductor pathway that allows current to flow along that volage gradient. At various points along the length of the resistor, the voltage gradually changes -- for a resistor that change in voltage follows a straight line. This fact is exploited to make variable resistors; see https://www.ohmite.com/power-controls-rheostats/ for example. (For diodes, the change is non-linear, see the Shockley equation.)

Voltage is not consumed by any of those components. However, Energy is released -- Energy is Voltage x Current, and Power is the time rate of Energy. Resistors, Diodes (including LED), Transistors, etc. will increase in temperature due to power being dissipated into the surrounding ambient environment. So your intuition that "something" is being consumed is correct, but it's actually energy not voltage or current that is consumed.

It's worth looking at Maxwell's Equations, https://en.wikipedia.org/wiki/Maxwell%27s_equations -- Maxwell more or less invented the idea of a "field" as both a visualization (inspired by the patterns that iron filings form when near a strong magnet) as well as a rigorous mathematical model. The math is kind of heavy, but these equations are the best and most accurate model we have of how all sorts of electrical and magnetic stuff behaves. Fortunately there are simpler and easier models, like DC lumped-constant circuit analysis, KVL + KCL + Ohms Law, so we don't usually have to work a bunch of partial differential equations to design everything... but for understanding the basic principles, it's a good foundation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.