Self-inductance. When the current in a coil changes, the flux it produces through itself changes, inducing a back-EMF that opposes the change. This property is self-inductance $L$, defined by $\varepsilon = -L\frac{dI}{dt}$. Its SI unit is the henry (H). A coil with large $L$ resists rapid changes in current — it behaves like electrical inertia. The energy stored in an inductor carrying current $I$ is $U = \frac{1}{2}LI^2$.
Mutual inductance. When two coils are close, a changing current in one (primary) induces an EMF in the other (secondary): $\varepsilon_2 = -M\frac{dI_1}{dt}$, where $M$ is the mutual inductance (also in henry). Mutual inductance is the working principle of the transformer.
Alternating current and RMS values. An alternating current reverses direction periodically: $I = I_0\sin\omega t$, where $I_0$ is the peak value and $\omega = 2\pi f$. Since AC averages to zero over a cycle, we use the root-mean-square (RMS) value, which gives the same heating as an equivalent DC: $I_{rms} = \frac{I_0}{\sqrt2} \approx 0.707\,I_0$ and similarly $V_{rms} = \frac{V_0}{\sqrt2}$. The $220\,\text{V}$ mains is an RMS value (peak $\approx 311\,\text{V}$).
AC through a resistor, inductor and capacitor.
- Pure resistor (R): voltage and current are in phase. $I_{rms} = V_{rms}/R$.
- Pure inductor (L): current lags the voltage by $90^\circ$. The opposition is the inductive reactance $X_L = \omega L$, which increases with frequency.
- Pure capacitor (C): current leads the voltage by $90^\circ$. The opposition is the capacitive reactance $X_C = \frac{1}{\omega C}$, which decreases with frequency.
Series LCR circuit and impedance. When R, L and C are in series, the net opposition is the impedance $Z = \sqrt{R^2 + (X_L - X_C)^2}$, and $I_{rms} = V_{rms}/Z$. The current and voltage differ in phase by $\phi$, where $\tan\phi = \frac{X_L - X_C}{R}$. A phasor diagram represents $V_R$, $V_L$ and $V_C$ as rotating vectors and adds them to find the resultant.
Resonance. When $X_L = X_C$, they cancel, so $Z = R$ is minimum and the current is maximum. This is series resonance, occurring at the resonant frequency $\omega_0 = \frac{1}{\sqrt{LC}}$ (i.e. $f_0 = \frac{1}{2\pi\sqrt{LC}}$). Resonant circuits are used to tune radios and TVs to a chosen station.
Power and power factor. The average power in an AC circuit is $P = V_{rms}I_{rms}\cos\phi$, where $\cos\phi$ is the power factor. For a pure resistor $\cos\phi = 1$ (maximum power), while for a pure inductor or capacitor $\cos\phi = 0$, so no power is consumed — the current that flows without dissipating energy is called wattless current.
The current in a coil falls from $5\,\text{A}$ to $1\,\text{A}$ in $0.2\,\text{s}$, inducing a back-EMF of $10\,\text{V}$. Find the self-inductance of the coil.
Solution- Step 1: Magnitude of self-induced EMF: $\varepsilon = L\dfrac{dI}{dt}$.
- Step 2: $\dfrac{dI}{dt} = \dfrac{5 - 1}{0.2} = \dfrac{4}{0.2} = 20\,\text{A/s}$.
- Step 3: $L = \dfrac{\varepsilon}{dI/dt} = \dfrac{10}{20}$.
- Step 4: $L = 0.5\,\text{H}$.
Answer: The self-inductance is $0.5\,\text{H}$.
The peak voltage of an AC source is $311\,\text{V}$. Find its RMS value.
Solution- Step 1: The RMS value relates to the peak by $V_{rms} = \dfrac{V_0}{\sqrt2}$.
- Step 2: $V_{rms} = \dfrac{311}{\sqrt2} = \dfrac{311}{1.414}$.
- Step 3: $V_{rms} \approx 220\,\text{V}$.
Answer: The RMS voltage is about $220\,\text{V}$ — the standard household mains value.
A $0.2\,\text{H}$ inductor is connected to a $50\,\text{Hz}$ AC source of $220\,\text{V}$. Find the inductive reactance and the RMS current.
Solution- Step 1: Angular frequency $\omega = 2\pi f = 2\pi \times 50 = 100\pi \approx 314\,\text{rad/s}$.
- Step 2: Inductive reactance $X_L = \omega L = 314 \times 0.2 = 62.8\,\Omega$.
- Step 3: RMS current $I_{rms} = \dfrac{V_{rms}}{X_L} = \dfrac{220}{62.8}$.
- Step 4: $I_{rms} \approx 3.5\,\text{A}$.
Answer: $X_L \approx 62.8\,\Omega$ and the RMS current is about $3.5\,\text{A}$.
A series LCR circuit has $R = 30\,\Omega$, $X_L = 70\,\Omega$ and $X_C = 30\,\Omega$. Find the impedance.
Solution- Step 1: Impedance $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
- Step 2: $X_L - X_C = 70 - 30 = 40\,\Omega$.
- Step 3: $Z = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500}$.
- Step 4: $Z = 50\,\Omega$.
Answer: The impedance of the circuit is $50\,\Omega$.
A series LCR circuit has $L = 2\,\text{H}$ and $C = 8\,\mu\text{F}$. Find its resonant frequency.
Solution- Step 1: Resonant angular frequency $\omega_0 = \dfrac{1}{\sqrt{LC}}$.
- Step 2: $LC = 2 \times 8 \times 10^{-6} = 16 \times 10^{-6}$, so $\sqrt{LC} = 4 \times 10^{-3}\,\text{s}$.
- Step 3: $\omega_0 = \dfrac{1}{4 \times 10^{-3}} = 250\,\text{rad/s}$.
- Step 4: $f_0 = \dfrac{\omega_0}{2\pi} = \dfrac{250}{6.28} \approx 39.8\,\text{Hz}$.
Answer: $\omega_0 = 250\,\text{rad/s}$, giving a resonant frequency of about $39.8\,\text{Hz}$.
In an AC circuit, $V_{rms} = 200\,\text{V}$, $I_{rms} = 4\,\text{A}$ and the phase angle is $60^\circ$. Find the average power and the power factor.
Solution- Step 1: Power factor $= \cos\phi = \cos 60^\circ = 0.5$.
- Step 2: Average power $P = V_{rms}I_{rms}\cos\phi$.
- Step 3: $P = 200 \times 4 \times 0.5$.
- Step 4: $P = 400\,\text{W}$.
Answer: The power factor is $0.5$ and the average power consumed is $400\,\text{W}$.