Q.A pure inductor of $25.0\ \text{mH}$ is connected to a source of $220\ \text{V}$. Find the inductive reactance and rms current in the circuit if the frequency of the source is $50\ \text{Hz}$.

Concept: Inductive Reactance and RMS Current in an AC Circuit

The opposition to current flow in an inductor is given by $X_L = \omega L = 2\pi f L$. Once $X_L$ is known, Ohm’s law for AC gives $I_{\text{rms}} = V_{\text{rms}} / X_L$.

Step 1 – Inductive reactance

$X_L = 2\pi f L = 2\pi (50)(25.0 \times 10^{-3})$

$X_L = 2\pi (1.25) = 2.5\pi \approx 7.85\ \Omega$

Step 2 – RMS current …

The key idea here is that a pure inductor does not dissipate power like a resistor — it stores and releases energy in its magnetic field. But it still opposes the flow of alternating current. This opposition is called inductive reactance ($X_L$), and it behaves like a frequency-dependent resistance.

For a DC circuit, an inductor is just a wire (zero resistance). But for AC, the changing current creates a changing magnetic field, which induces a back emf that fights the source voltage. The faster the current changes (higher frequency), the greater this opposition. That’s why $X_L$ depends directly on frequency.

Once we have $X_L$, Ohm’s law for AC circuits gives the rms current:

Let’s work through it step by step.

1. Convert inductance to henrys. The given inductance is $25.0\ \text{mH}$. Since $1\ \text{mH} = 10^{-3}\ \text{H}$:

2. Calculate inductive reactance. Frequency $f = 50\ \text{Hz}$. Using the formula:

First, $2\pi \times 50 = 100\pi \approx 314.16$. Then:

Rounding to three significant figures (matching the given data):

3. Find the rms current. The source voltage is $220\ \text{V}$ (rms value — AC voltmeters read rms). Using Ohm’s law for AC: …

Method: Inductive Reactance and RMS Current in a Pure Inductive Circuit

This problem uses the AC circuit analysis for a pure inductor — where the inductor opposes current flow due to self-inductance, and the opposition is called inductive reactance.

Step 1: Write the given data

• Inductance, $L = 25.0\ \text{mH} = 25.0 \times 10^{-3}\ \text{H}$
• Voltage (rms), $V_{\text{rms}} = 220\ \text{V}$
• Frequency, $f = 50\ \text{Hz}$

Step 2: Recall the formula for inductive reactance

Inductive reactance is given by:

This is the opposition offered by the inductor to AC current (measured in ohms, $\Omega$).

Step 3: Calculate $X_L$

Substitute the values:

First, $2 \pi \times 50 = 100 \pi \approx 314.16$

Then:

So, $X_L \approx 7.85\ \Omega$

Step 4: Recall Ohm’s law for AC circuits

For a pure inductor, the rms current is:

(Note: In a pure inductor, voltage leads current by $90^\circ$, but magnitude relation is the same as Ohm’s law.)

Step 5: Calculate $I_{\text{rms}}$ …

The Correct Solution First

Given:

• $L = 25.0\ \text{mH} = 25.0 \times 10^{-3}\ \text{H}$
• $V_{\text{rms}} = 220\ \text{V}$
• $f = 50\ \text{Hz}$

Step 1: Inductive reactance

Step 2: RMS current

Common Mistakes & How to Avoid Them

✗ Mistake 1: Forgetting to convert mH to H

• Students plug in $L = 25.0$ directly into $X_L = 2\pi f L$
• This gives $X_L = 2\pi \times 50 \times 25 = 7850\ \Omega$ — wrong by a factor of 1000

✓ How to avoid: Always write units explicitly. Convert milli (m) → $10^{-3}$ before substituting.

✗ Mistake 2: Using $V_{\text{rms}}$ in peak formulas

• Some students compute $I_0 = V_0 / X_L$ using $V_0 = \sqrt{2} \times 220$
• The question asks for rms current, so use rms voltage directly

✓ How to avoid: Read carefully — if the problem says "rms current" or just "current" in an AC circuit, use $I_{\text{rms}} = V_{\text{rms}} / X_L$.

✗ Mistake 3: Assuming power is dissipated

• A pure inductor has zero power dissipation (average power = 0)
• Some students try to compute $P = I^2 R$ or $P = VI \cos\phi$ and get confused

✓ How to avoid: Remember — in a pure inductor, voltage leads current by $90^\circ$, so $\cos\phi = 0$ and average power = 0. This question only asks for reactance and current, not power.

✗ Mistake 4: Using $X_L = \omega L$ but forgetting $\omega = 2\pi f$

• Writing $X_L = fL$ or $X_L = \frac{2\pi L}{f}$ instead of $2\pi f L$

✓ How to avoid: Memorise the exact formula: $X_L = 2\pi f L$. Write it down before substituting numbers.

✗ Mistake 5: Rounding too early …