Q.A $15.0\ \mu\text{F}$ capacitor is connected to a $220\ \text{V}$, $50\ \text{Hz}$ source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current?

The key idea is capacitive reactance $X_C = \frac{1}{\omega C}$, which opposes AC current. It depends inversely on frequency.

Step 1 — Capacitive reactance

Angular frequency $\omega = 2\pi f = 2\pi (50) = 100\pi\ \text{rad/s}$.

$C = 15.0\ \mu\text{F} = 15.0 \times 10^{-6}\ \text{F}$.

Step 2 — RMS current

By Ohm’s law for AC: $I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} = \frac{220}{212.2} \approx 1.037\ \text{A}$.

Step 3 — Peak current

$I_0 = \sqrt{2}\, I_{\text{rms}} \approx 1.414 \times 1.037 \approx 1.466\ \text{A}$.

Step 4 — Effect of doubling frequency …

Why Capacitive Reactance?

When you connect a capacitor to a DC source, it charges up and then blocks current — infinite resistance at steady state. But with AC, the voltage keeps changing, so the capacitor continuously charges and discharges. The faster the voltage changes (higher frequency), the less time the capacitor has to build up opposing charge, so the opposition to current — called capacitive reactance — drops.

This is not like resistance (which dissipates energy) or inductive reactance (which opposes current changes via magnetic fields). Capacitive reactance is purely a frequency-dependent opposition, given by:

where $f$ is frequency in hertz and $C$ is capacitance in farads. The unit is ohms ($\Omega$).

Once we have $X_C$, Ohm's law for AC capacitors works just like for resistors — but only for RMS values:

And peak values relate by $I_0 = \sqrt{2} \, I_{\text{rms}}$ (since voltage and current are sinusoidal).

Step-by-step solution

1. Write down the given data

• Capacitance: $C = 15.0\ \mu\text{F} = 15.0 \times 10^{-6}\ \text{F}$
• Voltage (rms): $V_{\text{rms}} = 220\ \text{V}$
• Frequency: $f = 50\ \text{Hz}$

2. Calculate capacitive reactance at $50\ \text{Hz}$

First compute the denominator:

So:

3. Find the RMS current

Using $V_{\text{rms}} = I_{\text{rms}} \, X_C$:

So $I_{\text{rms}} \approx 1.04\ \text{A}$.

4. Find the peak current

For a sinusoidal AC, $I_0 = \sqrt{2} \, I_{\text{rms}}$:

So $I_0 \approx 1.47\ \text{A}$. …

Method: Direct Formula Application (Capacitive Reactance & Ohm's Law for AC)

This problem uses the definition of capacitive reactance and Ohm's law for AC circuits.

Step 1 — Recall the formula for capacitive reactance

Capacitive reactance ($X_C$) is the opposition a capacitor offers to AC current:

Where:

• $f$ = frequency in Hz
• $C$ = capacitance in farads (F)

Step 2 — Convert units and plug in values

Given:

• $C = 15.0\ \mu\text{F} = 15.0 \times 10^{-6}\ \text{F}$
• $f = 50\ \text{Hz}$

First compute denominator:

So:

Capacitive reactance $\boxed{X_C \approx 212\ \Omega}$

Step 3 — Find RMS current using Ohm's law for AC

For AC circuits, Ohm's law uses RMS values:

Given $V_{\text{rms}} = 220\ \text{V}$:

RMS current $\boxed{I_{\text{rms}} \approx 1.04\ \text{A}}$

Step 4 — Find peak current

Peak current relates to RMS by:

Peak current $\boxed{I_0 \approx 1.47\ \text{A}}$

Step 5 — Effect of doubling frequency

If $f$ is doubled ($f' = 2f = 100\ \text{Hz}$):

• Capacitive reactance is inversely proportional to frequency: …

Here are the common mistakes students make on this exact problem, along with how to avoid each one.

1. Forgetting to Convert Microfarads ($\mu\text{F}$) to Farads (F)

The Mistake:

Plugging $C = 15.0$ directly into the formula $X_C = \frac{1}{2\pi f C}$ without converting. This gives a wildly wrong (and tiny) reactance.

How to Avoid:

Always write the conversion step explicitly.

$C = 15.0\ \mu\text{F} = 15.0 \times 10^{-6}\ \text{F}$.

Correct calculation:

2. Confusing RMS and Peak Values of Current

The Mistake:

Using the peak voltage ($V_0$) in $I = V / X_C$ (which gives RMS current) or vice versa. Many students forget that the standard formula $I = V / X_C$ uses RMS voltage unless stated otherwise.

How to Avoid:

• RMS current: $I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C}$
• Peak current: $I_0 = \frac{V_0}{X_C}$ where $V_0 = \sqrt{2} \cdot V_{\text{rms}}$

Correct values:

3. Thinking Capacitive Reactance Increases with Frequency

The Mistake:

Assuming $X_C$ behaves like resistance (which is constant) or like inductive reactance ($X_L \propto f$). Students often write $X_C \propto f$ instead of the correct inverse relation.

How to Avoid:

Memorise the formula and the reasoning:

Higher frequency → capacitor has less time to charge → less opposition → reactance decreases.

When frequency doubles ($f \to 100\ \text{Hz}$):

4. Forgetting That Current Changes Inversely with Reactance

The Mistake:

After correctly finding that $X_C$ halves, students sometimes say current also halves or stays the same.

How to Avoid:

Since $I = V / X_C$, if $X_C$ is halved, current doubles (for the same voltage).

Correct result when $f$ doubles:

5. Not Stating the "What Happens" Answer Clearly

The Mistake: …