3. Current Electricity

A circuit moves charge while transferring energy. This chapter connects that macroscopic picture to microscopic carrier motion, then uses conservation laws to analyse real sources and resistor networks.

Ohm established a quantitative relation between voltage and current for suitable conductors. Kirchhoff expressed circuit analysis through charge and energy conservation. The modern ampere is defined through the fixed elementary charge rather than a particular mechanical experiment.

A steady current is a continuing transfer of charge, not a supply of charge that gets used up.

A closed circuit needs mobile charge carriers and a maintained electric field. A cell uses chemical processes to transfer energy to charges. A resistor transfers that energy to its surroundings. The same amount of charge enters and leaves each element in a steady series circuit.

Electrostatic equilibrium in a metal has zero internal electric field. A wire carrying steady current is not in that equilibrium: a small field remains within it. Surface charges arrange themselves so that the field follows the circuit geometry.

A lamp converts electrical energy to heat and light. Does less charge return to the cell than leaves it?

Apply charge conservation to the lamp in a steady state. Energy changes form; charge does not disappear.

Result: No. The current entering and leaving the lamp is the same.

1. What must a cell maintain for current to continue in a resistive circuit?

Answers

1. A potential difference and hence a driving electric field in the closed circuit.

Current is the signed rate at which charge crosses a chosen surface.

For a steady transfer, $I=\Delta Q/\Delta t$; instantaneously $I=dQ/dt$. One ampere is one coulomb per second. Conventional current points in the direction positive charge would move. In a metal the drifting electrons move oppositely.

Current through a wire is treated as a signed circuit quantity. Current density is a vector: it describes current per area, with direction given by conventional flow. In electrolytes, oppositely moving positive and negative ions can contribute current in the same conventional direction.

A charge of 9 C passes a wire section in 30 s at a steady rate.

Divide transferred charge by elapsed time.

Result: I=0.3 A.

1. A steady 0.80 A current flows for 25 s. How much charge passes?

Answers

1. 20 C.

Random microscopic motion and a small directed drift coexist in a metal.

Conduction electrons already move rapidly and randomly even without a circuit field. The random velocity averages to zero. An applied electric field adds a small average drift velocity. Collisions transfer momentum to the material, preventing indefinite acceleration in the steady state.

For electron number density n, carrier charge magnitude e and uniform cross-section A, $I=neAv_d$ in magnitude. During time Δt, the drifting carriers in a cylinder of length $v_d\Delta t$ cross the section. Their number is $nAv_d\Delta t$, which yields the equation after multiplying by e and dividing by Δt.

A wire has n=8 × 10²⁸ m⁻³, A=1.0 mm² and current 1.28 A. Using e≈1.60 × 10⁻¹⁹ C, find electron drift speed.

Convert $1.0\ \mathrm{mm^2}=10^{-6}\ \mathrm{m^2}$ and solve $v_d=I/(neA)$.

Result: 0.1 mm/s, opposite conventional current.

A circuit can respond quickly even though individual electrons drift slowly. The electromagnetic influence propagates through the circuit much faster than an individual carrier travels around it.

1. At fixed I and n, what happens to drift speed if wire area doubles?

Answers

1. It halves.

An ohmic element has a straight voltage-current relation under fixed physical conditions.

Ohm’s law is $V=IR$ when R is constant over the range considered. V is the drop in the current direction through a passive resistor. An I–V graph is a straight line through the origin for an ideal ohmic resistor. If V is on the vertical axis its slope is R; if I is on the vertical axis its slope is 1/R.

For a uniform wire, $R=\rho L/A$, where ρ is resistivity in Ω m. Conductivity is $\sigma=1/\rho$. The local form is $\mathbf J=\sigma\mathbf E$ for an isotropic ohmic material. A wire’s resistance depends on shape; its material resistivity does not depend on its length or area.

A wire has ρ=2.0 × 10⁻⁸ Ω m, length 3.0 m and area 0.50 mm². Find R and the voltage required for 2.0 A.

Use SI area, then V=IR.

Result: R=0.12 Ω; V=0.24 V.

1. Two wires of the same material and area have lengths L and 3L. Compare their resistance.

Answers

1. The longer wire has three times the resistance.

The collision model explains why steady drift is proportional to electric field.

For an electron, acceleration between collisions is $\mathbf a=-e\mathbf E/m$. Let τ be the mean relaxation time. Averaging the directed motion gives $\mathbf v_d=-e\tau\mathbf E/m$. Since $\mathbf J=-ne\mathbf v_d$, we obtain $\mathbf J=(ne^2\tau/m)\mathbf E$. Therefore $\sigma=ne^2\tau/m$ and $\rho=m/(ne^2\tau)$.

This simple model is useful for connecting microscopic and macroscopic quantities. It does not describe every detail of metallic conduction. The relaxation time represents the average effect of scattering rather than a perfectly periodic sequence of identical collisions.

Carrier mobility is drift speed per field magnitude: $\mu=v_d/E=e\tau/m$ for this model. Its unit is $\mathrm{m^2\,V^{-1}\,s^{-1}}$. For one carrier type, $\sigma=ne\mu$.

A carrier mobility is 0.040 m²/(V s) in a field of 0.030 V/m. Find drift speed.

Use $v_d=\mu E$.

Result: 1.2 mm/s.

1. In the simple model, what happens to conductivity when τ doubles at fixed n?

Answers

1. It doubles.

Ohm’s law is an experimentally limited material relation, not a universal rule for every device.

An element may have a nonlinear I–V curve, different behaviour for reversed polarity, or a current that depends on earlier conditions. A heated filament changes resistance as its temperature changes. A diode is strongly asymmetric. Electrolytic and semiconductor devices can require more detailed descriptions.

The ratio V/I can still be calculated at a nonzero operating point, but it may not be a constant material resistance. A local slope dV/dI describes small changes near that point and need not equal V/I.

A device carries 0.20 A at 2 V and 0.30 A at 4 V. Is its V/I constant?

Calculate the ratio at both operating points.

Result: The ratios are 10 Ω and 13.3333 Ω. It is not ohmic over this interval.

1. Why specify constant temperature when testing a wire for Ohm’s law?

Answers

1. Heating can change its resistivity during the measurement.

Resistivity allows fair comparison between materials of different shapes.

Metals usually have low resistivity because many carriers can move. Insulators have very high resistivity under ordinary conditions. Semiconductors lie between these broad classes and are strongly affected by temperature and impurities. All such comparisons require specified conditions.

Alloys are often used in resistors and heaters because suitable compositions provide higher resistivity and relatively small temperature changes in resistance. A useful conductor for a wire has low resistivity, but mechanical strength, stability and cost also matter in a real design.

A material has resistivity 4.0 × 10⁻⁷ Ω m. Find its conductivity.

Use $\sigma=1/\rho$.

Result: $2.5\times10^6\ \mathrm{S\,m^{-1}}$.

1. Does cutting a uniform wire in half change the material’s resistivity?

Answers

1. No; each half has half the resistance but the same resistivity at the same temperature.

Temperature changes both scattering and, in some materials, the carrier population.

Over a limited temperature interval, $\rho(T)=\rho_0[1+\alpha(T-T_0)]$. For unchanged dimensions the same approximation applies to R. The coefficient α is defined relative to the stated reference temperature and has units K⁻¹ or °C⁻¹ for a temperature difference.

For many metals near room temperature, increased lattice motion increases scattering and resistivity rises. In many semiconductors, increased carrier concentration can dominate so resistivity falls. A straight-line formula is a local approximation and must not be extrapolated across all temperatures or phase changes.

A wire has R₀=8 Ω at 20°C and α=0.004 K⁻¹. Estimate its resistance at 70°C.

Use the 50 K change, not the absolute Celsius temperature.

Result: 9.6 Ω.

1. Why can semiconductor resistivity decrease when temperature rises?

Answers

1. More charge carriers become available, which can outweigh increased scattering.

Voltage supplies energy per charge; current tells how much charge passes per second.

A charge ΔQ crossing a potential drop V transfers energy $V\Delta Q$. Dividing by time gives $P=VI$. For an ohmic resistor, substitute V=IR to obtain $P=I^2R=V^2/R$. Energy over a constant-power interval is $W=Pt$.

A watt is a joule per second. A kilowatt-hour is an energy unit, not a power unit. Resistive heating can be useful in a heater and unwanted in a supply wire. At fixed delivered power, higher transmission voltage permits lower current, reducing an I²R line loss.

A 12 Ω resistor carries 0.50 A for 4 min. Find power and transferred energy.

Use $P=I^2R$ and convert minutes to seconds.

Result: P=3 W; W=720 J.

1. At fixed V, doubling R changes resistor power how?

Answers

1. It halves, since P=V²/R.

A real cell has an energy supply and an internal voltage drop.

The emf $\mathcal E$ is energy supplied by the source per unit charge around its internal process. Despite its name, it is a voltage, not a mechanical force. With internal resistance r and discharge current I, terminal voltage is $V=\mathcal E-Ir$. For an external load R, $I=\mathcal E/(R+r)$.

Open circuit means I=0 in this ideal model, so terminal voltage equals emf. Under load, part of the energy is converted to heat inside the source. During charging, current enters the positive terminal and the terminal voltage is greater than the emf: V=ℰ+Ir with I taken as a positive charging magnitude.

A cell has emf 9 V and internal resistance 0.50 Ω. It supplies a 4 Ω load. Find current and terminal voltage.

Include internal resistance in the total circuit resistance.

Result: I=2 A; V=8 V.

1. Why is short-circuiting a real cell dangerous?

Answers

1. The current is limited mainly by a small internal resistance, causing large heating and possible damage.

Cell arrangements trade available voltage against internal resistance.

For identical cells connected in series with aiding polarity, $\mathcal E_{\rm eq}=n\mathcal E$ and $r_{\rm eq}=nr$. Reversing a cell reverses its emf contribution but not its positive internal resistance.

For identical cells in parallel with matching polarity, $\mathcal E_{\rm eq}=\mathcal E$ and $r_{\rm eq}=r/n$. Each carries an equal share of the load current in this symmetric ideal model. Unequal cells should not be assumed to share equally: they can drive circulating currents.

Three identical 2 V cells each have r=0.30 Ω. Compare series and parallel equivalents.

For series, add emfs and internal resistances. For parallel, keep emf and divide internal resistance by three.

Result: Series: 6 V and 0.9 Ω. Parallel: 2 V and 0.1 Ω.

For two unequal cells in parallel with the same polarity, Kirchhoff’s equations give ℰeq=(ℰ₁/r₁+ℰ₂/r₂)/(1/r₁+1/r₂) and req=r₁r₂/(r₁+r₂). This ideal result assumes fixed emf values and internal resistances. It is an analysis model, not permission to connect arbitrary rechargeable cells together.

1. Which arrangement of identical cells reduces equivalent internal resistance?

Answers

1. Parallel.

Kirchhoff’s rules express conservation of charge and energy in a lumped DC circuit.

At a junction, $\sum I_{\rm in}=\sum I_{\rm out}$. This follows from no continuing accumulation of charge in a steady state. Around a closed loop, the signed potential changes sum to zero for this circuit model.

• Choose a current direction for each branch; the choices can be guesses.
• When traversing a resistor in the chosen current direction, write −IR; against it, write +IR.
• Crossing an ideal cell from − to + gives +ℰ; from + to − gives −ℰ.
• Write enough independent junction and loop equations, then solve simultaneously. A negative current reverses its guessed direction.

At a junction, 1.8 A enters from one branch and 0.7 A leaves through a second. Find the third current.

Choose the third as leaving and apply charge conservation.

Result: 1.1 A leaves.

A loop contains an aiding 10 V source, an opposing 4 V source, and resistors 2 Ω and 3 Ω. Find its current.

Net emf is 10−4 V. The loop equation is $10-4-I(2+3)=0$.

Result: 1.2 A in the direction driven by the 10 V source.

1. Why must resistor signs change when a loop is traversed against current?

Answers

1. Potential rises rather than falls when moving against current through a passive resistor.

A balanced bridge compares resistance using a zero-current condition.

Label bridge arms P and Q in one branch and R and S in the other, with the galvanometer between their intermediate junctions. At balance those junctions have equal potential and the galvanometer current is zero. The two branches then give $I_1P=I_2R$ and $I_1Q=I_2S$, so $P/Q=R/S$.

The null condition is useful because the balance ratio does not require the galvanometer’s resistance. Its sensitivity still affects how precisely you can locate zero. A bridge that is not balanced requires a general circuit calculation; the simple ratio cannot be applied then.

A bridge has P=120 Ω, Q=180 Ω and R=200 Ω. Find S for balance.

Solve $S=RQ/P$.

Result: 300 Ω.

1. Does zero galvanometer current mean zero current in every bridge arm?

Answers

1. No. The two main branches can carry nonzero currents.

Use $I=dQ/dt$, $I=neAv_d$, $V=IR$ for ohmic elements, and $R=\rho L/A$ for a uniform wire.

Source emf and terminal voltage differ by the internal drop during discharge. Junction equations conserve charge; loop equations conserve energy in the stated circuit model. A balanced bridge has equal midpoint potentials.

• Resistance depends on geometry; resistivity describes a material at specified conditions.
• Current is conserved through series components while energy is transferred.
• Choose signs consistently and let negative answers correct guessed directions.

Try these newly written questions before reading the answers. Show a diagram, the governing equation, substitutions and units where appropriate.

3.1 A charge of 24 C passes in 80 s. Find current.

3.2 A steady current is 1.5 A for 3 min. Find transferred charge.

3.3 What direction do electrons drift in a metal relative to conventional current?

3.4 A wire’s area doubles at fixed current and carrier density. What happens to drift speed?

3.5 A wire has resistance 6 Ω. Find current at 9 V.

3.6 A uniform wire is stretched to twice its length without volume change. Neglect resistivity change. Find the resistance factor.

3.7 A material has conductivity 2 × 10⁶ S/m. Find resistivity.

3.8 Why is carrier drift much slower than a circuit’s electrical response?

3.9 Mobility is 0.025 m²/(V s), field 0.08 V/m. Find drift speed.

3.10 A device has V/I values 5 Ω and 8 Ω at two operating points. Is a constant-R model supported?

3.11 A metal has R=20 Ω at 25°C and α=0.003 K⁻¹. Estimate R at 75°C.

3.12 Why can a semiconductor have a negative temperature coefficient?

3.13 A 30 Ω resistor carries 0.20 A. Find power.

3.14 A 15 W device operates for 120 s. Find energy.

3.15 Convert 0.50 kWh to joules.

3.16 A 6 V cell with r=1 Ω supplies R=5 Ω. Find current and terminal voltage.

3.17 Explain the difference between emf and terminal voltage during discharge.

3.18 Two identical 1.5 V cells, each with 0.20 Ω internal resistance, are in series. Find their equivalent.

3.19 Find the equivalent of the same cells in parallel.

3.20 At a junction 2.4 A and 0.9 A enter while 1.1 A leaves. Find the other outgoing current.

3.21 A 12 V source opposes a 3 V source in a 6 Ω loop. Find current.

3.22 A solved branch current is −0.4 A. What does that mean?

3.23 A balanced bridge has P=80 Ω, Q=120 Ω, R=100 Ω. Find S.

3.24 Does a balanced bridge need the galvanometer resistance in its ratio equation?

3.25 Why do long transmission lines benefit from lower current at a fixed delivered power?

3.26 A wire is cut into three equal pieces and the pieces connected in parallel. Compare equivalent resistance with the original R.

Answers

3.1 0.3 A.

3.2 270 C.

3.3 Opposite.

3.4 It halves.

3.5 1.5 A.

3.6 Four: area halves and length doubles.

3.7 0.5 μΩ m.

3.8 The response propagates electromagnetically; it does not wait for one electron to traverse the circuit.

3.9 2 mm/s.

3.10 No.

3.11 23 Ω.

3.12 Heating can increase its free-carrier population substantially.

3.13 1.2 W.

3.14 1800 J.

3.15 1800000 J.

3.16 I=1 A; V=5 V.

3.17 Emf is source energy per charge; terminal voltage is smaller by Ir.

3.18 3 V and 0.4 Ω.

3.19 1.5 V and 0.10 Ω.

3.20 2.2 A.

3.21 1.5 A, driven by the larger source.

3.22 The actual current is 0.4 A opposite the assumed direction.

3.23 150 Ω.

3.24 No; its current is zero at balance.

3.25 Their resistive loss I²R is reduced.

3.26 Each piece is R/3; three in parallel give R/9.