Question Analysis: In an RC series circuit connected to an AC source, $V_1$ is the voltage across the resistor ($V_R$) and $V_2$ is the voltage across the capacitor ($V_C$).

Step 1 (Governing Law & Phase Relationship): [0.5 Mark]

In an AC circuit containing a resistor and a capacitor:

• The current $I$ and the voltage across the resistor ($V_R = V_1$) are always in phase (phase difference $\phi = 0$).
• The current $I$ leads the voltage across the capacitor ($V_C = V_2$) by a phase angle of $\frac{\pi}{2}$ radians (or $90^\circ$).

Step 2 (Step-by-Step Logical Deduction): [0.25 Mark]

Let the current be represented as $I = I_0 \sin(\omega t)$.

• Voltage across resistor: $V_1 = V_{R0} \sin(\omega t)$
• Voltage across capacitor: $V_2 = V_{C0} \sin(\omega t - \frac{\pi}{2})$

Comparing the phases, $V_1$ has a phase of $\omega t$ and $V_2$ has a phase of $(\omega t - \frac{\pi}{2})$. Therefore, $V_1$ leads $V_2$ by $\frac{\pi}{2}$.

Step 3 (Final Concluded Answer): [0.25 Mark]

Since $V_1$ leads $V_2$ by $90^\circ$, the correct statement is: (b) The voltage shown in $V_1$ is ahead in phase with voltage shown in $V_2$.

💡 EL10 Concept Insight: Think of the resistor as a "follower" that stays perfectly in sync with the current, while the capacitor is a "slacker" that always falls exactly a quarter-turn behind the current; because the resistor is ahead of the current and the capacitor is behind it, the resistor is naturally ahead of the capacitor.