Q.A $44\ \text{mH}$ inductor is connected to $220\ \text{V}$, $50\ \text{Hz}$ ac supply. Determine the rms value of the current in the circuit.

The key idea is that in a purely inductive AC circuit, the current lags the voltage by $90^\circ$ and the opposition to current is given by inductive reactance $X_L$, not resistance.

Step 1: Compute the inductive reactance.

Step 2: Apply Ohm's law for an AC circuit. For a pure inductor, the rms current is the rms voltage divided by the inductive reactance.

Step 3: Calculate the result.

Why This Works — The Concept

When you connect an inductor to an AC supply, something interesting happens. Unlike a resistor, an inductor doesn't just "resist" current — it opposes changes in current. This opposition is called inductive reactance ($X_L$), and it depends on both the frequency of the supply and the inductance itself.

The key idea: For a pure inductor (no resistance), the voltage and current are out of phase by exactly $90^\circ$, with the current lagging behind the voltage. But for calculating the magnitude of the current (the rms value), we can treat the inductor just like a resistor — using Ohm's law for AC circuits:

The beauty is that rms values follow the same arithmetic as DC values, as long as we use the correct "resistance" (reactance) for the component.

Step-by-Step Solution

1. Identify the given quantities

We have:

• Inductance, $L = 44\ \text{mH} = 44 \times 10^{-3}\ \text{H}$
• Supply voltage (rms), $V_{\text{rms}} = 220\ \text{V}$
• Frequency, $f = 50\ \text{Hz}$

2. Calculate the inductive reactance

The inductive reactance tells us how much the inductor "resists" the AC current:

Substitute the values:

If you want a numerical value: $X_L \approx 4.4 \times 3.1416 \approx 13.82\ \Omega$

3. Apply Ohm's law for AC circuits

For a pure inductor, the rms current is simply:

Simplify: $220/4.4 = 50$, so:

4. Get the numerical value

Why No Phase Angle in the Answer?

You might wonder: shouldn't we account for the $90^\circ$ phase difference? The answer is no — because the question specifically asks for the rms value of the current. RMS values are magnitudes only; they don't carry phase information. The phase angle matters when you're combining components or calculating instantaneous power, but for a single inductor's current magnitude, it's just $V_{\text{rms}} / X_L$.

Method: Inductive Reactance & Ohm’s Law for AC Circuits

This method uses the concept that an ideal inductor opposes AC current through inductive reactance ($X_L$), which behaves like resistance in Ohm’s law for AC circuits.

Steps

Step 1: Recall the formula for inductive reactance

where:

• $f$ = frequency in Hz
• $L$ = inductance in henry (H)

Step 2: Convert units and substitute values

Given:

• $L = 44\ \text{mH} = 44 \times 10^{-3}\ \text{H}$
• $f = 50\ \text{Hz}$
• $V_{\text{rms}} = 220\ \text{V}$

Step 3: Apply Ohm’s law for AC circuits

For a purely inductive circuit:

Step 4: Compute the numerical value

Final Answer

Key Concept Check

• The inductor does not dissipate power (ideal case) — it only stores and returns energy.
• The current lags the voltage by $90^\circ$ in a pure inductor.
• The opposition is reactance ($X_L$), not resistance — hence no power dissipation, only reactive power.

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

Mistake 1: Forgetting to Convert Inductance to Henry

The mistake:

Students see $44\ \text{mH}$ and plug in $44$ directly into the formula, forgetting the milli prefix.

Why it happens:

The formula $X_L = 2\pi f L$ expects $L$ in henry (H), not millihenry.

How to avoid:

Always write the conversion step explicitly:

Mistake 2: Using the Wrong Formula for Impedance

The mistake:

Students treat the inductor like a resistor and write $I = V / R$, using $R$ instead of $X_L$.

Why it happens:

They confuse resistive circuits with inductive AC circuits.

How to avoid:

Remember: For a pure inductor, there is no resistance — only inductive reactance:

Then use Ohm’s law for AC:

Mistake 3: Using Peak Voltage Instead of RMS Voltage

The mistake:

Students take $220\ \text{V}$ as peak voltage and use $V_0$ in the formula.

Why it happens:

They forget that the problem explicitly says "220 V, 50 Hz ac supply" — this is the rms value by convention in Indian exams.

How to avoid:

In AC circuit problems, unless stated as "peak voltage" or "$V_0$", assume the given voltage is rms.

Here:

Mistake 4: Forgetting the Factor of $2\pi$ in $X_L$

The mistake:

Students write $X_L = fL$ or $X_L = \pi f L$, missing the $2$.

Why it happens:

They memorise the formula incorrectly.

How to avoid:

Write the full formula every time:

And remember: $\pi \approx 3.14$, so $2\pi \approx 6.28$.

Mistake 5: Calculation Errors with $\pi$

The mistake:

Using $\pi = 3.14$ but making arithmetic mistakes, or rounding too early.

How to avoid:

Do the calculation step-by-step:

1. $X_L = 2 \times 3.14 \times 50 \times 0.044$
2. First: $2 \times 50 = 100$
3. Then: $100 \times 0.044 = 4.4$
4. Finally: $4.4 \times 3.14 = 13.816\ \Omega$

Then:

Final answer: $\boxed{15.92\ \text{A}}$ (or approximately $16\ \text{A}$)

Quick Checklist to Avoid All Mistakes

• Convert mH → H ($\times 10^{-3}$)
• Use $X_L = 2\pi f L$, not $R$
• Use $V_{\text{rms}} = 220\ \text{V}$ (not peak)
• Include $2\pi$, not just $\pi$ or $fL$
• Do arithmetic carefully, avoid early rounding

Master these, and you’ll never lose marks on this problem again.