$A$ current of $\frac{25}{\pi} \text{ Hz}$ frequency is passing through an $AC$ circuit having a series combination of $R=100 \ \Omega$ and $L=2 \text{ H}$. The phase difference between voltage and current is . . . . . . . (in $^{\circ}$)

$A$ sinusoidal voltage of peak value $283\,V$ and frequency $50\,Hz$ is applied to a series $LCR$ circuit in which $R = 3\,\Omega$,$L = 25.48\,mH$,and $C = 796\,\mu F$. Find the impedance of the circuit in $\Omega$.

In a series $L-C-R$ circuit,the frequency of a $10 \, V$ $a.c.$ voltage source is adjusted in such a fashion that the reactance of the inductor measures $15 \, \Omega$ and that of the capacitor $11 \, \Omega$. If $R = 3 \, \Omega$,the potential difference across the series combination of $L$ and $C$ will be.....$V$.

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

In a series $R-C$ circuit shown in the figure,the applied voltage is $10\, V$ and the voltage across the capacitor is found to be $8\, V$. Then,the voltage across $R$ and the phase difference between the current and the applied voltage will respectively be

$A$ sinusoidal voltage of peak value $283 \, V$ and angular frequency $320 \, rad/s$ is applied to a series $LCR$ circuit. Given that $R = 5 \, \Omega$,$L = 25 \, mH$,and $C = 1000 \, \mu F$. The total impedance and the phase difference between the voltage across the source and the current will respectively be: