For a $L-C-R$ circuit, the power transferred from the driving source to the driven oscillator is $P=I^2 Z \cos \phi$.
When an $A C$ voltage of 220 V is applied to the capacitor $C$
The line that draws power supply to your house from street has
If a L-C circuit is considered analogous to a harmonically oscillating springblock system, which energy of the L-C circuit would be analogous to potential energy and which one analogous to kinetic energy?
If we consider a L-C circuit analogous to a harmonically oscillating springblock system. The electrostatic energy $\frac{1}{2} C V^2$ is analogous to potential energy and energy associated with moving charges (current) that is magnetic energy $\left(\frac{1}{2} L I^2\right)$ is analogous to kinetic energy.
Draw the effective equivalent circuit of the circuit shown in figure, at very high frequencies and find the effective impedance.
We know that inductive reactance $X_L=2 \pi f L$
and capacitive reactance $X_C=\frac{1}{2 \pi f C}$
For very high frequencies $(f \rightarrow \infty), X_L \rightarrow \infty$ and $X_C \rightarrow 0$
When reactance of a circuit is infinite it will be considered as open circuit. When reactance of a circuit is zero it will be considered as short circuited.
So, $C_1, C_2 \rightarrow$ shorted and $L_1, L_2 \rightarrow$ opened.
So, effective impedance $=R_{\text {eq }}=R_1+R_3$