Alternating Current, on the exam screen.

Series resonance occurs at $f_0 = \dfrac{1}{2\pi\sqrt{LC}}$, which is independent of the resistance. With $L = 1\ \text{mH} = 10^{-3}\ \text{H}$ and $C = 0.1\ \mu\text{F} = 10^{-7}\ \text{F}$, $f_0 = \dfrac{1}{2\pi\sqrt{10^{-10}}} = \dfrac{1}{2\pi \times 10^{-5}} \approx 15.9\ \text{kHz}$ (option (4)). The resistance value is a distractor since it does not enter the resonance formula.

With $i = i_{peak}\sin(\omega t)$ and $\omega = 2\pi f = 120\pi\ \text{rad/s}$, the current first equals its peak when $\sin(\omega t) = 1$, i.e. $\omega t = \pi/2$. This gives $t = \dfrac{1}{4f} = \dfrac{1}{4 \times 60} = \dfrac{1}{240}\ \text{s}$ (a quarter of the period).

With $\omega = 2 \times 10^3\ \text{rad s}^{-1}$, the reactances are $X_L = \omega L = 20\ \Omega$ and $X_C = \frac{1}{\omega C} = 20\ \Omega$. Since $X_L = X_C$ the circuit is at RESONANCE, the reactances cancel and the impedance reduces to $Z = R = 100\ \Omega$. The current amplitude is then $I_0 = \frac{V_0}{Z} = \frac{220}{100} = 2.2\ \text{A}$. NCERT Class 12, Alternating Current (LCR series circuit and resonance).

For the RC circuit, tan($\pi$/3) = Xc/R; for the RL circuit, tan($\pi$/3) = XL/R. Hence XL = Xc in the complete series circuit, so it is at resonance and its power factor is cos 0 = 1. Chapter: Alternating Current.

For a capacitor, I = V(2$\pi$fC) = $200 \times 2\pi \times 50 \times 40 \times 10^{-6} \approx 2.51$ A, matching 2.5 A. Chapter: Alternating Current.