Q.A $60\ \mu\text{F}$ capacitor is connected to a $110\ \text{V}$, $60\ \text{Hz}$ ac supply. Determine the rms value of the current in the circuit.

The key idea is that in a purely capacitive AC circuit, the current leads the voltage by $90^\circ$, and the rms current is given by $I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C}$, where $X_C$ is the capacitive reactance.

Step 1: Compute the capacitive reactance.

Step 2: Simplify the denominator.

$2\pi \times 60 \times 60 \times 10^{-6} = 2\pi \times 3600 \times 10^{-6} = 2\pi \times 3.6 \times 10^{-3} \approx 0.02262$ …

Concept and Intuition

When an AC voltage is applied across a capacitor, the capacitor charges and discharges alternately. Unlike a resistor, a capacitor does not dissipate energy — it stores and returns it. But it does oppose the flow of charge, and this opposition is called capacitive reactance ($X_C$).

The key idea: For a sinusoidal AC supply, the relationship between rms voltage and rms current in a capacitor looks just like Ohm's law — but with resistance replaced by reactance:

Here $f$ is the frequency in hertz, $C$ is the capacitance in farads. The reactance $X_C$ has units of ohms ($\Omega$).

A common mistake is to forget converting microfarads to farads, or to use the peak voltage instead of rms. The problem directly gives rms voltage, so no conversion is needed there.

Step-by-Step Solution

1. Write down the given data

• Capacitance: $C = 60\ \mu\text{F} = 60 \times 10^{-6}\ \text{F}$
• RMS voltage: $V_{\text{rms}} = 110\ \text{V}$
• Frequency: $f = 60\ \text{Hz}$

2. Calculate the capacitive reactance

The angular frequency is $\omega = 2\pi f$, so:

Substitute the values:

First compute the denominator:

Using $\pi \approx 3.1416$:

Thus:

$X_C \approx \frac{1}{0.02262} \approx 44.21\ \Omega$ …

Method: Capacitive Reactance Approach (Ohm's Law for AC Circuits)

This method treats the capacitor as a frequency-dependent resistor in an AC circuit.

Steps

1. Recall the formula for capacitive reactance The opposition to current flow in a capacitor is given by:

where $f$ is the frequency in Hz and $C$ is the capacitance in farads.

2. Convert units

   Given: $C = 60\ \mu\text{F} = 60 \times 10^{-6}\ \text{F}$, $V_{\text{rms}} = 110\ \text{V}$, $f = 60\ \text{Hz}$.

3. Calculate $X_C$

4. Apply Ohm's law for AC circuits For a purely capacitive circuit, the RMS current is: $I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C}$ …

Common Mistakes & How to Avoid Them

Mistake 1: Using the DC formula for power/current

The error: Students often try to use $P = \frac{V^2}{R}$ or $I = \frac{V}{R}$, forgetting that in a pure capacitive AC circuit, there is no resistance — only capacitive reactance.

How to avoid:

• Recognise that for a pure capacitor in AC, the opposition to current is capacitive reactance $X_C$, not resistance $R$.
• The correct formula is:

where

Mistake 2: Forgetting to convert units

The error: Using $C = 60$ instead of $C = 60 \times 10^{-6}\ \text{F}$.

How to avoid:

• Always write the unit conversion explicitly:

• Double-check: $\mu$ means $10^{-6}$, so never plug in 60 directly.

Mistake 3: Confusing peak and rms values

The error: Using $V_0 = 110\ \text{V}$ (peak) instead of $V_{\text{rms}} = 110\ \text{V}$.

How to avoid:

• The problem clearly states "110 V, 60 Hz ac supply" — in standard AC problems, this voltage is always rms unless specified as "peak" or "maximum".
• For a pure capacitor:

No need to convert to peak values here.

Mistake 4: Misplacing the $2\pi$ factor in $X_C$

The error: Writing $X_C = 2\pi f C$ or $X_C = \frac{2\pi}{f C}$.

How to avoid:

• Memorise the correct formula:

• Think: larger $f$ or $C$ → smaller $X_C$ (easier for current to flow). This helps catch sign errors.

--- …