I find it helpful to think of capacitors and inductors to be complimentary.
simulate this circuit – Schematic created using CircuitLab
Figure 1. Ideal and imperfect components.
- Capacitors store energy in an electric field. Inductors store energy in a magnetic field.
- A capacitor holds energy when open circuit. An inductor holds energy when short circuited.
- Capacitors lose energy through parallel leakage resistance. Inductors lose energy through series resistance.
- Capacitors "like" to keep the voltage across them constant. Inductors like to keep the current through them constant.
- When a capacitor is short circuited the resultant current is very high. When an inductor is open-circuited the resultant voltage is very high.
... when an inductor is disconnected the magnetic field starts collapsing inducing a very high voltage, wouldn't this voltage breaks down any transistors used in the switching from the charging to the discharging circuit?
Yes it would but there's a simple solution:
Figure 2. A simple buck converter. Source: All About Circuits.
In Figure 2 S is the transistor switch similar to that mentioned in your question. When it is switched the inductor tends to maintain current in the direction of I. Since the right side of L is "held" by C and current is to keep going then the left side of L goes negative to try to maintain current. Whe the voltage reaches -0.7 V D starts to conduct and maintains the current through L keeping it "happy" and avoiding a transient high voltage.
You will see this arrangement more commonly in snubber diodes on relay coils.
Figure 3. A typical relay control circuit. Without D1 the inductance of the relay coil would generate a large negative voltage on switch off. This would be likely to destroy Q1 . The diode limits the negative excursion on Q1 collector to -0.7 V.