Above, we considered two basic concepts in electrical engineering - an ideal voltage generator and an ideal current generator.
An ideal voltage generator produces a given voltage U (pressure in the water-pipe analogy) at any load (resistance of an external circuit).
Moreover, in accordance with Ohm's law, I = U / R, even if R tends to zero, and the current increases to infinity.
The internal resistance of an ideal voltage generator is 0.
An ideal current generator produces a given current I (flow in the water-pipe analogy), even if the resistance of the external circuit tends to infinity. In this case, the voltage across the load also tends to infinity U = I * R.
The internal resistance of an ideal current generator is ∞.
Here you can see a certain symmetry, dualism.
We considered a capacitor C which can accumulate a charge (therefore it is called capacitance) C = Q / U. The larger the capacitance, the slower the voltage (pressure) increases when the charge U = Q / C is pumped into the capacitor.
If the charge capacity is very large (tends to infinity), then such a capacitor of infinite capacity will be an ideal voltage generator . It will never be discharged and at the same time it can produce a current of any magnitude, and the voltage across it will remain constant.
The symmetrical (dual) element to the capacitor will be inductance . Inductance is designated by the letter L (see diagram below).
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The picture is wrong. In the correct version, when the source was disconnected, a resistor was connected and the circuit remained closed.
Consider the following chain
Backfill question: What will be the voltage across the inductance at the first moment after switching the key S from the upper position to the lower one?
Hint: You don't have to put your brain out trying to figure out what sign the EMF of self-induction will have and what will happen to it next. It is necessary to apply a simple rule: The current in the inductor at the first moment of time after switching remains unchanged. Further apply Ohm's law.