Electricity and energy in context

Start with the boundary and units

When we say a motor “consumes energy”, we do not mean that the energy disappears. Energy enters the equipment, may increase stored energy and leaves in several forms: mechanical work, heating, sound or other transfers. The boundary must be chosen: a motor alone does not have the same balance as motor, source and cables together.

Power is the rate of energy transfer: average P = ΔE/Δt. One watt means one joule per second. Energy transferred at constant power is E = PΔt. If power varies, we add small PΔt contributions or use E = ∫P(t)dt. Rated power multiplied by time is appropriate only when it reasonably represents operation over that interval.

A common energy unit is the kilowatt-hour: 1 kWh = 1000 W × 3600 s = 3,6 × 10⁶ J. The symbol includes multiplication by time; it is not “kilowatts per hour”. The distinction matters: two devices can have the same power but transfer very different energies if they run for different durations.

Voltage, current and efficiency in a direct-current circuit

Potential difference U corresponds to energy per unit charge. If charge Δq crosses a component during Δt, average current is I = Δq/Δt. Electrical energy transferred is UΔq when U is constant, and power is P = UI. The product V × A has unit J/C × C/s = J/s, confirming that this is power.

For an ohmic resistor, U = RI. Substituting in the power expression gives P = I²R or P = U²/R. These are useful forms of the same relationship but depend on the resistive model used. A motor also has electromechanical conversion and electromotive force associated with motion; it is not appropriate to always treat it as a constant-resistance resistor.

Energy efficiency is η = useful energy/energy received, or the ratio of the corresponding average powers under suitable operating conditions. We have 0 ≤ η ≤ 1 for this simple definition of conversion. What counts as useful depends on the goal: heating is useful in a heater and may be a loss in a motor. An energy loss for the task does not mean energy is destroyed; it means energy is transferred into forms that do not meet the chosen goal.

What an ampere-hour rating allows us to conclude

Capacity expressed in ampere-hours is an amount of charge: 1 Ah = 3600 C. By itself, it is not energy. To estimate the electrical energy supplied by a battery whose voltage is approximately constant, we can multiply voltage by capacity: E ≈ U × capacity in Ah, giving Wh. This approximation ignores changes in voltage, discharge conditions, temperature and other limits of real operation.

A 12 V, 5 Ah battery corresponds, in this model, to about 60 Wh, or 216000 J. If a conversion delivers 80% of that energy to a constant 6 W load, the useful energy would be 48 Wh and the idealized operating time would be 8 h. The result is an estimate, not a guarantee of operating time. Available capacity and efficiency can change during use.

Instantaneous current must not be confused with total capacity either. A battery labelled 5 Ah does not necessarily supply 5 A in every circuit: current depends on the connected system and its limits. The calculations in this lesson analyze models; they are not instructions for assembling or modifying batteries, protective devices or electrical installations.

Why a zero-average current can cause heating

In direct current, the conventional direction does not alternate; in a simple steady state, its value is also constant. In alternating current, direction changes periodically. A sine wave can be written i(t) = I₀ sin(ωt), with amplitude I₀, frequency f and angular frequency ω = 2πf. Over a complete period, the algebraic average of this current is zero.

In a resistor, instantaneous dissipated power is p(t) = Ri²(t). Even when i is negative, its square is positive: energy can continue to be transferred to the resistor. For a sine wave, the average of sin² over a period is 1/2, so average P = RI₀²/2. This is why average current is not the appropriate value for comparing heating effects.

The effective value, also called RMS, is defined for a waveform as the square root of its mean square over the appropriate interval. The relationship amplitude divided by √2 is specific to a sine wave. For an ideal rectangular wave alternating between +I₀ and −I₀, the square is always I₀² and the RMS value is I₀. Knowing the peak without knowing the waveform may therefore be insufficient.

Active power, apparent power and power factor

In an ideal resistor supplied by sinusoidal voltage, current and voltage are in phase: they reach corresponding zeros and maxima at the same instants. Average power is RMS U × RMS I. Capacitors and inductors can store energy and return it to the circuit, shifting the phase of current relative to voltage. The product of RMS values is then not necessarily the average power converted into other forms.

If u(t) = U₀ sin(ωt) and i(t) = I₀ sin(ωt − φ), their product can be split into a mean term and an oscillating term at twice the frequency. When averaging over complete periods, the oscillating term disappears, leaving P = RMS U × RMS I × cos φ. This formula assumes the sinusoidal waveforms and periodic conditions being considered.

Apparent power S is the product RMS U × RMS I, expressed in volt-amperes, VA. Active power P is measured in W. For this sinusoidal case, P/S = cos φ is the power factor. With distorted waveforms, power factor can still be defined as P/S, but must not automatically be identified with the cosine of a single angle. For a passive load with positive active power, the factor is between 0 and 1.

A lower power factor implies a higher current to transfer the same active power at the same RMS voltage. This can increase resistive losses in cables. It does not mean that reactive energy has been entirely converted to heat in the component: some is exchanged back with the source during the cycle.

A transformer changes voltage through induction

An ideal transformer has two windings associated with a common varying magnetic flux. By Faraday's law, the magnitude of induced electromotive force is proportional to the number of turns and the rate of change of flux. If the same flux passes through every turn of both windings, the voltage ratio is U_s/U_p = N_s/N_p, using corresponding values, such as amplitudes or RMS values.

The subscripts p and s identify primary and secondary; N counts turns. A secondary with fewer turns allows the voltage to be reduced. In the ideal model, average input power equals output power. For resistive loads, this implies U_pI_p = U_sI_s, so the current ratio is the inverse of the voltage ratio. A transformer does not increase voltage and current simultaneously by creating power.

A changing flux is needed to induce voltage continuously. Applying a steady DC voltage does not reproduce how an AC transformer works. Electronic devices can first switch a DC source and then use induction at high frequency; that is a broader system. Real transformers have winding and core losses, and limits involving frequency, temperature and saturation.

Why raising voltage can reduce losses

In a resistive line model, dissipated power is P_loss = I²R_line. To transfer active power P at power factor 1 and voltage U, we have I = P/U. Substitution gives P_loss = P²R_line/U². Keeping the assumptions unchanged, multiplying voltage by ten divides current by ten and loss by one hundred.

This reasoning helps explain the use of high voltages in transmission. Reducing losses does not eliminate conductor resistance or mean that voltage can be increased without limits. Insulation, geometry, electric fields, equipment, protection and infrastructure costs also matter; this expression is not a design for a real grid.

Voltage can be transformed at different stages to suit transmission and use. At every real stage, the balance must include transformer and line losses. One prediction is comparable to another only when the boundaries, reference power and duration are the same.

Turn numbers into justified conclusions

To analyze a device, first identify what was measured: input power, useful output, average or peak value? Then ask whether the data refer to the same interval. An intermittent appliance may operate only part of the time, and multiplying maximum power by all the hours may overestimate energy.

Write the units before substituting numbers and check the order of magnitude. A calculation in Wh cannot be added directly to another in J. If a cost per kWh is used, it must be stated as problem data; a fictional exercise tariff is not current information about real prices. Also distinguish lower energy consumption from higher efficiency: they can change for different reasons.

A responsible conclusion states the conditions: “in this constant-power model”, “assuming approximately constant voltage” or “under sinusoidal operating conditions”. Assumptions make a conclusion useful because they say where it can fail. For understanding real installations, the models in this lesson provide analytical foundations; practical procedures and grid work require appropriate training and conditions.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Mistakes worth avoiding

kWh is power or power per hour.

It is energy: power multiplied by the interval. One kWh equals 3,6 million joules.

RMS value is always amplitude divided by √2.

That holds for a sine wave. The general definition depends on the waveform's mean square.

RMS U times RMS I is always active power.

For sinusoidal voltage and current with a phase difference, the factor cos φ is needed. For general waves use the average instantaneous product.

A transformer increases voltage and current freely.

Ideally it conserves power; in reality there are losses. Voltage and current ratios cannot create energy.

Ah is already a measure of energy.

Ah measures charge. Estimating energy requires knowing voltage and operating conditions.