Electric Current In Conductors

• Electric current (i):

  • Scalar quantity with direction (not a vector)
  • Measured in amperes (A)
  • Represents rate of charge flow: $i = \frac{dQ}{dt}$
  • Does not add like vectors

• Current density (j):

  • True vector quantity
  • Measured in A/m²
  • Direction matches charge motion for positive charges; opposite for negative charges
  • Relation to current: $j = \frac{i}{A}$ for uniform current perpendicular to area
  • For non-perpendicular areas: $j = \frac{i}{A\cos\theta}$
  • General relationship: $i = \int \vec{j} \cdot d\vec{S}$

$j = nevd$

• Term origin and meaning:

  • $n$: Number of free electrons per unit volume (carriers/m³)
  • $e$: Elementary charge of electron ($1.6 \times 10^{-19}$ C)
  • $v_d$: Average drift velocity of electrons (m/s)

• Microscopic explanation:

  • Free electrons move randomly in material
  • Applied electric field creates biased drift motion opposite to field direction
  • Drift velocity remains small (~mm/s) despite high field due to frequent collisions
  • Number of electrons crossing area A in time Δt = nAvdΔt
  • Charge crossing = nAvdΔt·e
  • Current = charge/time = nAvd·e
  • Current density = current/area = nevd

• Example: In copper with $n = 8.5 \times 10^{28}$ electrons/m³, a current of 1A through 2mm² yields drift velocity of only 0.036 mm/s

Series combination:

• Formula: $R_{eq} = R_1 + R_2 + R_3 + ...$
• Derivation:
  • Same current i through all resistors
  • Total voltage: $V = V_1 + V_2 + V_3 = iR_1 + iR_2 + iR_3$
  • Therefore: $\frac{V}{i} = R_1 + R_2 + R_3$

Parallel combination:

• Formula: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ...$
• Derivation:
  • Same voltage V across all resistors
  • Total current: $i = i_1 + i_2 + i_3 = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3}$
  • Therefore: $\frac{i}{V} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

Strategy for complex networks:

1. Identify simple series and parallel sections
2. Reduce each section to its equivalent resistance
3. Repeat process with new equivalent resistances
4. If network cannot be reduced further, apply Kirchhoff's laws
5. For delta-wye networks, use delta-wye transformation formulas

Special cases:

• Two resistors in parallel: $R_{eq} = \frac{R_1R_2}{R_1+R_2}$
• N identical resistors in series: $R_{eq} = NR$
• N identical resistors in parallel: $R_{eq} = \frac{R}{N}$
• Balanced Wheatstone bridge: No current through diagonal branch

The equivalent resistance of a uniform conductor is given by:

R = ρl/A

Where:

• R = resistance (ohms, Ω)
• ρ = resistivity of the material (ohm·meters, Ω·m)
• l = length of the conductor (meters, m)
• A = cross-sectional area (square meters, m²)

Key relationships:

• Resistance is directly proportional to length
• Resistance is inversely proportional to cross-sectional area
• Resistivity is an intrinsic property of the material (e.g., copper: 1.72 × 10⁻⁸ Ω·m)
• Conductivity (σ) is the reciprocal of resistivity: σ = 1/ρ

Example: A 2m copper wire with 1mm² cross-section has approximately 0.0344 Ω resistance at room temperature.

• Normal metals: Resistivity gradually increases with temperature in a continuous curve, never reaching zero
• Superconductors: Resistivity maintains a normal metal-like behavior above critical temperature (Tc), but abruptly drops to exactly zero when cooled below Tc
• This zero-resistance state below Tc allows superconductors to maintain persistent currents without energy loss
• Example: Mercury becomes superconducting at 4.2K, enabling applications like powerful electromagnets used in MRI machines and particle accelerators

During discharge of a battery:

• A non-electrostatic force (often called the "battery force") drives positive charges from the negative terminal to the positive terminal against the electric field inside the battery
• This force converts chemical energy into electric potential energy
• The work done per unit charge by this non-electrostatic force equals the emf of the battery
• This mechanism maintains the potential difference between terminals by continuously separating charges
• When connected to a circuit, electrons enter the positive terminal, and the battery force pushes positive charges toward this terminal to maintain the potential difference

Example: In a 1.5V alkaline battery, chemical reactions between zinc and manganese dioxide create this non-electrostatic force that maintains the potential difference regardless of current flow (until chemical reactants are depleted).

The terminal voltage of a battery with internal resistance decreases during discharge according to:

$V = ℰ - ir$

Where:

• V = terminal voltage (voltage measured across battery terminals)
• ℰ = electromotive force (emf) of the battery
• i = current flowing through the battery
• r = internal resistance of the battery

This relationship shows that:

• Terminal voltage is always less than the emf during discharge
• The voltage drop (ir) increases with higher current draw
• As internal resistance increases (e.g., with battery age), the voltage drop becomes more significant
• For an ideal battery where r = 0, V = ℰ

Example: A 12V battery with 0.5Ω internal resistance delivering 2A will have terminal voltage of 12V - (2A × 0.5Ω) = 11V

To analyze a circuit with two parallel batteries connected to resistor R:

1. Apply Kirchhoff's loop law to each loop:

   • For the loop with battery 1: Ri + r₁i₁ - ℰ₁ = 0
   • For the loop with battery 2: Ri + r₂(i-i₁) - ℰ₂ = 0

2. Solve the system of equations to find:

   • Total current: i = (ℰ₁r₂ + ℰ₂r₁)/[(r₁+r₂)R + r₁r₂]
   • Current through battery 1: i₁ = (ℰ₁r₂ - ℰ₂r₁)/(r₁r₂) + i·r₂/(r₁+r₂)

3. Current distribution principle:

   • If ℰ₁ > ℰ₂ and r₁ = r₂, battery 1 supplies more current
   • If ℰ₁ = ℰ₂ and r₁ < r₂, battery 1 supplies more current
   • A battery may even receive current if its emf is significantly lower than the other

Note: The circuit behaves as if there were a single battery with emf ℰ₀ and internal resistance r₀.

The equivalent resistance of 10Ω and 30Ω resistors in parallel is:

Req = (R₁ × R₂)/(R₁ + R₂) = (10Ω × 30Ω)/(10Ω + 30Ω) = 300Ω/40Ω = 7.5Ω

This value is less than either individual resistor because:

• Parallel connections provide additional paths for current flow
• The total cross-sectional area available for current increases
• The effective resistance to current flow decreases
• Mathematically, since 1/Req = 1/R₁ + 1/R₂, adding terms makes 1/Req larger, making Req smaller

This is analogous to adding more lanes to a highway - traffic flows more easily with multiple paths.

Series Configuration

• Equivalent emf: $\mathcal{E}_0 = \mathcal{E}_1 + \mathcal{E}_2$
• Equivalent internal resistance: $r_0 = r_1 + r_2$
• Current through external resistance $R$: $i = \frac{\mathcal{E}_1 + \mathcal{E}_2}{R + (r_1 + r_2)}$
• Special case: If polarity of one battery is reversed, $\mathcal{E}_0 = |\mathcal{E}_1 - \mathcal{E}_2|$

Parallel Configuration

• Equivalent emf: $\mathcal{E}_0 = \frac{\mathcal{E}_1 r_2 + \mathcal{E}_2 r_1}{r_1 + r_2}$
• Equivalent internal resistance: $r_0 = \frac{r_1 r_2}{r_1 + r_2}$
• Current distribution: More current flows through the battery with lower internal resistance

Special Case - Identical Batteries

When $\mathcal{E}_1 = \mathcal{E}_2 = \mathcal{E}$ and $r_1 = r_2 = r$:

• Series: $\mathcal{E}_0 = 2\mathcal{E}$ and $r_0 = 2r$ (doubles both)
• Parallel: $\mathcal{E}_0 = \mathcal{E}$ and $r_0 = \frac{r}{2}$ (maintains emf, halves resistance)

Note: Choose series connections to increase voltage, parallel connections to maintain voltage while decreasing internal resistance (increasing available current).