$A$ bulb and a capacitor are in series with an $ac$ source. On increasing the frequency,how will the glow of the bulb change?

Calculate the capacitive reactance of a capacitor in order to run a bulb rated at $10 \, W, 60 \, V$ when connected in series to an $a.c.$ source of $100 \, V$.

For an $LCR$ circuit driven at frequency $\omega $,the equation reads $L\frac{di}{dt} + Ri + \frac{q}{C} = V_i = V_m \sin \omega t$.
$(a)$ Multiply the equation by $i$ and simplify where possible.
$(b)$ Interpret each term physically.
$(c)$ Cast the equation in the form of a conservation of energy statement.
$(d)$ Integrate the equation over one cycle to find that the phase difference between $V$ and $i$ must be acute.

$A$ $100 \;\mu F$ capacitor in series with a $40 \;\Omega$ resistance is connected to a $110 \;V, 12 \;kHz$ supply.
$(a)$ What is the maximum current in the circuit?
$(b)$ What is the time lag between the current maximum and the voltage maximum?
Hence,explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a $dc$ circuit after the steady state.

In an $A.C.$ circuit,a resistance $R = 40 \ \Omega$ and an inductance $L$ are connected in series. If the phase angle between voltage and current is $45^{\circ}$,then the value of the inductive reactance is $(\tan 45^{\circ} = 1)$. (in $\Omega$)

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?