Current Electricity

Current Electricity

Outline

• Electric current
• Potential difference
• Resistance and resistivity
• Ohm's law, Ohmic and non-ohmic conductors
• Resistors in series and in parallel
• Electromotive force and circuit
• Electrical energy
• Electrical power and efficiency
• RC Circuits (Charging and Discharging Capacitors)

Electric Current

The electric current (I) is defined as the charge (ΔQ) passing through a given cross-sectional area A of a wire per unit time.

I = ΔQ / Δt

The unit of current is Coulomb per second, which is called ampere (A).

In differential form:

I = dQ / dt

The total charge Q that passes through area A in a time t is:

Q = qnAvdt

where n = number of charges q per unit volume and v_d = drift velocity of the charges.

Current Density

The current density (J) is defined as electric current per unit cross-sectional area. For a uniform current flow:

J = I / A = qnvd

The S.I. unit of J is ampere per metre square (Am^-2).

Example 1: Copper Wire Calculation

A copper wire has a square cross section, 2.0 mm on a side. It carries a current of 10 A. If the density of free electrons is 8 x 10²⁸ m^-3, calculate the current density in the wire and the drift speed.

• Area A = (2.0 x 10^-3 m)² = 4 x 10^-6 m²
• Current density J = I / A = 10 A / (4 x 10^-6 m²) = 2.5 x 10⁶ Am^-2
• Drift speed v_d = J / (nq) = (2.5 x 10⁶ Am^-2) / (8 x 10²⁸ m^-3 x 1.6 x 10^-19 C) = 1.953 x 10^-4 m/s

Ohm's Law

Ohm's law states that under constant physical conditions, the current (I) in a conducting wire is proportional to the potential difference (V) applied to its ends.

I α V
I = V / R

where R is the resistance.

Resistance and Resistivity

Resistance (R) is found experimentally to be proportional to the length (L) of the wire and inversely proportional to the cross-sectional area (A).

R α L / A
R = ρL / A

where ρ is the constant of proportionality known as resistivity of the material. Resistivity depends only on the material and the temperature. The unit of resistivity ρ is ohm-metre (Ωm).

Conductivity

Conductivity (σ) is the reciprocal of the resistivity.

σ = 1 / ρ (Ωm)-1

Ohmic and Non-Ohmic Conductors

Ohmic Conductors

Ohmic conductors are conductors that obey Ohm's law. Their characteristic or I-V graph is a straight line passing through the origin.

• Examples: Copper, Tungsten

Non-Ohmic Conductors

Non-ohmic conductors are those which do not obey Ohm's law. Their characteristic or I-V graph may be a curve instead of a straight line, or it may not pass through the origin.

• Examples: Junction diode, neon gas, diode valve, dilute H₂SO₄ (with platinum electrodes)

Resistors in Series and Parallel

Resistors in Series

• Definition: Resistors are in series if they are connected one after the other so the current is the same in all of them.
• Equivalent Resistance (R_eq): The sum of the individual resistances.

  Req = R1 + R2 + R3 + ...

• Characteristics:
  1. The same current flows through all resistors in series.
  2. Total potential difference = sum of individual potential differences (V = V₁ + V₂ + V₃).
  3. Individual potential differences are directly proportional to individual resistances (V₁ = IR₁, V₂ = IR₂, V₃ = IR₃).
• Note: Series resistors have a total resistance LARGER than the largest individual resistance present.

Resistors in Parallel

• Definition: Resistors are in parallel if they are connected so that the potential difference across them must be the same.
• Equivalent Resistance (R_eq): The reciprocal of the equivalent resistance is the sum of the reciprocals of the individual resistances.

  1/Req = 1/R1 + 1/R2 + 1/R3 + ...

• Characteristics:
  1. Total current (I) is equal to the sum of individual currents flowing through the resistors in parallel (I = I₁ + I₂ + I₃).
  2. The potential difference across each resistor is the same and is equal to the full voltage V.
  3. Individual current is inversely proportional to individual resistances (V = I₁R₁, V = I₂R₂, V = I₃R₃).
• Note: Parallel resistors have a total resistance SMALLER than the smallest individual resistance present.

Series and Parallel Combinations

Resistors can be connected in combinations of series and parallel.

Electromotive Force (e.m.f.) and Circuit

The electromotive force or e.m.f. (E) is the potential difference across the terminals of a battery (or any other generator) on open circuit, i.e., when no current is flowing.

• The e.m.f. of a battery depends on the nature of the chemicals used and not on its size.
• The internal resistance (r) of the battery depends on the size of the battery.

When a current I flows through an external resistance R, the terminal potential difference (V) across the battery is:

I = E / (R + r)
V = IR = E - Ir

Thus, the terminal potential difference V can also be expressed as:

V = ER / (R + r)

Example 2: Terminal Potential Difference

A battery of e.m.f. 4V and internal resistance 2 Ω is joined to a resistor of 8 Ω. Calculate the terminal p.d.

• Current I = E / (R + r) = 4V / (8Ω + 2Ω) = 4V / 10Ω = 0.4 A
• Terminal p.d. V = IR = 0.4 A x 8Ω = 3.2V

Electrical Power and Efficiency

Electrical Work

The electrical work (W) required to transfer a charge q through a potential difference V is given by:

W = qV

Electrical Power

Electrical Power (P) is the rate at which electrical work is done.

P = Electrical Work / Time = W / t = qV / t = VI

The unit of power is watt (W) (1W = 1J/s).

The e.m.f. E of a battery or any other generator is the total energy per coulomb it delivers round a circuit joined to it.

Total electrical power generated is:

Pgen = W / t = EI

Efficiency

Efficiency (η) of a circuit is the ratio of the power output to the power generated.

η = Power output / Power generated
η = Pout / Pgen = IV / IE = V / E

Using V = IR and E = I(R+r):

η = IR / (I(R+r)) = R / (R + r)

Equivalent Resistance (Examples)

The document illustrates how to calculate equivalent resistance and current/power in circuits with combinations of resistors, including examples for series and parallel light bulbs.

Example 4: Light bulbs in series

• Two 2Ω resistors in series with an 8V battery (r=0).
• R_equivalent = 2Ω + 2Ω = 4 Ohms
• Current I = 8V / 4Ω = 2 A
• Total Power = I²R_equivalent = (2A)² x 4Ω = 16 Watts (8 Watts for each bulb)

Example 5: Light bulbs in parallel

• Two 2Ω resistors in parallel with an 8V battery (r=0).
• 1/R_equivalent = 1/2Ω + 1/2Ω = 1Ω^-1 => R_equivalent = 1 Ohm
• Total Current I_total = 8V / 1Ω = 8 A
• Total Power = I_total²R_equivalent = (8A)² x 1Ω = 64 Watts (32 Watts for each bulb)

Adding Capacitors to DC circuits (RC circuits)

RC circuits include Batteries (Voltage sources!), Resistors, Capacitors, and Switches. They involve TIMING considerations for charging and discharging.

Charging a Capacitor

When a switch is closed in an RC circuit with an uncharged capacitor and a battery:

• The charge on the capacitor increases exponentially with time toward a final value Q_f = Cε.
• The current in the circuit decreases exponentially with time from an initial value I₀ = ε/R to zero.

Time Constant (τ)

The time constant (τ) for an RC circuit is given by:

τ = RC

• τ increases with R: Larger resistors decrease current, less charge/time arrives at capacitor, takes longer to fill up.
• τ increases with C: Larger capacitors have more capacity, takes longer to fill up.
• In one time constant (τ):
  • Current drops to 1/e (about 36.8%) of its initial value.
  • Charge on capacitor plates rises to ~63.2% of maximum value (Q_f).

Differential Equation for Charging

Applying Kirchhoff's voltage law to a charging RC circuit:

E - iR - q/C = 0

Since i = dq/dt:

E - (dq/dt)R - q(t)/C = 0

This is a differential equation involving charge q.

Boundary Conditions for Charging

• At t=0: q(t) = 0 (uncharged capacitor)
• At t=∞ (when capacitor is full): q = Q_max = Cε, i(t) = 0

General Solution for Charging

The general solution for charge as a function of time:

q(t) = Qmax (1 - e-t/RC)

The general solution for current as a function of time:

i(t) = dq/dt = (Qmax/RC) e-t/RC = I0 e-t/RC

where I₀ = E/R.

Example: Charging a Capacitor Calculation

10 MΩ resistor connected in series with 1.0 μF uncharged capacitor and a battery with emf of 12.0 V.

• Time constant (τ): τ = RC = (10 x 10⁶ Ω) x (1.0 x 10^-6 F) = 10 seconds.
• Fraction of final charge after 46 seconds:

  q(t) = Qmax (1 - e-t/RC)
  q(46)/Qmax = (1 - e-46/10) = 1 - e-4.6 ≈ 1 - 0.010 = 0.99 (or 99%)

• Fraction of initial current (I₀) flowing then:

  i(t) = I0 e-t/RC
  i(46)/I0 = e-46/10 = e-4.6 ≈ 0.010 (or 1%)

Discharging a Capacitor

When the battery is disconnected and a charged capacitor is allowed to discharge through a resistor (capacitor "drains"):

• The charge on the capacitor decreases exponentially with time from an initial value Q₀ to zero.
• The current in the circuit decreases exponentially with time from an initial value I₀ to zero (often considered negative as direction is opposite to charging).

Time Constant (τ)

The time constant (τ = RC) is still the same for discharging.

• In one time constant (τ):
  • Charge on plates drops to ~36.8% of its initial value (a 63.2% drop).
  • Current through resistor drops to ~36.8% of its initial value (a 63.2% drop).

Differential Equation for Discharging

Applying Kirchhoff's voltage law to a discharging RC circuit (with no EMF source):

iR + q/C = 0

Since i = dq/dt:

(dq/dt)R + q(t)/C = 0

Boundary Conditions for Discharging

• At t=0: q(t) = Q_max (initially charged capacitor)
• At t=∞: q = 0, i(t) = 0 (capacitor is empty)

General Solution for Discharging

The general solution for charge as a function of time:

q(t) = Qmax e-t/RC

The general solution for current as a function of time:

i(t) = dq/dt = -(Qmax/RC) e-t/RC = -I0 e-t/RC

where I₀ = Q_max/(RC).

Example: Discharging a Capacitor Calculation

Same circuit as before (10 MΩ, 1.0 μF), but the battery is disconnected. Assume at t = 0, Q(0) = 5.0 μC.

• Time constant (τ): τ = RC = 10 seconds (still!).
• When will charge = 0.50 μC?

  q(t) = Qmax e-t/RC
  0.50 μC = 5.0 μC * e-t/10
  0.1 = e-t/10
  ln(0.1) = -t/10
  t = -10 * ln(0.1) = -10 * (-2.30) ≈ 23 seconds (2.3 τ)

• What is current then (at t = 23 s)?

  I(t) = -Qmax/(RC) * e-t/RC
  Initial current I0 = Qmax/RC = (5.0 x 10-6 C) / (10 s) = 0.5 x 10-6 A
  I(2.3τ) = -I0 * e-2.3 ≈ -I0 * 0.1 = -(0.5 x 10-6 A) * 0.1 = -5.0 x 10-8 Amps

Assignments

The document includes a list of problems (13.1 to 13.19) covering all topics discussed for practice and submission.