Can anyone explain how does the current amplitude go to infinity, at
resonance?
The brief answer is that the AC steady state current isn't infinite but, rather, the circuit has no AC steady state solution.
Recall that one of the assumptions justifying AC (phasor) analysis is that the circuit is in AC steady state, i.e., that all transients have decayed.
For the circuit given, the time domain solution for the current is proportional to
$$i(t) \propto t \cos\left( \frac{t}{\sqrt{LC}}\right),\, t \ge 0$$
The amplitude of the current starts at zero and grows linearly with time once the switch is closed but for any value of time \$t\$, the current is finite, i.e., the current is never infinite.
Note that this solution has no sinusoidal steady state - the amplitude does not approach a constant as \$t \rightarrow \infty\$ so this solution has no phasor representation and, thus, we should not be surprised that applying phasor analysis to this problem produces an undefined division by zero result.
Given the solution for the current, one can solve for the voltages across the inductor and capacitor as well as the energy stored in each as a function of time.
Different initial conditions will have different initial energies but will not affect the main result that the amplitude of the current will grow without bound once the switch is closed.
Can you please list the steps that lead to the derivation of the
expression i(t)∝ tcos(t/√LC),t≥0 ?
By KVL, we have
$$v_S = v_L + v_C = L\frac{di}{dt} + \frac{1}{C}\int_0^ti(\tau)d\tau$$
(we assume zero initial voltage across the capacitor).
Differentiating both sides with respect to time and dividing through by \$L\$ yields
$$\frac{d^2i}{dt^2} + \frac{1}{LC}i = \frac{1}{L}\frac{dv_S}{dt}$$
Assuming \$v_S = V\cos\omega_0 t\$ yields the following non-homogeneous 2nd order ODE:
$$\frac{d^2i}{dt^2} + \frac{1}{LC}i = -\frac{\omega_0 V}{L}\sin \omega_0 t $$
where
$$\omega_0 = \frac{1}{\sqrt{LC}} $$
Assume a solution of the form
$$i(t) = t\left(A \cos \omega_0 t + B \sin \omega_0 t \right) $$
Substitute this \$i(t)\$ into the ODE to find
$$A = \frac{V}{2L}, B = 0 $$
The resonant frequency of an LC circuit is given by
$$ f = \frac {1}{2 \pi \sqrt {LC}} $$
and plugging in the values you have chosen gives
$$ f = \frac {1}{2 \pi \sqrt {0.1 \cdot 100 \mu}} = \frac {1}{2 \pi \cdot 0.00316 } = 50.3~Hz$$
So your resonant frequency is almost exactly on the mains frequency of 50 Hz.
- Why does resonance increases the voltage?
It's like pushing a swing or pendulum. If you do it at the natural frequency of oscillation and at the right point in the cycle then every push will boost a little bit more. In the case of the pendulum or swing air resistance will eventually limit the amplitude of the oscillation. In the LC circuit the internal resistance will cause losses which increase with increasing amplitude until an equilibrium is reached.
- Why the voltage does not increase at a frequency of say 2 KHz?
See the formula.
- If I made this experiment practically, Will the transformer wear out?
Transformers lifespan is mostly determined by temperature which is usually a result of too-much current or insulation breakdown which is a result of too high a voltage being applied. You would need to know the winding insulation breakdown voltage to figure that out. At 50 V you are unlikely to have a problem.
Best Answer
If the passive resonant circuit has a high 'Q' then energy can build up in the circuit over many cycles. It will similarly die down naturally over many cycles.
The intensity is not "ever increasing", however. As the intensity increases, so do the losses and at some point the losses equals the input power, so you have an equilibrium.
Imagine a tuning fork. If you keep exciting it, the vibrations will build up (as will the losses) and at some point the metal shape might change if the excitation is powerful enough, but it's unlikely. Most real circuits have a rather lower Q than the mechanical Q of a tuning fork.
In a real LC resonant circuit, the losses are usually due to inductor resistance, to core losses (if a core is used) and to electromagnetic radiation, especially at higher frequencies. Capacitor dielectric losses contribute too. Superconducting circuits can have huge Q's (in the thousands), resonant circuits made with parts from your favorite distributor, much more disappointing.