The switch stores energy but cannot close

What happens if a switch S is closed 4 years ago?

When the switch S was closed four years ago, the initial charge of capacitor C1 was q0 and its capacitance was C1. Capacitor C2 initially had no charge (0) and a capacitance of C2. During the charging process of capacitor C1 charging capacitor C2, some energy was dissipated in the resistor and transferred to heat.

What happens when you close a switch?

Here are my thoughts: Before closing the switch, there will be a steady finite current I in the circuit.The moment we close the switch, there can be no current passing through R, else there will be some finite voltage on the capacitor ⟶ V ˙ C will be infinite ⟶ I C =C ⋅ V ˙ C will be infinite which cannot be.

What happens if a circuit is open?

But as soon as the switch is opened, the current would become 0, which make the magnetic field lines disappear suddenly, which according to Faraday's law must induce an emf. But as the circuit is open no current will flow in it (according to my teacher, charge can never accumulate in a circuit.

What happens if an inductor sparks a switch?

What happens in the real world is that the inductor creates enough emf to form a spark in the switch. This means the switch no longer acts like an ideal switch. In the real world, we call this effect "flyback.". It can damage components, so we typically design circuits to prevent this from occuring.

How inductive energy is dissipated?

The inductive energy is dissipated by producing a spark at the switch terminals. The core of the spark is a thread of very hot, ionized gas which produces light and noise with some of the energy, and heat in the gas with the rest of the energy. Thus, energy is conserved.