Q.A device Y is connected across an AC source of emf $e = e_0\sin\omega t$. The current through Y is given as $i = i_0\sin(\omega t + \frac{\pi}{2})$.

[!TLDR] Since i leads e by $\pi/2$, device Y is a pure CAPACITOR, with reactance $X_C=1/(\omega C)$, decreasing hyperbolically with frequency; its phasor $i_0$ sits $90^\circ$ ahead of $e_0$. [!ANSWER] Y is …

(a) The given current is $i=i_0\sin\left(\omega t+\dfrac{\pi}{2}\right)$, i.e. it LEADS the applied emf $e=e_0\sin\omega t$ by exactly $\pi/2$ ($90^\circ$). From section 13.5.3, this is exactly the signature behaviour of a PURE CAPACITOR (an inductor instead makes current LAG). So device Y is a capacitor of some capacitance C, and its reactance is $X_C=\dfrac{1}{\omega C}=\dfrac{1}{2\pi fC}$.\n\n(b) Over one cycle, plotted against $\omega t$ on the horizontal axis: e is a plain sine curve, zero at $\omega t=0$, peaking at $e_0$ at $\omega t=\pi/2$. The current i is the SAME sine shape but shifted a quarter cycle EARLIER (to the left), so it already peaks at $i_0$ when $\omega t=0$ and crosses zero (going downward) at $\omega t=\pi/2$, exactly where e is at its own peak -- i.e. i's peaks and zero-crossings each occur a quarter cycle before e's corresponding points.\n\n(c) $X_C=\dfrac{1}{2\pi fC}$ is INVERSELY proportional to frequency f: plotted with $X_C$ on the vertical axis and f on the horizontal axis, the graph is a downward-sloping hyperbola-like curve ($X_C\propto1/f$), very large (tending to infinity) as $f\to0$ and falling towards zero as f becomes very large.\n\n(d) On t …

Identify the device from the phase relationship (current leading by $\pi/2$ = capacitor), state its reactance formula, and describe the e-i graph, the $X_C$-vs-f graph, and th …

Confusing 'current leads' with 'current lags' and misidentifying Y as an inductor instead of a capacitor -- the two elements have exactly opposite phase signatures …