Q.If L is inductance, C is capacitance and R is resistance, then show that L/CR has the unit of ohm.

The capacitive reactance of an ideal capacitor is given by

Rearranging this as $X_C = \frac{1}{C} \cdot \frac{1}{\omega}$, we see that $X_C$ is directly proportional to $\frac{1}{\omega}$. When we plot $X_C$ versus $\frac{1}{\omega}$, we get a straight line passing through the origin with slope equal to $\frac{1}{C}$.

The key insight: a steeper line means a larger slope, which means a larger value of $\frac{1}{C}$, which in turn means a smaller capacitance. The graph shows two such lines for capacitors $C_1$ and $C_2$, making angles of 45° and 30° respectively with the horizontal axis.

1. Find the slope of line $C_1$: The line makes an angle of 45° with the $\frac{1}{\omega}$ axis. The slope is

Since slope $= \frac{1}{C_1}$, we have

2. Find the slope of line $C_2$: The line makes an angle of 30° with the $\frac{1}{\omega}$ axis. The slope is

Since slope $= \frac{1}{C_2}$, we have …

The key to this problem is understanding what capacitive reactance actually means physically. A capacitor in an AC circuit doesn't "resist" current the way a resistor does — instead, it opposes changes in voltage by storing and releasing charge. The faster the voltage changes (higher frequency) or the larger the capacitor (more charge storage per volt), the easier it is for current to flow. That's why reactance $X$ is inversely proportional to both capacitance $C$ and frequency $\nu$.

Let's work through the change step by step.

1. Write the standard formula for capacitive reactance. For a capacitor of capacitance $C$ connected to an AC source of frequency $\nu$, the reactance is:

This is a direct relationship — no tricks, just the definition.

2. Identify the new values.

   The capacitance is doubled: $C' = 2C$

   The frequency is tripled: $\nu' = 3\nu$

3. Substitute these into the formula for the new reactance $X'$.

4. Simplify the denominator.

5. Recognize the original reactance in the expression. Since $X = \frac{1}{2\pi \nu C}$, we have: $X' = \frac{X}{6}$ …

The core idea here is simple: reactance is not a fixed property — it depends on frequency. An inductor opposes changes in current, and the faster the current changes (higher frequency), the more it opposes. A capacitor, on the other hand, stores and releases charge; at higher frequencies, it has less time to charge up, so it offers less opposition.

Let’s see exactly how each behaves when frequency is lowered.

1. Inductive reactance is given by

Here $f$ is the frequency and $L$ is the inductance (a constant for a given inductor). Since $X_L$ is directly proportional to $f$, decreasing $f$ makes $X_L$ smaller. So the inductor’s opposition weakens.

2. Capacitive reactance is given by

$C$ is the capacitance (constant). Here $X_C$ is inversely proportional to $f$ — as $f$ goes down, $X_C$ goes up. So the capacitor’s opposition strengthens. …