Electromagnetic oscillations and alternating current

Connect a charged capacitor to a coil. When the charge reaches zero, the discharge does not stop: current continues and charges the plates with opposite signs. What plays the role of inertia here, and where does the electrical energy go temporarily?

When the ideal circuit is closed, the capacitor drives a current through the coil. Its charge falls while the current rises: electric-field energy becomes magnetic-field energy. At q=0 the current is greatest. Self-induction prevents it from vanishing instantly, so it continues to transfer charge and charges the capacitor with reversed polarity. Like a mass passing through equilibrium, the system passes through zero q while the energy exchange continues.

To follow this cycle, q is the signed charge on the chosen plate; qₘ is its greatest magnitude, both in coulombs. I and Iₘ are current and its amplitude in amperes, C is capacitance in farads, and L is inductance in henries. The subscript m means maximum, not mass. Choosing the other plate reverses q but changes neither energy nor qₘ. The signs of current and charge change must agree with the chosen direction around the circuit.

Greater capacitance allows more charge to be transferred at the same voltage; greater inductance more strongly opposes a change in current. The exchange becomes slower: T=2\pi\sqrt{LC} gives the period in seconds. Multiplying L or C by four doubles T. This describes natural oscillations in a closed ideal circuit with constant L and C, no resistance and no source. Real losses require a damped model.

In the ideal case the charge varies harmonically: q=q_m\sin(\omega t+\varphi_0), where ω=2π/T in rad/s and φ₀ is in radians. Current is related to the rate of change of charge. At an extreme value of q the graph is momentarily horizontal, and I=0. At q=0 the graph is steepest, so the current magnitude is greatest. The graph therefore explains why the maxima of charge and current cannot occur together.

Transferring the same charge in a shorter part of the cycle requires more current: I_m=q_m\omega. Doubling qₘ at unchanged frequency doubles Iₘ; doubling ω at unchanged qₘ also doubles it. Maximum |I| occurs at q=0, a quarter-period away from maximum |q|. The units C·s⁻¹ give amperes. The possible sign between I and dq/dt depends on the chosen directions.

At maximum charge the current is zero and all the energy is in the capacitor's electric field. Thus W_0=q_m^2/(2C)=CU_m^2/2, where Uₘ=qₘ/C is the voltage amplitude in volts and W₀ is in joules. Doubling qₘ at fixed C quadruples the energy. When comparing different values of C, specify whether charge or voltage stays fixed: the equivalent expressions predict different changes because the conditions differ.

A quarter-cycle later q=0: the electric energy vanishes and the magnetic energy reaches its maximum, W_0=LI_m^2/2. At an intermediate instant, W_0=q^2/(2C)+LI^2/2 uses the current values of q and I. These are two forms of the same energy store, so their maxima must not be added. Doubling Iₘ at unchanged L quadruples the total energy. With resistance, some energy becomes heat in the conductors each cycle.

Alternating current changes its value and direction periodically. Iₘ, Uₘ and εmax are the greatest magnitudes of current, voltage and EMF; the latter two are measured in volts. In a real freely oscillating circuit, resistance reduces energy and amplitude. An external periodic source can replace the losses and sustain forced oscillations. Periodic behaviour therefore does not mean that no energy input is needed.

The signed mean of a sinusoidal current over a period is zero. Why does a resistor still heat up? Heating power depends on I² and is positive in either direction. The RMS value produces the same average heating as an equal direct current: for a sine wave, I_{\text{д}}=I_m/\sqrt2, U_{\text{д}}=U_m/\sqrt2. This is a root-mean-square measure, not a signed mean; the factor √2 is not universal for other waveforms.

When current changes faster, a coil produces a greater self-induced EMF. Its inductive reactance to sinusoidal current, X_L=2\pi\nu L, therefore increases with frequency ν in hertz and inductance L. X_L is measured in ohms. As ν→0 an ideal coil no longer opposes steady direct current. This is reactance: energy is stored temporarily and returned; the ideal coil itself does not turn it into heat.

A capacitor behaves differently. More frequent voltage reversals recharge the plates more often, giving greater current at the same voltage amplitude: X_C=1/(2\pi\nu C) in ohms. Doubling frequency or C halves X_C. At constant voltage, there is no steady current through an ideal capacitor. Current and voltage differ in phase, so X_L and X_C cannot simply be added like two resistances that dissipate heat.

The two transformer windings share a changing magnetic flux. Alternating current in the primary produces core flux whose change induces EMF in the secondary. The wires need not be electrically connected: the field transfers energy. Steady direct current does not produce continuous induction in the secondary winding. Switching on may produce a brief response, but not continuous voltage transformation.

The same flux creates the same EMF per turn in ideal windings. Thus U_1/U_2=n_1/n_2, where n₁ and n₂ are dimensionless turn counts and U₁ and U₂ are RMS voltages in volts. Doubling the secondary turns at unchanged input voltage doubles the output voltage. The formula assumes ideal magnetic coupling and no losses; it does not determine the current without information about the load.

Where does the extra energy come from when voltage increases? An ideal transformer creates none. For a resistive load, U_1I_1=U_2I_2, hence U_1/U_2=I_2/I_1. I₁ and I₂ are RMS currents in amperes; losses and no-load current are neglected. At the same transferred power, tripling the voltage reduces the current to a third. A load with a phase shift requires an additional factor when calculating active power.

An ideal transformer has 200 primary turns and 50 secondary turns. The input is 240 V; the resistive load draws 2 A at the output. Find the output voltage and input current.

The secondary has 50 turns rather than 200, one quarter as many. The shared flux produces equal EMF per turn, so the output voltage is one quarter of the input.

U₂ = U₁n₂/n₁ = 240 × 50/200 = 60 В. All voltages in this problem are RMS values.

The resistive load receives P₂ = U₂I₂ = 60 × 2 = 120 Вт. In the ideal model, the same power enters the transformer.

I₁ = P₁/U₁ = 120/240 = 0,5 А. Check: 240 × 0,5 = 60 × 2 = 120 Вт; the lower output voltage is accompanied by a greater current.

U₂ = 60 В, I₁ = 0,5 А. The transformer reduces the voltage to a quarter and transfers 120 W to the load without losses in this model.

Self-induction allows current to continue recharging the capacitor; electric and magnetic energy reach their maxima in turn.

The RMS value describes heating and differs from both amplitude and signed mean current.

A transformer uses changing flux to trade voltage against current while conserving power in the ideal model.

Next we will examine how oscillations travel through space.