Chapter 7 · Introduction

Chapter 7 — Alternating Current

The current from a wall socket reverses direction a hundred times a second. Resistors, coils and capacitors each respond to that reversal in their own way, and putting them together gives tuned circuits, power factors and the transformers that carry electricity across a country.

The socket in an Indian home supplies $230$ V at $50$ Hz: the voltage swings from about $+325$ V to $-325$ V and back again fifty times every second, and the current in a heater or a fan reverses with it. Almost every power grid in the world works this way, at $50$ Hz or $60$ Hz. Why alternating current and not the steady current of a battery? The main reason is the transformer. With AC, a pair of coils on an iron core can raise the voltage of a power station to $400\,000$ V for the long journey across the country and then lower it again to $230$ V outside your house. Transmitting at high voltage means a small current, and a small current wastes very little energy heating the wires.

Alternating current also behaves in ways a steady current cannot. A capacitor blocks a steady current but lets AC through, and it passes high frequencies more easily than low ones. A coil does the reverse: it lets a steady current through but holds back AC, more strongly at high frequency. Combine a resistor, a coil and a capacitor and you get a circuit that responds strongly to just one frequency — the principle behind tuning a radio. In this chapter we describe AC with rms values and rotating arrows called phasors, find how $R$, $L$ and $C$ respond to it, analyse the series LCR circuit and its resonance, work out the power an AC circuit really uses, and end with the transformer.

$v=v_m\sin\omega t$→rms values→$R$, $L$, $C$ & phasors→Impedance $Z$→Resonance & $Q$→Power factor→Transformer
🎯 By the end of this chapter you can…
  • describe an alternating voltage or current by its peak value, frequency and phase, and convert between peak and rms values;
  • find the current through a resistor, an inductor or a capacitor connected to an AC source, with its phase, using the reactances $X_L=\omega L$ and $X_C=1/\omega C$;
  • draw phasor diagrams and use them to find the impedance, current and phase angle of a series LCR circuit;
  • find the resonant frequency of a series LCR circuit and describe the sharpness of resonance by its bandwidth and quality factor;
  • calculate the average power in an AC circuit, explain the power factor and the wattless current;
  • explain how a transformer works, use the turns ratio, and show why power is transmitted at high voltage;
  • solve board, JEE Main, JEE Advanced and NEET problems on all of the above.
Teaser — One Source, Three Components

An AC source of $10$ V (rms) is connected to a single component: a $100\ \Omega$ resistor, a $0.30$ H coil or a $10\ \mu\text{F}$ capacitor. Change the frequency (or press Play to sweep it) and watch the current. The resistor does not care about frequency; the coil passes less current as the frequency rises, the capacitor passes more. Section 7.1 explains why.

50
frequency $\nu$—
opposition $V/I$ (Ω)—
rms current $I$—
current compared with voltage—

Sections in this chapter

7.1

AC Voltage Applied to a Resistor, an Inductor and a Capacitor

Peak and rms values, phasors, and how a resistor, a coil and a capacitor respond to a sinusoidal voltage: reactance and phase.

$I=\dfrac{i_m}{\sqrt2}$, $\ X_L=\omega L$, $\ X_C=\dfrac{1}{\omega C}$
Open section →7.2

Phasors, the Series LCR Circuit and Resonance

Adding voltages with phasors, impedance and phase angle, resonance, bandwidth and the quality factor.

$Z=\sqrt{R^2+(X_L-X_C)^2}$, $\ \omega_0=\dfrac{1}{\sqrt{LC}}$
Open section →7.3

Power in AC Circuits and the Transformer

Average power, the power factor and wattless current; the ideal transformer, its losses, and power transmission at high voltage.

$P=VI\cos\phi$, $\ \dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}$
Open section →
✅ Before you start
You need Ohm's law and electrical power from Chapter 3 (Section 3.1 and Section 3.2), and capacitance, $q=CV$ and the energy $\tfrac12CV^2$ from Section 2.3 and Section 2.4. From Chapter 6 — Electromagnetic Induction you need the self-induced emf $-L\,di/dt$ of a coil, the energy $\tfrac12Li^2$ it stores, mutual induction (for the transformer), and the AC generator that produces $v=v_m\sin\omega t$. From Class 11 Physics, Chapter 13 — Oscillations, the language of SHM (amplitude, angular frequency $\omega=2\pi\nu$, phase) is used throughout. From mathematics you need $\sin$ and $\cos$, the identity $\sin^2\theta=\tfrac12(1-\cos2\theta)$, and the derivatives of $\sin\omega t$ and $\cos\omega t$.
← 6.3 Self and Mutual Inductance and the AC Generator7.1 AC Voltage Applied to a Resistor, an Inductor and a Capacitor →