Unit 23: DC Circuits

Electric Current and Drift Velocity

Electric current (I) is defined as the rate of flow of charge:

where Q is charge in coulombs and t is time in seconds. The SI unit is the ampere (A).

In a conductor, the current is carried by charge carriers (usually electrons) moving with an average drift velocity v_d:

where n is the number density of charge carriers (carriers per m³), A is the cross‑sectional area of the conductor (m²), and e is the elementary charge (1.602×10⁻¹⁹ C).

Example 1: A copper wire of radius 0.5 mm carries a current of 2 A. Given the free‑electron density of copper n ≈ 8.5×10²⁸ m⁻³, calculate the drift velocity.

1. Compute area: A = πr² = π(0.5×10⁻³)² ≈ 7.85×10⁻⁷ m².
2. Insert values: v_d = I/(n A e) = 2 / (8.5×10²⁸ × 7.85×10⁻⁷ × 1.602×10⁻¹⁹) ≈ 1.86×10⁻⁴ m/s.

The drift velocity is therefore of order 10⁻⁴ m/s, showing that the actual motion of electrons is very slow despite the rapid establishment of the electric field.

Ohm's Law and Resistance

Ohm’s law states that, at constant temperature, the potential difference V across a conductor is directly proportional to the current I flowing through it:

where R is the resistance (Ω). Resistance can also be expressed as:

The intrinsic property of the material is resistivity ρ:

with L the length of the conductor and A its cross‑sectional area. Conductivity σ is the reciprocal:

Units: resistance – ohm (Ω; resistivity – ohm·meter; conductivity – siemens per metre (S/m).

Example 2: A nichrome wire 2 m long and 0.3 mm diameter has a measured resistance of 10 Ω. Find its resistivity.

1. Radius r = 0.15 mm = 1.5×10⁻⁴ m; area A = πr² ≈ 7.07×10⁻⁸ m².
2. ρ = R A / L = 10 × 7.07×10⁻⁸ / 2 ≈ 3.54×10⁻⁷ Ω·m.

Current‑Voltage Relations

When V is plotted against I:

• Ohmic devices give a straight line through the origin (constant R). Examples: metals, carbon resistors.
• Non‑Ohmic devices show a curved V‑I graph. Examples: semiconductor diode, filament lamp, thermistor.

Example 3: A silicon diode has the approximate relation I = I₀(e^{V/(nV_T)} – 1). Sketch the V‑I curve and comment on its non‑ohmic nature.

Because the current grows exponentially with voltage, the ratio V/I is not constant – the device is non‑ohmic.

Resistances in Series and Parallel

Series combination: the same current flows through each resistor; the equivalent resistance is the sum

Parallel combination: voltage across each resistor is the same; the reciprocal of the equivalent resistance equals the sum of reciprocals:

Example 4: Three resistors 2 Ω, 3 Ω and 5 Ω are connected (a) in series, (b) in parallel. Find R_eq in each case.

1. Series: R_eq = 2 + 3 + 5 = 10 Ω.
2. Parallel: 1/R_eq = 1/2 + 1/3 + 1/5 = (15+10+6)/30 = 31/30 → R_eq = 30/31 ≈ 0.967 Ω.

Potential Divider

A potential divider (voltage divider) uses two resistors in series to obtain a fraction of the input voltage:

Applications: volume controls, sensor biasing, reference voltages.

Example 5: A divider with R₁ = 4 kΩ and R₂ = 6 kΩ is fed from a 12 V supply. Calculate V_out.

1. V_out = (6 kΩ / (4 kΩ+6 kΩ)) × 12 V = (6/10)×12 = 7.2 V.

EMF and Internal Resistance

The electromotive force (EMF, ε) is the energy supplied per unit charge by a source (e.g., a battery). When a current I flows, the terminal voltage V is reduced by the internal voltage drop:

where r is the internal resistance of the source.

Maximum power transfer theorem: the power delivered to an external load R_L is maximised when R_L equals the internal resistance r.

Example 6: A battery has ε = 9 V and internal resistance r = 0.5 Ω. Find the terminal voltage when it supplies a current of 2 A, and the load resistance for maximum power.

1. Terminal voltage: V = 9 – (2)(0.5) = 9 – 1 = 8 V.
2. For maximum power, set R_L = r = 0.5 Ω.

Work and Power in Electrical Circuits

Work done (energy transferred) when a charge Q moves through a potential difference V is W = QV. Since Q = I t, we obtain:

Power is the rate of doing work:

Unit: watt (W) = volt·ampere.

Example 7: A 12 V car battery supplies a current of 5 A to a headlamp for 30 minutes. Compute the energy consumed and the power dissipated.

1. Power: P = V I = 12 × 5 = 60 W.
2. Time t = 30 min = 1800 s. Energy: W = P t = 60 × 1800 = 108 000 J = 108 kJ.

Summary

This chapter has introduced the microscopic picture of electric current, derived the drift velocity expression, explained Ohm’s law and material resistivity, distinguished ohmic from non‑ohmic behaviour, analysed series and parallel resistor networks, shown how a potential divider works, incorporated EMF and internal resistance, and related work, power and energy in DC circuits. The worked numerical problems illustrate the application of each concept.