In the $AC$ circuit shown,$E = E_0 \sin(\omega t + \phi)$ and $i = i_0 \sin(\omega t + \phi + \frac{\pi}{4})$. Then,the box contains:

An electric lamp connected in series with a capacitor and an $A.C.$ source is glowing with a certain brightness. On increasing the value of capacitance,the brightness of the lamp

In an $ac$ circuit,the current lags behind the voltage by $\pi /3$. The components in the circuit are

$A$ capacitor and a resistor of resistance $100 \sqrt{3} \Omega$ are connected in series to an $AC$ source of voltage $V = 100 \sin(200t) \text{ V}$,where $t$ is time in seconds. If the phase difference between the voltage and the current in the circuit is $30^{\circ}$,then the capacitance of the capacitor is: (in $\mu \text{F}$)

$A$ resistor of $200 \; \Omega$ and a capacitor of $15.0 \; \mu F$ are connected in series to a $220 \; V, 50 \; Hz$ $ac$ source.
$(a)$ Calculate the current in the circuit.
$(b)$ Calculate the voltage $(rms)$ across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes,resolve the paradox.

If a dielectric slab is inserted in the capacitor of the given $RC$ circuit,then the brightness of the bulb will: