The Workbook

\(i = i_0\sin\omega t\), with \(\omega = 2\pi f\). Mains: \(\omega = 2\pi \times 50 = 100\pi \approx 314\) rad/s. Peak \(= \sqrt2 \times 220 \approx 311\) V.

1. \(T = 20\) ms, \(f = \dfrac{1}{T} = 50\) Hz, \(\omega = 2\pi f = 100\pi\) rad/s.

2. \(v = 10\sin(100\pi t)\) V, with \(t\) in seconds.

3. \(100\pi \times 0.0025 = \dfrac{\pi}{4}\): \(v = 10\sin 45^\circ = 5\sqrt2 \approx 7.1\) V.

The rms value is the steady current that would produce the same heating in a resistor as the alternating current: \(I_{\text{rms}} = \sqrt{\langle i^2\rangle}\). Heating depends on \(i^2\), which does not average to zero, so rms measures what the current can do. For a sinusoid \(I_{\text{rms}} = \dfrac{i_0}{\sqrt2} = \dfrac{2}{\sqrt2} \approx 1.41\) A.

\(i^2 = i_0^2\sin^2\omega t\). Over a cycle \(\sin^2\omega t\) averages to \(\dfrac{1}{2}\) (Q24), so \(\langle i^2\rangle = \dfrac{i_0^2}{2}\) and \(I_{\text{rms}} = \sqrt{\dfrac{i_0^2}{2}} = \dfrac{i_0}{\sqrt2}\).

Over a full cycle the positive and negative halves cancel: mean \(= 0\). Over the half cycle \(0\) to \(\dfrac{T}{2}\): \(\bar i = \dfrac{2}{T}\int_0^{T/2}i_0\sin\omega t\,dt = \dfrac{2i_0}{\pi} \approx 0.64i_0\).

1. \(i^2 = 25\) at every instant, so \(\langle i^2\rangle = 25\) and \(I_{\text{rms}} = 5\) A.

2. The current is \(+5\) for half the time and \(-5\) for the other half: mean \(= 0\).

1. The current rises uniformly from \(0\) to \(6\) A, so its mean is \(3\) A.

2. With \(i = \dfrac{6t}{T}\): \(\langle i^2\rangle = \dfrac{1}{T}\int_0^T\dfrac{36t^2}{T^2}dt = 12\), so \(I_{\text{rms}} = \sqrt{12} = 2\sqrt3 \approx 3.5\) A.

Mean: the sine averages to zero, leaving \(3\) A. \(i^2 = 9 + 24\sin\omega t + 16\sin^2\omega t\), whose average is \(9 + 0 + 8 = 17\). \(I_{\text{rms}} = \sqrt{17} \approx 4.1\) A.

\(2\sin\omega t + 2\cos\omega t = 2\sqrt2\sin\left(\omega t + \dfrac{\pi}{4}\right)\): a sinusoid of peak \(2\sqrt2 \approx 2.8\) A. Rms \(= \dfrac{2\sqrt2}{\sqrt2} = 2\) A.

Mean: the positive half contributes \(\dfrac{2i_0}{\pi}\) over its own half, and the other half is zero: \(\bar i = \dfrac{i_0}{\pi} = \dfrac{4}{\pi} \approx 1.3\) A. Rms: \(\langle i^2\rangle\) is half that of the full sine, \(\dfrac{i_0^2}{4}\), so \(I_{\text{rms}} = \dfrac{i_0}{2} = 2\) A.

Heat \(\propto I_{\text{rms}}^2\). DC: \(4R\) per unit time. AC of peak \(2\) A has \(I_{\text{rms}} = \sqrt2\) A, giving \(2R\): half as much heat.

A phasor is a vector rotating anticlockwise at the angular frequency \(\omega\). Its length is the peak value and its angle at any instant is the phase \(\omega t + \phi\). The instantaneous value is the projection of the phasor on the vertical axis, \(v_0\sin(\omega t + \phi)\).

1. The current phasor is \(90^\circ\) behind: \(i = i_0\sin\left(\omega t - \dfrac{\pi}{2}\right) = -i_0\cos\omega t\).

2. A pure inductor, in which the current lags the voltage by a quarter cycle.

In a resistor \(i = \dfrac{v}{R} = 4\sin\omega t\) A, in phase with \(v\). The two phasors point along the same line, of lengths \(20\) V and \(4\) A, and rotate together.

\(X_L = \omega L = 100\pi \times 0.1 = 10\pi \approx 31\) Ω. \(I_{\text{rms}} = \dfrac{220}{31.4} \approx 7\) A. The current lags the voltage by \(90^\circ\).

\(X_C = \dfrac{1}{\omega C} = \dfrac{1}{100\pi \times 10^{-4}} = \dfrac{100}{\pi} \approx 32\) Ω. \(I_{\text{rms}} = \dfrac{220}{31.8} \approx 6.9\) A. The current leads the voltage by \(90^\circ\).

1. Phasors add like vectors: \(V = \sqrt{30^2 + 40^2} = 50\) V.

2. \(\tan\phi = \dfrac{30}{40} = 0.75\), so \(\phi = 37^\circ\) ahead of the \(40\) V phasor.

\(X_L = \omega L\) grows in proportion to the frequency; \(X_C = \dfrac{1}{\omega C}\) falls as the frequency rises. For direct current \(\omega = 0\): an inductor has zero reactance and passes DC freely, while a capacitor has infinite reactance and blocks it.

\(I_{\text{rms}} = \dfrac{P}{V_{\text{rms}}} = \dfrac{100}{220} \approx 0.45\) A. Peak \(= \sqrt2 \times 0.45 \approx 0.64\) A.

AC meters read rms values. Peak \(= \sqrt2 \times 2 \approx 2.8\) A.

\(V_{\text{rms}} = \dfrac{100}{\sqrt2} \approx 70.7\) V. \(\omega = 314 = 2\pi f\), so \(f = 50\) Hz. The first peak is at \(\omega t = \dfrac{\pi}{2}\): \(t = \dfrac{T}{4} = 5\) ms.

The coil's torque reverses every half cycle. Its inertia is far too large to follow \(50\) reversals a second, so it responds only to the average current, which is zero. A hot-wire or rectifier instrument is needed for AC.

The phasors are \(3\) along the reference and \(4\) at \(90^\circ\) ahead. Resultant \(\sqrt{9 + 16} = 5\) A, at \(\tan^{-1}\dfrac{4}{3} = 53^\circ\) ahead: \(i = 5\sin(\omega t + 53^\circ)\).

\(\sin^2\omega t = \dfrac{1 - \cos 2\omega t}{2}\). Over a full cycle the cosine term averages to zero, since it spends equal times positive and negative. What remains is \(\dfrac{1}{2}\). Equivalently, \(\sin^2 + \cos^2 = 1\) and the two have equal averages, so each is \(\dfrac{1}{2}\).

\(Z = \sqrt{R^2 + (X_L - X_C)^2}\) and \(\tan\phi = \dfrac{X_L - X_C}{R}\). In the phasor diagram the current phasor is the reference. \(V_R\) lies along it, \(V_L\) is \(90^\circ\) ahead and \(V_C\) is \(90^\circ\) behind. The applied voltage is the vector sum of the three.

1. \(Z = \sqrt{30^2 + (80 - 40)^2} = \sqrt{900 + 1600} = 50\) Ω. \(I = \dfrac{200}{50} = 4\) A.

2. \(V_R = 4 \times 30 = 120\) V, \(V_L = 4 \times 80 = 320\) V, \(V_C = 4 \times 40 = 160\) V.

3. \(\tan\phi = \dfrac{40}{30} = \dfrac{4}{3}\), so \(\phi = 53^\circ\). \(X_L > X_C\), so the current lags the voltage.

Inductive means \(X_L > X_C\): the inductor dominates and the current reaches its peak after the voltage does. Lowering the frequency, or reducing \(C\) so that \(X_C\) grows past \(X_L\), would make the circuit capacitive with the current leading.

1. \(Z\) is the hypotenuse: \(\sqrt{30^2 + 40^2} = 50\) Ω.

2. \(\cos\phi = \dfrac{R}{Z} = \dfrac{30}{50} = 0.6\).

\(Z = \sqrt{40^2 + 30^2} = 50\) Ω. \(I = \dfrac{100}{50} = 2\) A. \(V_R = 80\) V, \(V_L = 60\) V; check \(\sqrt{80^2 + 60^2} = 100\) V. \(\tan\phi = \dfrac{30}{40} = 0.75\), so the current lags by \(37^\circ\).

\(Z = \sqrt{60^2 + 80^2} = 100\) Ω. \(\tan\phi = \dfrac{80}{60} = \dfrac{4}{3}\), so \(\phi = 53^\circ\). With a capacitor the current leads the voltage.

The three voltages peak at different times: \(V_L\) and \(V_C\) are opposite in phase and partly cancel, and \(V_R\) is \(90^\circ\) from both. Only their phasor sum equals the supply: \(\sqrt{120^2 + (320 - 160)^2} = \sqrt{14400 + 25600} = 200\) V. Adding the numbers directly ignores the phases.

Resonance: \(X_L = X_C\), that is \(\omega L = \dfrac{1}{\omega C}\), so \(\omega_0 = \dfrac{1}{\sqrt{LC}}\). Here \(\omega_0 = \dfrac{1}{\sqrt{0.1 \times 10^{-5}}} = \dfrac{1}{10^{-3}} = 1000\) rad/s and \(f_0 = \dfrac{1000}{2\pi} \approx 159\) Hz.

The reactances cancel, so \(Z = R = 10\) Ω, its smallest value. \(I = \dfrac{20}{10} = 2\) A, the largest possible. \(\phi = 0\): the circuit behaves as a pure resistor.

1. \(Q = \dfrac{\omega_0L}{R} = \dfrac{1000 \times 0.1}{10} = 10\).

2. \(\Delta\omega = \dfrac{\omega_0}{Q} = \dfrac{1000}{10} = 100\) rad/s, the width between the points where the current is \(\dfrac{1}{\sqrt2}\) of its peak.

\(V_L = IX_L = 2 \times 1000 \times 0.1 = 200\) V, ten times the \(20\) V supply. \(V_C\) is also \(200\) V but opposite in phase, so the two cancel and the supply only has to drive the current through \(R\). The ratio \(\dfrac{V_L}{V} = Q\).

The capacitance is varied, which moves \(\omega_0\) to the frequency of the wanted station. A high \(Q\) gives a narrow, tall resonance peak, so the circuit responds strongly to that station and hardly at all to neighbouring frequencies.

\(Q = \dfrac{\omega_0L}{R}\): \(10\) for \(R = 10\) Ω and \(2.5\) for \(R = 40\) Ω. The bandwidth \(\dfrac{\omega_0}{Q}\) is four times larger for the second, so its resonance is four times broader and its peak current four times smaller.

\(P = V_{\text{rms}}I_{\text{rms}}\cos\phi\), with the power factor \(\cos\phi = \dfrac{R}{Z}\). For Q2: \(P = 200 \times 4 \times 0.6 = 480\) W. Check: \(I^2R = 16 \times 30 = 480\) W. All the power goes into the resistor.

The current can be split into \(I\cos\phi\) in phase with the voltage and \(I\sin\phi\) at \(90^\circ\) to it. The second part delivers no average power and is called wattless: \(4 \times \sin 53^\circ = 4 \times 0.8 = 3.2\) A.

1. \(p\) is positive for half of each cycle and equally negative for the other half, so the average power is zero.

2. While \(p > 0\) the source feeds energy into the inductor's magnetic field as the current grows. While \(p < 0\) the field collapses and returns the energy to the source. Nothing is dissipated.

\(I = \dfrac{P}{V\cos\phi}\): with \(\cos\phi = 0.5\), \(I = \dfrac{1000}{100} = 10\) A; with \(\cos\phi = 1\), \(I = 5\) A. The same useful power needs twice the current at the poorer power factor, and the \(I^2R\) loss in the transmission lines is four times larger.

Apparent power \(= V_{\text{rms}}I_{\text{rms}} = 200 \times 4 = 800\) VA. True power \(= 800 \times 0.6 = 480\) W. The difference is reactive power that shuttles back and forth.

A capacitor of suitable size is connected in parallel with the load. Its current leads the voltage by \(90^\circ\) and cancels the lagging wattless current of the inductive load. The total current then comes closer to being in phase with the voltage, and the power factor rises towards \(1\).

A choke is a coil of large inductance and very small resistance. Its reactance \(\omega L\) limits the alternating current just as a resistor would. Since it dissipates almost no \(I^2R\) heat, the power drawn from the supply is nearly all used in the lamp.

\(X_L = 2\pi \times 50 \times 1 = 314\) Ω. \(Z = \sqrt{100^2 + 314^2} \approx 330\) Ω. \(I = \dfrac{220}{330} \approx 0.67\) A. Lamp: \(I^2R = 0.44 \times 100 \approx 44\) W. Choke: \(I^2r \approx 0\).

The lamp needs the same \(0.67\) A, so the total resistance must be \(\dfrac{220}{0.67} \approx 330\) Ω and the added resistor \(\approx 230\) Ω. It wastes \(I^2R = 0.44 \times 230 \approx 103\) W, more than twice the lamp's own power.

Thick wire keeps the resistance, and so the \(I^2R\) loss, small. The iron core multiplies the inductance, giving a large reactance with few turns. Laminating the core cuts the eddy currents that a solid core would carry, so the core itself wastes little energy.

\(v = v_0\sin\omega t\) and \(i = i_0\sin\left(\omega t - \dfrac{\pi}{2}\right) = -i_0\cos\omega t\). \(p = vi = -v_0i_0\sin\omega t\cos\omega t = -\dfrac{v_0i_0}{2}\sin 2\omega t\). A sine of double frequency averages to zero over a cycle, so \(\bar P = 0\).