7.3 Power in AC Circuits and the Transformer

Large factories in many countries pay a penalty on their electricity bill if their "power factor" is too low — even when they use exactly the same number of kilowatt-hours as a neighbour. Why should a utility care how the energy is used? Because an electric motor full of coils draws a current that is partly out of step with the voltage. That part of the current delivers no energy on average, yet it flows through every cable, transformer and generator between the power station and the factory and heats them all. The cure — banks of capacitors — follows directly from the phasors of Section 7.2.

In this section we find the average power in an AC circuit, $P=VI\cos\phi$, define the power factor $\cos\phi$ and the wattless current, and then turn to the device that makes AC the world's choice for power networks: the transformer. We see how a pair of coils on an iron core steps a voltage up or down, what limits its efficiency, and why sending power at high voltage cuts the energy wasted in the transmission lines.

Power in an AC Circuit

Let the source voltage be $v=v_m\sin\omega t$ and the current $i=i_m\sin(\omega t+\phi)$. The instantaneous power delivered by the source is $$p=vi=v_mi_m\sin\omega t\,\sin(\omega t+\phi).$$ Using $2\sin A\sin B=\cos(A-B)-\cos(A+B)$, $$p=\frac{v_mi_m}{2}\left[\cos\phi-\cos(2\omega t+\phi)\right].$$ The second term oscillates at twice the supply frequency and averages to zero over a cycle. The first term is constant.

The four important cases:

• Pure resistor: $\phi=0$, $\cos\phi=1$, $P=VI$ — the largest possible.
• Pure inductor or capacitor: $\phi=\pm\pi/2$, $\cos\phi=0$, $P=0$ although a current flows (Section 7.1).
• Series LCR circuit: $\cos\phi=R/Z$, $P=I^2R$.
• At resonance: $Z=R$, $\cos\phi=1$, $P=V^2/R$ — the most power the circuit can take from the source.

The instantaneous power $p$ is negative for part of every cycle whenever $\phi\ne0$: during those intervals the coil or capacitor hands stored energy back to the source.

Wattless Current and the Power Factor

Draw the current phasor $\mathbf I$ at angle $\phi$ to the voltage phasor $\mathbf V$ and split it into two perpendicular components. The part $I\cos\phi$ along $\mathbf V$ is in phase with the voltage and carries all the power, $V(I\cos\phi)$. The part $I\sin\phi$ is $90^\circ$ out of phase with the voltage — like the current in a pure inductor or capacitor — and carries none.

Wattless current is not harmless. It flows through the supply cables and the generator windings and heats them ($I^2R$ depends on the whole current), and for a given power a small $\cos\phi$ means a large current $I=P/(V\cos\phi)$. Electric motors, welding sets and old fluorescent fittings are inductive and can have power factors of $0.6$–$0.8$. Utilities therefore ask large customers to keep $\cos\phi$ close to $1$, usually by connecting capacitors across the supply (see the 🚀 box).

The Transformer

A transformer is two coils wound on the same core of soft iron: the primary, with $N_p$ turns, connected to the AC supply, and the secondary, with $N_s$ turns, connected to the load. The alternating current in the primary sets up an alternating magnetic flux in the core, and the iron guides almost all of that flux through the secondary as well. By mutual induction (Chapter 6) the changing flux induces an emf in every turn of both coils.

Suppose (i) all the flux $\Phi$ through each turn of the primary also passes through each turn of the secondary (no flux leakage), and (ii) the coils have negligible resistance. Each turn then has the same induced emf $-d\Phi/dt$, so $$\varepsilon_p=-N_p\frac{d\Phi}{dt},\qquad \varepsilon_s=-N_s\frac{d\Phi}{dt}.$$ With no resistance the primary emf balances the applied voltage, $v_p=-\varepsilon_p$, and on open circuit the secondary voltage is $v_s=-\varepsilon_s$. Dividing, $\dfrac{v_s}{v_p}=\dfrac{N_s}{N_p}$ — at every instant, and therefore for rms values too. If, in addition, (iii) no energy is lost, the power drawn from the supply equals the power delivered to the load: $V_pI_p=V_sI_s$.

A transformer is not a source of free energy: a step-up transformer raises the voltage only by lowering the current. The secondary current is set by the load ($I_s=V_s/Z_\text{load}$), and the primary then draws just enough current to supply that power.

Energy losses in real transformers. Large transformers reach efficiencies above $95\%$, but some energy is always lost:

• Flux leakage — some flux of the primary misses the secondary. Reduced by winding the two coils one over the other.
• Resistance of the windings — $I^2R$ heating ("copper loss"). Reduced by using thick wire, especially for the low-voltage, high-current coil.
• Currents induced in the iron core (eddy currents), which heat it. Reduced by building the core from thin insulated sheets — a laminated core.
• Hysteresis — energy used in magnetising the core back and forth every cycle (Chapter 5). Reduced by using soft iron, which is easy to magnetise and demagnetise.

The efficiency of a transformer is $\eta=\dfrac{\text{output power}}{\text{input power}}=\dfrac{V_sI_s}{V_pI_p}$ (for loads with power factor close to 1).

Transmitting Power at High Voltage

A power station must send a power $P$ to a distant town along lines of total resistance $R_\text{line}$. The current in the lines is $I=P/V$, where $V$ is the transmission voltage (taking the power factor as $1$), and the power lost as heat in the lines is $I^2R_\text{line}$.

This is the reason for alternating current. Generators produce a few kilovolts to a few tens of kilovolts; step-up transformers at the power station raise this to $220$ kV, $400$ kV or $765$ kV for long-distance lines; a chain of step-down transformers then lowers it to $33$ kV, $11$ kV and finally $230$ V for homes. A d.c. system cannot change its voltage so simply, which is why the AC grid won the "war of the currents" at the end of the nineteenth century.

Exercises