$A$ resistor of resistance $40 \Omega$,a capacitor of capacitive reactance $20 \Omega$,and an inductor of inductive reactance $50 \Omega$ are connected in series to an $ac$ source of $100 \ V$. The current through the circuit is (in $A$)

Obtain an analytical solution for the relation of phase between instantaneous current and voltage for an $LCR$ series $AC$ circuit.

For the circuit shown in the figure,the current through the inductor is $0.9\,A$ while the current through the capacitor is $0.4\,A$. Then

In the circuit shown,$C = \frac{\sqrt{3}}{2} \times 10^{-3} \, F$,$R_2 = 20 \, \Omega$,$L = \frac{\sqrt{3}}{10} \, H$,and $R_1 = 10 \, \Omega$. The current in the $L-R_1$ branch is $I_1$ and in the $C-R_2$ branch is $I_2$. The voltage of the $A.C.$ source is given by $V = 200\sqrt{2} \sin(100t) \, V$. The phase difference between $I_1$ and $I_2$ is:

$A$ series $LCR$ circuit has $R=200 \Omega$, $L=663 \text{ mH}$, and $C=26.5 \mu F$. The applied alternating voltage has an amplitude of $50 \text{ V}$ and a frequency of $60 \text{ Hz}$ such that $X_{L}=250 \Omega$ and $X_{C}=100 \Omega$. The peak current is: (in $\text{ A}$)

$A$ series $AC$ circuit containing an inductor $(20\,mH),$ a capacitor $(120\,\mu F)$ and a resistor $(60\,\Omega)$ is driven by an $AC$ source of $24\,V$ and $50\,Hz.$ The energy dissipated in the circuit in $60\,s$ is