Electronic – Transient Current in an LC circuit with a DC supply

capacitorinductortransient

enter image description hereHere is a LC circuit with a DC supply.

When the switch is closed at t=0 capacitor behaves as a short circuit while the inductor behaves as an open circuit as the voltage across the inductor immediately jumps to battery voltage.

Now we know that the current in inductor increases while in a capacitor current exponentially decreases with respect to time.

I wonder then how the current will overall behave in that circuit as when capacitor is short circuit inductor is open circuit at t=0 then when current in capacitor decreases then at the same time in inductor it is expected to increase. So what will be the overall shape of the current with respect to time. How will its graph look like and whats the theoratical explaination for that ?

Best Answer

schematic

simulate this circuit – Schematic created using CircuitLab

I would like to provide you the analytical approach of getting the result as mentioned by others already!

In the circuit shown, assuming the the initial conditions of the inductor current and capacitor voltages being:

$$ i_L(0^-)= 0 A \\ v_c(0^-)= 0 V $$ From the topology, $$ V_s = L \frac{di_L(t)}{dt} + v_c(t)$$ $$ v_c(t) = \frac{1}{C}\int i_L(t) dt$$ $$ V_s = L \frac{di_L(t)}{dt} + \frac{1}{C}\int i_L(t) dt $$ Now writing the laplace transform for the above we end up with $$V_s(s) = L[sI_L(s) - i_L(0^-)] + \frac{I_L(s)}{Cs} -v_c(0^-)$$ The variables in capital letters denote frequency domain representation and those in small letters denote it's time domain counterparts. Knowing the initial conditions for the state variables respectively, $$ i_L(0^-)= 0 A \ , v_c(0^-)= 0 V $$ $$V_s(s) = LsI_L(s) + \frac{I_L(s)}{Cs}$$ For a dc voltage source or step input excitation voltage of magnitude Vs, $$V_s(s)=\frac{V_s}{s}$$ $$\frac{V_s}{s} = \frac{LCs^2+1}{Cs}I_L(s)$$ $$I_L(s) = \frac{\frac{V_s}{L}}{s^2+\frac{1}{LC}}$$ Laplace transform of sin(at) is $$\frac{a}{s^2+a^2}$$ Similarly we get, $$i_L(t) = \sqrt\frac{C}{L}V_s \sin(\frac{t}{\sqrt{LC}})$$ Substituting iL(t) in other equations , one can get the cap voltages as, $$v_c(t) = V_s[1- \cos\frac{t}{\sqrt{LC}}]$$