Q.A 25 $\mu F$ capacitor, a 0.10 H inductor and a 25 $\Omega$ resistor are connected in series with an AC source whose emf is given by $e = 310\sin(314t)$ (volt). What is the frequency, reactance, impedance, current and phase angle of the circuit?

[!TLDR] $\omega=314\Rightarrow f=50$Hz; $X_L=31.4\,\Omega$, $X_C\approx127.4\,\Omega$, reactance $|X_L-X_C|\approx95.98\,\Omega$; $Z=\sqrt{25^2+95.99^2}\approx99.19\,\Omega$; $i_{rms}=e_{rms}/Z\approx2.211$ A; $\phi=\tan^{-1}(95.99/25)\approx1.316$ rad. [!ANSWER] f = …

Given $e=310\sin(314t)$ V, so $e_0=310$ V, $\omega=314$ rad/s; $C=25\,\mu F$, $L=0.10$ H, $R=25\,\Omega$. Frequency: $f=\dfrac{\omega}{2\pi}=\dfrac{314}{6.284}\approx50$ Hz.\n\n$X_L=\omega L=314\times0.10=31.4\,\Omega$. $X_C=\dfrac{1}{\omega C}=\dfrac{1}{314\times25\times10^{-6}}=\dfrac{1}{7.85\times10^{-3}}\approx127.39\,\Omega$.\n\nNet reactance: $|X_L-X_C|=|31.4-127.39|=95.99\,\Omega\approx95.98\,\Omega$. Impedance: $Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{25^2+95.99^2}=\sqrt{625+9214.1}=\sqrt{9839.1}\approx99.19\,\Omega$.\n\nPeak current: $i_0=e_0/Z=310/99.19\approx3.125$ A; rms current $i_{rms}=i_0/\sqrt2\approx2.210$ A ($\approx2.211$ A). Phase angle: $\tan\phi=\dfrac{X_C-X_L}{R}=\dfrac{ …

Get f from $\omega=2\pi f$; compute $X_L$, $X_C$ and the net reactance; get Z from the impedance formula; get $i_{rms}$ from $i_0/\sqrt2$ with $i_0=e_0/Z$; g …

Reporting the peak current instead of rms current (irms is what the problem implicitly expects, matching the book's answer), or leaving the p …