The average power
A resistor takes power on every part of the cycle; an inductor or a capacitor takes none on average. A real circuit lies between the two. Let the supply be \(v = v_0\sin\omega t\), and let the current lag it by the phase angle \(\phi\): \(i = i_0\sin(\omega t - \phi)\). The power delivered at an instant is \(p = vi\).
\[ p = v_0 i_0\sin\omega t\,\sin(\omega t - \phi) \]
Why. multiply the voltage and the current at the same instant.
\[ p = \dfrac{v_0 i_0}{2}\left[\cos\phi - \cos(2\omega t - \phi)\right] \]
Why. the identity \(2\sin A\sin B = \cos(A - B) - \cos(A + B)\), with \(A - B = \phi\).
\[ \langle\cos(2\omega t - \phi)\rangle = 0 \]
Why. a cosine of twice the frequency averages to zero over a whole cycle.
\[ P = \dfrac{v_0 i_0}{2}\cos\phi = V_{rms}I_{rms}\cos\phi \]
Read it. only the constant term survives; \(v_0 i_0/2 = (v_0/\sqrt2)(i_0/\sqrt2)\).
The power factor of an AC circuit is \(\cos\phi\), where \(\phi\) is the phase angle between the supply voltage and the current. The average power is \(P = V_{rms}I_{rms}\cos\phi\). For a series circuit \(\cos\phi = R/Z\).
Play each chip of Fig 6.1. For the resistor, \(p\) never goes below zero and averages \(800\,\text{W}\). For the inductor and the capacitor, the brown and teal lobes are equal, and the average is zero. For the circuit of lesson 4, with \(\phi = 53°\), the brown lobes are larger. The average is \(200 \times 4 \times 0.6 = 480\,\text{W}\). The same \(200\,\text{V}\) and \(4\,\text{A}\) deliver anything from \(800\,\text{W}\) down to nothing, as the phase changes.
Put \(\cos\phi = R/Z\) and \(V = IZ\) into the formula. Then \(P\) \(= IZ \times I \times R/Z\) \(= I^2R\). So all the average power goes into the resistance. For the lesson-4 circuit, \(I^2R\) \(= 16 \times 30\) \(= 480\,\text{W}\), the same as before. The product \(V_{rms}I_{rms} = 800\,\text{VA}\) is called the apparent power. It is written in volt-amperes, VA, to keep it apart from the true power in watts.
The series circuit of lesson 4 has \(R = 30\,\Omega\), \(X_L = 80\,\Omega\) and \(X_C = 40\,\Omega\) on \(200\,\text{V}\). Find the power factor, the average power, the apparent power, and the power at resonance on the same supply.
\(Z = 50\,\Omega\) and \(I = 4\,\text{A}\), so \(\cos\phi\) \(= R/Z\) \(= 30/50\) \(= 0.6\).
\(P\) \(= 200 \times 4 \times 0.6\) \(= 480\,\text{W}\). Check: \(I^2R\) \(= 16 \times 30\) \(= 480\,\text{W}\).
The apparent power is \(200 \times 4 = 800\,\text{VA}\).
At resonance \(Z = R\) and \(\cos\phi = 1\), so \(I = 200/30 = 6.7\,\text{A}\) and \(P\) \(= 200^2/30\) \(= 1330\,\text{W}\).
Values used: \(\cos 53° = 0.6\), \(\sin 53° = 0.8\), as the sheets take them.
A motor takes \(5\,\text{A}\) rms from the \(220\,\text{V}\) mains with a power factor of \(0.8\). Find the average power it takes, in watts.
Which statements about the power factor are true?
The current that does no work
The current phasor can be split into two parts. One part, \(I\cos\phi\), lies along the voltage phasor. The other, \(I\sin\phi\), is a quarter turn from it. The part along the voltage carries all the power, \(V \times I\cos\phi\). The part at right angles carries none on average, exactly as the current in a pure inductor does. It is called the wattless current.
The wattless current carries no power to the load, but it still flows in the wires. So it heats the cables, the transformers and the lines of an \(11\,\text{kV}\) feeder by \(I^2R\) all the same. Take a load that needs \(1000\,\text{W}\) at \(200\,\text{V}\). At a power factor of \(1\) it draws \(1000/200 = 5\,\text{A}\). At a power factor of \(0.5\) it draws \(1000/(200 \times 0.5) = 10\,\text{A}\). The loss in the lines, \(I^2R\), is then four times as large.
Motors, fans and tube-light chokes are inductive loads, with a lagging current. A capacitor connected across such a load draws a current that leads the voltage by \(90°\). That leading current cancels the lagging wattless current of the load. The line then carries only the current the load needs for power. So factories connect banks of capacitors across their motors. Electricity boards charge large consumers extra when the power factor falls below a set value.
A load draws \(10\,\text{A}\) rms, lagging the supply voltage by \(37°\). Find its wattless current, in amperes. Take \(\sin 37° = 0.6\).
Two factories take the same power at the same voltage. One has a power factor of \(0.5\) and the other of \(1\). Compare the loss of power in the lines that feed them.
The choke coil
Often a load must take less current than the supply would drive through it. A tube light is the everyday case. A resistor in series would hold the current down, but it would turn power into heat. A coil does the same job without the heat. A choke coil is a coil of large inductance and very small resistance. Its reactance limits the current. Its current is a quarter cycle behind the voltage across it, so it takes almost no average power.
A lamp of resistance \(100\,\Omega\) is connected to the \(220\,\text{V}\), \(50\,\text{Hz}\) mains through a choke of \(1\,\text{H}\) and negligible resistance. Find the reactance of the choke, the impedance, the current, and the power in the lamp and in the choke.
The lamp is the resistance, \(R = 100\,\Omega\). The choke is a pure inductor, \(L = 1\,\text{H}\), on \(\omega = 314\,\text{rad/s}\).
\(X_L = \omega L\), \(Z = \sqrt{R^2 + X_L^2}\), \(I = V/Z\), \(P_{lamp} = I^2R\).
\(X_L = 314\,\Omega\). \(Z\) \(= \sqrt{100^2 + 314^2}\) \(= 330\,\Omega\). \(I\) \(= 220/330\) \(= 0.67\,\text{A}\). \(P_{lamp}\) \(= 0.667^2 \times 100\) \(= 44.5\,\text{W}\).
The lamp takes \(44.5\,\text{W}\) and the choke nothing. The whole \(44.5\,\text{W}\) drawn from the mains goes into the lamp. The power factor is only \(100/330 = 0.30\).
A lamp of resistance \(100\,\Omega\) must carry \(1\,\text{A}\) from the \(220\,\text{V}\), \(50\,\text{Hz}\) mains. Find the inductance of the pure choke needed in series, in henry.
Why not a resistor?
Choose the second chip of Fig 6.3. To give the lamp the same \(0.67\,\text{A}\), the total resistance must be the same \(330\,\Omega\). So a resistor of about \(230\,\Omega\) must be added. It turns \(0.667^2 \times 230 = 102\,\text{W}\) into heat, more than twice the lamp's own \(44.5\,\text{W}\). The mains then supplies \(147\,\text{W}\) to light a \(44.5\,\text{W}\) lamp. The choke gives the same light for \(44.5\,\text{W}\).
A good choke needs a large inductance and a small resistance. Thick copper wire keeps the resistance, and so the \(I^2R\) loss, small. An iron core multiplies the inductance, so few turns give a large reactance. The core is built of thin insulated sheets, or laminations. A solid core would carry eddy currents and waste energy, as the chapter Eddy Currents, Generators and Inductance showed.
Why does a choke limit an alternating current with almost no loss of power, while a resistor doing the same job gets hot?
A home inverter is rated \(900\,\text{VA}\). It runs fans and tube lights whose overall power factor is about \(0.8\). Estimate the largest total power, in watts, of the loads it can run.
An old tube-light fitting has a heavy black box in series with the tube: the choke. When the starter opens, the choke's \(L\,dI/dt\) strikes the tube, as the chapter Eddy Currents, Generators and Inductance explained. Once the tube glows, the choke's second job begins. The glowing gas would draw a large and growing current on its own. The choke's reactance holds the current near the rated value with little heat. Its lagging current gives the fitting a low power factor, which is why many fittings also carry a capacitor across the mains. The old fan regulator worked like a resistor and grew warm; many newer step regulators use capacitors in series with the fan and stay cool.
- Writing the average power as \(V_{rms}I_{rms}\) for every circuit.That is the apparent power, in VA. The true power is \(V_{rms}I_{rms}\cos\phi\).
- Thinking the wattless current does no harm.It carries no average power, but it flows in the lines and heats them by \(I^2R\).
- Finding a choke by \(X_L = Z - R\).Resistance and reactance add at right angles: \(X_L = \sqrt{Z^2 - R^2}\).
- Expecting a choke to limit a direct current.On direct current \(\omega = 0\), so a choke has only its small resistance and limits almost nothing.