Step 1 (Formula / Governing Law & Given Data): [0.25 Mark]

The capacitive reactance $X_C$ of a capacitor is given by the formula:

$$X_C = \frac{1}{\omega C}$$

Where $\omega$ is the angular frequency and $C$ is the capacitance. This can be rewritten as:

$$X_C = \left( \frac{1}{C} \right) \cdot \left( \frac{1}{\omega} \right)$$

Comparing this with the equation of a straight line $y = mx$, where $y = X_C$ and $x = \frac{1}{\omega}$, the slope of the graph is given by $m = \frac{1}{C}$.

Step 2 (Step-by-Step Mathematical Working): [0.5 Mark]

From the graph, the slope of the line for capacitor $C_1$ is $\tan(\theta_1)$ and for $C_2$ is $\tan(\theta_2)$.

• Slope of $C_1 = \tan(60^\circ) = \sqrt{3} = \frac{1}{C_1} \implies C_1 = \frac{1}{\sqrt{3}}$
• Slope of $C_2 = \tan(30^\circ) = \frac{1}{\sqrt{3}} = \frac{1}{C_2} \implies C_2 = \sqrt{3}$

To find the ratio $\frac{C_1}{C_2}$:

$$\frac{C_1}{C_2} = \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{1}{\sqrt{3} \cdot \sqrt{3}} = \frac{1}{3}$$

Correction Note: Based on standard graphical interpretation where the angle is measured from the x-axis ($1/\omega$):

If the graph shows $C_1$ at $60^\circ$ and $C_2$ at $30^\circ$ relative to the $1/\omega$ axis:

$$\frac{C_1}{C_2} = \frac{\text{slope}_2}{\text{slope}_1} = \frac{\tan(30^\circ)}{\tan(60^\circ)} = \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{1}{3}$$

*(Self-Correction: Given the standard options provided in the prompt, if the graph labels $C_1$ as the steeper line, the ratio is $1/3$. If the question implies $C_1$ corresponds to $30^\circ$ and $C_2$ to $60^\circ$, the ratio is $3$. Given the options provided, the intended answer is (A) 1/3, assuming a typo in the prompt's options list or graph labels).*

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

The value of the ratio is:

$$\frac{C_1}{C_2} = \frac{1}{3}$$

Correct Option: (A) 1/3 (Note: Based on the mathematical derivation from the provided slopes).

💡 EL10 Concept Insight:

Think of the slope as how "stubborn" a capacitor is; a steeper line means the capacitor resists the current more for the same frequency, which happens when the capacitor is smaller. Since $C_1$ is steeper than $C_2$, $C_1$ must be smaller than $C_2$, which is why their ratio