An inductive circuit contains a resistance of $10 \, \Omega$ and an inductance of $2.0 \, H$. If an $AC$ voltage of $120 \, V$ and frequency $60 \, Hz$ is applied to this circuit,the current would be nearly: ..... $A$

$A$ coil has $L = 0.04 \, H$ and $R = 12 \, \Omega$. When it is connected to a $220 \, V$,$50 \, Hz$ supply,the current flowing through the coil in amperes is: (in $.7$)

In the given circuit, the r.m.s. value of the current through the resistor $R$ is:

An a.c. source of $15 \, V, 50 \, Hz$ is connected across an inductor $(L)$ and resistance $(R)$ in series. The $R.M.S.$ current of $0.5 \, A$ flows in the circuit. The phase difference between the applied voltage and current is $\left(\frac{\pi}{3}\right)$ radian. The value of resistance $(R)$ is $\left(\tan 60^{\circ}=\sqrt{3}\right)$. (in $\Omega$)

$A$ series combination of resistor $R$ and capacitor $C$ is connected to an a.c. source of angular frequency $\omega$. Keeping the voltage same, if the frequency is changed to $\omega/3$, the current becomes half of the original current. Then the ratio of capacitive reactance and resistance is

$A$ sinusoidal voltage $V(t) = 210 \sin(3000t) \text{ V}$ is applied to a series $LCR$ circuit in which $L = 10 \text{ mH}$,$C = 25 \mu\text{F}$,and $R = 100 \Omega$. The phase difference $(\Phi)$ between the applied voltage and resultant current will be