$A$ lamp is connected in series with a capacitor and an $AC$ source. What happens if the capacitance of the capacitor is reduced?

In an $LCR$ series circuit,an inductor $30 \, {mH}$ and a resistor $1 \, \Omega$ are connected to an $AC$ source of angular frequency $300 \, {rad/s}$. The value of capacitance for which the current leads the voltage by $45^{\circ}$ is $\frac{1}{x} \times 10^{-3} \, {F}$. Then the value of $x$ is ..... .

In a $L-R$ circuit of $4 \text{ mH}$ inductance and $3 \text{ } \Omega$ resistance, e.m.f. $E = \cos(1000t) \text{ V}$ is applied. The amplitude of current is (in $\text{ A}$)

$A$ 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 $.7$)

$A$ $400 \Omega$ resistor, a $250 \text{ mH}$ inductor, and a $2.5 \mu \text{F}$ capacitor are connected in series with an $AC$ source of peak voltage $5 \text{ V}$ and angular frequency $\omega = 2000 \text{ rad/s}$. What is the peak value of the electrostatic energy of the capacitor (in $\mu \text{J}$)?

$A$ 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?