This mini-lesson covers Eduqas Component 3, Option A: the rotating coil and Faraday's law, peak and rms values, mean power, the oscilloscope, the phase relationships for inductors and capacitors, reactance, phasors, impedance, RCL series resonance and the Q factor.
⚠️ This is an OPTION. Eduqas Component 3 Section B offers four options — A Alternating Currents, B Medical Physics, C The Physics of Sports, D Energy and the Environment. You study exactly ONE of them. Only work through this mini-lesson if Option A: Alternating Currents is the one your school teaches. Everyone also needs the Component 3 core (Sections 1–10), which has its own mini-lesson.
Press Start when you are ready.
(a)–(c) The rotating coil
Where alternating e.m.f. comes from
Spin a flat coil of N turns and area A in a uniform field B at angular velocity ω. The angle between the coil's normal and the field is θ = ωt, so the flux linkage is
NΦ = BAN cos ωtFaraday: the induced e.m.f. is the rate of change of flux linkage — i.e. minus the derivative of the above.
V = ωBAN sin ωtSo the peak e.m.f. is V₀ = ωBAN. Note where the maxima are: the e.m.f. is largest when the flux linkage is zero (coil in the plane of the field), because that is where the flux is changing fastest.
The counter-intuitive result: maximum flux → zero e.m.f.; zero flux → maximum e.m.f. Faraday's law depends on the rate of change, not the value.
Calculate
Your turn — peak e.m.f. of a generator
1A flat coil of 200 turns and area 0.010 m² rotates at 50 Hz in a uniform field of flux density 0.20 T. Calculate the peak e.m.f.
An alternating current is constantly changing, so its mean value over a cycle is zero — useless. What matters is the power it delivers, and power depends on I². The root-mean-square value is defined as the steady d.c. value that would dissipate the same mean power in a resistor.
mean P = IrmsVrms = Irms²R = Vrms² / RUse rms values in power calculations — never peak values. UK mains is quoted as 230 V rms, so its peak is 230 × √2 ≈ 325 V.
Using a CRO: the time-base (s/div) gives the period from the width of one cycle (and then f = 1/T); the Y-gain (V/div) gives the peak voltage from the height of the trace above the centre line. For a.c. the trace is a sine wave; for d.c. it is a displaced horizontal line.
Calculate
Your turn — rms voltage
2An alternating supply has a peak voltage of 325 V. Calculate its rms voltage.
V
Hint: V(rms) = V₀ / √2 = 325 ÷ 1.414.
Calculate
Your turn — mean power
3An alternating supply of 12 V rms is connected across a 6.0 Ω resistor. Calculate the mean power dissipated.
W
Hint: Mean P = V(rms)² / R = 12² ÷ 6.0. (Using the peak value here would give the wrong answer — this is exactly what rms is for.)
Quick check
What rms actually means
?What does the rms value of an alternating current tell you?
(h)–(k) Inductors and capacitors in a.c.
Phase, reactance and zero mean power
In a pure resistor, current and p.d. are in phase. Add an inductor or a capacitor and they no longer are:
Inductor: the current lags the p.d. by 90°. (The back-e.m.f. opposes any change in current, so the current is always "catching up".)
Capacitor: the current leads the p.d. by 90°. (Charge has to flow before the p.d. across the plates can build up.)
XL = ωL · XC = 1 / ωCReactance = V(rms) / I(rms), measured in ohms. XLrises with frequency; XCfalls with frequency. At f = 0 (d.c.) an inductor is a plain wire and a capacitor is an open circuit.
Zero mean power: because the current and p.d. are exactly 90° out of phase, an ideal inductor or capacitor dissipates no net power — it stores energy for a quarter cycle and gives it all back the next. Only the resistance in a circuit dissipates energy.
Calculate
Your turn — inductive reactance
4Calculate the reactance of a 0.25 H inductor at a frequency of 50 Hz.
?What is the mean power dissipated in an ideal capacitor in a sinusoidal a.c. circuit?
Sort it
Resistor, inductor or capacitor?
Tap a property, then tap the component it belongs to.
🟦 Resistor
🌀 Inductor
🔋 Capacitor
(l)–(m) Phasors and impedance
Adding p.d.s with phasors
Because the p.d.s across R, L and C in series are out of phase with one another, you cannot simply add them arithmetically. Represent each as a rotating vector — a phasor — and add them as vectors:
VR along the current direction (in phase with I).
VL at +90° (p.d. leads the current).
VC at −90° (p.d. lags the current).
V = √(VR² + (VL − VC)²)and dividing throughout by I(rms): Z = √(R² + (XL − XC)²)
Z = Vrms / Irms · tan φ = (XL − XC) / RImpedance Z (in ohms) is the a.c. equivalent of resistance; φ is the phase angle between the supply p.d. and the current.
Right-angled triangle — so Pythagoras gives the impedance and the tangent gives the phase angle.Calculate
Your turn — impedance
6A series circuit has R = 60 Ω and an inductive reactance of X(L) = 80 Ω (no capacitor). Calculate the impedance.
At resonance the impedance is a minimum and equals just R, so the current is a maximum and is in phase with the supply p.d. (φ = 0). This is exactly how a radio is tuned: adjust C until f₀ matches the station.
Q = VL / VR = VC / VR (at resonance)A high Q means a tall, sharp resonance peak — good selectivity. A low Q gives a broad, flat peak.
The surprise: at resonance VL and VC can each be far larger than the supply voltage (that is precisely what a Q factor above 1 means) — yet they are 180° out of phase with each other, so they cancel and the phasor sum still equals the supply p.d.
Calculate
Your turn — resonant frequency
7A series RCL circuit has L = 0.10 H and C = 2.0 µF. Calculate the resonant frequency.