In a simple $L-R$ circuit with an $A.C.$ source,the potential difference at any instant across the inductor and the resistor are $V_L$ and $V_R$ respectively,and the $A.C.$ source has a potential difference $V_{AC}$ at the same instant. Then:

In an $L-C-R$ circuit,the potential difference across the inductor is $60\,V$,across the capacitor is $30\,V$,and across the resistor is $40\,V$. The supply voltage will be: (in $,V$)

$A$ circuit element $X$ when connected to an a.c. supply of peak voltage $100\,V$ gives a peak current of $5\,A$ which is in phase with the voltage. $A$ second element $Y$ when connected to the same a.c. supply also gives the same value of peak current which lags behind the voltage by $\frac{\pi}{2}$. If $X$ and $Y$ are connected in series to the same supply,what will be the rms value of the current in ampere?

In the circuit shown below,the $ac$ source has voltage $V = 20\cos (\omega t)$ volts with $\omega = 2000 \,rad/sec$. The amplitude of the current will be nearest to:

$A$ pure inductor of $25.48 \ mH$ and a pure resistor of $8 \ \Omega$ are connected in series with an $AC$ source of frequency $50 \ Hz$. The phase difference between current $(I)$ and voltage $(V)$ in this circuit is . . . . . . . (in $^{\circ}$)

$A$ coil of inductance $0.50 \; H$ and resistance $100 \; \Omega$ is connected to a high-frequency supply $(240 \; V, 10 \; kHz)$.
$(a)$ What is the maximum current in the coil?
$(b)$ What is the time lag between the voltage maximum and the current maximum?
Hence,explain the statement that at very high frequency,an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a $DC$ circuit after the steady state?