〰️

Topic 11 of 15

Inductance & AC Circuits

An inductor stores energy in a magnetic field, just as a capacitor stores energy in an electric field. Together in an AC circuit they create reactances that depend on frequency — making it possible to build filters, tuners, and resonant circuits. The interplay of resistance, inductance, and capacitance in RLC circuits underlies everything from AM radio receivers to medical MRI machines.

Key Concepts

Self-Inductance

An inductor of inductance

L

(henries) opposes changes in current:

\mathcal{E} = -L\,dI/dt

. The faster the current changes, the larger the back-EMF. A solenoid of

n

turns/m, length

\ell

, and area

A

has

L = \mu_0 n^2 A \ell

Energy Stored in an Inductor

An inductor carrying current

I

stores energy in its magnetic field:

U = \frac{1}{2}LI^2

. This is the magnetic analogue of

U = \frac{1}{2}CV^2

for a capacitor.

RL Circuits

When a switch is closed in an

RL

circuit, current builds exponentially with time constant

\tau = L/R

. After one time constant, current reaches

\approx63\%

of its final value

I_f = \mathcal{E}/R

Reactance in AC Circuits

Inductive reactance

X_L = \omega L

opposes AC current and increases with frequency. Capacitive reactance

X_C = 1/(\omega C)

opposes AC and decreases with frequency. Unlike resistance, reactances do not dissipate energy.

RLC Series Circuit and Resonance

Impedance

Z = \sqrt{R^2 + (X_L - X_C)^2}

; peak current

I_0 = V_0/Z

. At resonance

\omega_0 = 1/\sqrt{LC}

, we have

X_L = X_C

Z = R

(minimum impedance, maximum current).

Transformers

An ideal transformer steps voltage up or down via the turns ratio:

V_2/V_1 = N_2/N_1

. Power is conserved:

V_1 I_1 = V_2 I_2

, so current steps inversely:

I_2/I_1 = N_1/N_2

Key Equations

Inductor back-EMF

\mathcal{E} = -L\frac{dI}{dt}

Induced EMF in an inductor of inductance L (henries) opposing the rate of current change.

Energy in an inductor

U = \tfrac{1}{2}LI^2

Energy stored in the magnetic field of an inductor carrying current I.

Reactances

X_L = \omega L \qquad X_C = \frac{1}{\omega C}

Inductive reactance increases with frequency; capacitive reactance decreases. Units: ohms.

RLC impedance

Z = \sqrt{R^2 + (X_L - X_C)^2}

Total impedance of a series RLC circuit. Peak current I₀ = V₀/Z.

Resonant frequency

\omega_0 = \frac{1}{\sqrt{LC}}

At ω₀, X_L = X_C and Z = R (minimum impedance). Resonant frequency f₀ = ω₀/(2π).

Transformer turns ratio

\frac{V_2}{V_1} = \frac{N_2}{N_1}

Voltage ratio equals turns ratio for an ideal transformer. Power is conserved: V₁I₁ = V₂I₂.

Worked Example

RLC Series Circuit

Problem

A series RLC circuit has $R = 100\,\Omega$ , $L = 50$ mH, $C = 20\,\mu$ F, connected to an AC source with peak voltage $V_0 = 120$ V at $f = 200$ Hz. Find $X_L$ , $X_C$ , $Z$ , and peak current $I_0$ .

Solution

Angular frequency: $\omega = 2\pi f = 2\pi(200) = 1257$ rad/s.

X_L = \omega L = (1257)(0.050) = 62.8\,\Omega

X_C = \frac{1}{\omega C} = \frac{1}{(1257)(20\times10^{-6})} = \frac{1}{0.02513} = 39.8\,\Omega

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{100^2 + (62.8 - 39.8)^2} = \sqrt{10000 + 529} = 102.6\,\Omega

I_0 = \frac{V_0}{Z} = \frac{120}{102.6} = 1.17\text{ A}

Answer X_L = 62.8 Ω; X_C = 39.8 Ω; Z = 102.6 Ω; I₀ = 1.17 A.

Practice

Exercises

7 problems

1 of 7

What is the inductive reactance (in Ω) of an $L = 100$ mH inductor at $f = 50$ Hz?

Your answer

2 of 7

What is the capacitive reactance (in Ω) of a $C = 10\,\mu$ F capacitor at $f = 100$ Hz?

Your answer

3 of 7

An inductor with $L = 200$ mH carries a current of $I = 3.0$ A. How much energy (in J) is stored in it?

Unlock Exercise 3