In an $L-R$ circuit,the inductive reactance is equal to the resistance $R$ in the circuit. An emf $E = E_0 \cos \omega t$ is applied to the circuit. The power consumed in the circuit is

In an $L-R$ circuit,the value of $L$ is $(\frac{0.4}{\pi}) \, H$ and the value of $R$ is $30 \, \Omega$. If an alternating e.m.f. of $200 \, V$ at $50 \, Hz$ is connected to the circuit,what are the impedance and the current in the circuit?

An inductor of $\left(\frac{100}{\pi}\right) mH$,a capacitor of capacitance $\left(\frac{10^{-3}}{2 \pi}\right) F$,and a resistor of $10 \Omega$ are connected in series with an $AC$ voltage source of $110 \text{ V}, 50 \text{ Hz}$. The tangent of the phase angle $\phi$ between the voltage and the current is:

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:

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 $\Omega$)

$A$ capacitor of reactance $4 \sqrt{3} \Omega$ and a resistor of resistance $4 \Omega$ are connected in series with an $AC$ source of peak value $8 \sqrt{2} \text{ V}$. The power dissipation in the circuit is . . . . . $\text{W}$.