Electrical work and circuits

Identical lamps can have different brightnesses in different connections. To understand why, follow the energy transfer: how much energy each coulomb receives and how many coulombs pass each second. Current itself is not used up.

Voltage U gives energy per coulomb; current I gives coulombs per second. Their product is power, P=UI: energy transferred per second. Its unit is the watt: 1\,\text{Вт}=1\,\text{В}\cdot\text{А}. At the same U, doubling current doubles the charge passing each second and hence doubles power. Here we consider energy delivered to a load in a direct-current circuit. When the quantities vary, their product gives the instantaneous power.

For constant current, charge q = It passes in time t. At constant U, the work is therefore A=Uq=UIt=Pt. Doubling the time doubles the energy transferred at the same power. The units agree: 1\,\text{Дж}=1\,\text{Вт}\cdot\text{с}=1\,\text{В}\cdot\text{Кл}. If voltage or current varies, multiplying selected values by the entire time need not give the correct result. Instead, add the energy transfers over intervals; A/t gives average power.

For an ohmic resistor, substitute either U = IR or I = U/R into P = UI to obtain P=I^2R=U^2/R. For constant quantities, A=I^2Rt=(U^2/R)t. These are different forms of the same energy transfer relation using Ohm’s law. Choose the form that matches what is known and held constant. An electric motor also converts energy into work and produces a back EMF, so it cannot automatically be represented by a single ohmic resistor.

In a resistor, the directed motion of charge carriers transfers energy to the material, increasing its internal energy. For constant I and R, Joule’s law gives the electrical heat generated Q: Q=I^2Rt. Doubling current makes it four times as large; doubling time doubles it. Q in joules need not equal the observed increase in internal energy: the hot resistor may simultaneously transfer energy to its surroundings. Its temperature also depends on heat capacity and heat transfer.

Replace the resistor with a motor lifting a load. Part of the incoming electrical energy becomes mechanical work; part heats the windings and other components. Thus UIt cannot all be called heat without checking the energy conversions. The expression I^2Rt describes heating losses in the winding resistance, not the motor’s entire useful output. In an ordinary stationary resistor without other conversions, electrical energy becomes internal energy; its subsequent transfer away is a separate process.

The power expressions seem to disagree: increasing R increases I^2R but decreases U^2/R. The contradiction disappears once the fixed quantity is named. At a given current, the source must provide a higher voltage across a larger R, so power rises. At a given voltage, current through a larger R falls, so power falls. Before comparing, imagine changing the resistance and ask how the source responds. Related quantities I and U cannot both be treated as independent.

In a series connection, charge has one continuous path. If more charge arrived through the first resistor each second than left through the second, charge would accumulate between them. This does not happen in a steady state, so I_1=I_2=I. Voltages add, U=U_1+U_2, because energy per coulomb is transferred in both successive sections. Equal currents do not imply equal energy transfers: the voltages can differ.

With a common current, U=IR_1+IR_2=I(R_1+R_2), hence R=R_1+R_2. Adding a series element increases total resistance and reduces current at the same source voltage. A larger R has a larger drop Uᵢ = IRᵢ and a larger power at the common current. For n identical resistors, R=nR_1, where n is the number of elements. Doubling their number doubles the total resistance if each individual resistance stays unchanged.

Parallel branches start at one common junction and end at another. Therefore U_1=U_2=U: each branch spans the same two potentials. Current in the common wire divides among branches, I=I_1+I_2, by charge conservation at the junction. Add a branch of finite resistance at the same U: it carries extra current while the existing ideal branches retain their currents. The source must supply the increased total charge flow; otherwise its voltage may change.

Substitute branch currents U/Rᵢ into their sum and cancel the common voltage: 1/R=1/R_1+1/R_2. Conductances add: they describe how readily current flows at a given U. For two branches, R=R_1R_2/(R_1+R_2), less than either individual resistance. Check the limit: if the added branch has a very large resistance, its current is almost zero and the result is close to the original R₁. An extra path cannot increase resistance in this ideal parallel circuit.

For n identical branches, each resistor carries U/R₁ and total current is n times as large. The equivalent resistance is therefore R=R_1/n. Doubling the number of branches at the same U doubles total current and power, but not current in an individual branch. Compare series connections: adding elements lengthens a single path there, whereas here it adds independent paths. This physical check tells us whether to multiply R₁ by n or divide.

Why does heating a metal filament change its current? In an ordinary metal, a temperature rise generally increases carrier scattering by thermal vibrations of the structure, increasing resistance. Over a limited range we introduce a temperature coefficient α in K⁻¹. In this notation R₀ is specifically the resistance at 0 °C; α gives the relative slope referenced to this R₀. It characterizes the chosen material and temperature range, rather than being a universal rule for all conductors.

The linear approximation is R=R_0(1+\alpha t_C), where t_C is the numerical temperature in °C. At tC = 0 it returns R₀; for positive α, heating raises R. The dimensionless quantity αtC describes relative change. A different reference temperature requires its corresponding resistance, temperature difference and coefficient; changing only R₀ is incorrect. Far beyond the applicable range, the straight line may give a meaningless result and must not be extended arbitrarily.

The source itself can heat up. In the model with EMF \mathcal E and internal resistance r, it supplies power \mathcal EI. Using \mathcal E=I(R+r) gives the distribution: the external load receives I^2R, and the internal resistance receives I^2r. Their sum is the source power. At a common current, a larger r means a larger share is released internally. The source both maintains current and participates in the energy balance.

For nonzero current, the fraction of source power delivered to the external resistor is \eta=I^2R/[I^2(R+r)]=R/(R+r). This dimensionless ratio applies to a steady circuit with one source. For R much larger than r it is close to unity; for very small R, energy is mainly released internally. However, high efficiency does not guarantee high useful power: as R → ∞, current and load power tend to zero. In an open circuit the power ratio is undefined.

Electrons carry charge in a metal wire, while positive and negative ions move in a directed way in an electrolyte. Both signs contribute to current while moving in opposite directions. Reactions occur at the electrodes: ions gain or lose electrons and substances may be produced. Charge transfer is therefore related to the amount of reaction product. Unlike simple resistor heating, this electrical energy transfer can change the chemical state of matter.

For a given electrode process, each amount of charge corresponds to a definite mass of product. The electrochemical equivalent k, in kg/C, gives mass per coulomb, so m=kq=kIt for constant current. Doubling I or time t doubles the mass for the same process. This formula assumes that all the counted current contributes to the intended reaction; side reactions reduce the yield. Here k is not Coulomb’s constant: its name and unit determine its meaning.

Resistors of 3 ohms and 6 ohms are connected in parallel across a constant 6 V. Find the total current and power.

Both parallel branches have the same voltage, 6 V. Use Ohm’s law to find the current in each.

I₁=6/3=2 А; I₂=6/6=1 А. Total current: I=2+1=3 А.

Total power: P=UI=6×3=18 Вт.

Check: the branch powers, 12 W and 6 W, add to the same 18 W. The equivalent resistance, 6/3=2 Ом, is less than 3 ohms.

I=3 А, P=18 Вт.

Power UI is energy per coulomb multiplied by coulombs per second. Heating is not always equal to all the electrical work.

Series circuits share a current and their voltages add. Parallel branches share a voltage and their currents add.

Before comparing power, establish whether I or U is fixed. High source efficiency does not guarantee high load power.

Next we study the magnetic effects of electric current.