11.2 Current Electricity & Circuits

11.2 Current Electricity & Circuits

Quick Answer: Steady current I = Q/t obeys Ohm's law V = IR in ohmic devices. Combine resistors with series and parallel rules, compute power from P = VI = I²R = V²/R, and use Kirchhoff's junction and loop rules when networks are too branched for simple series–parallel reduction.

Aircraft and ground-support systems are rich in DC and AC circuits, but the written exam expects intermediate physics: definitions, ohmic behaviour, network reduction, power, and introductory Kirchhoff analysis.

Electric Current

Electric current is the rate of flow of charge:

$I = \frac{Q}{t}$

SI unit: ampere (A) = C/s. In metallic conductors, free electrons drift slowly opposite to the conventional current arrow; exams usually use conventional current (positive-flow direction).

Current density (useful for conceptual questions): $J = I/A$ for uniform current through cross-section A.

Drift speed is tiny (mm/s order), yet the electric field propagates near light speed — that is why a lamp lights "instantly" when a switch closes.

Ohm's Law and Resistance

For many metals at fixed temperature:

$V = IR$

A material is ohmic if V–I is linear through the origin. Resistance of a uniform wire:

$R = \rho \frac{L}{A}$

where ρ is resistivity (Ω·m), L length, A cross-sectional area. Increasing length increases R; increasing thickness decreases R.

Worked example — resistivity. A copper wire (ρ = 1.7 × 10⁻⁸ Ω·m) is 2.0 m long with cross-section 0.50 mm². Find R.

$A = 0.50 \times 10^{-6}\ \mathrm{m}^2 = 5.0 \times 10^{-7}\ \mathrm{m}^2$.

$R = (1.7 \times 10^{-8})\frac{2.0}{5.0 \times 10^{-7}} = 0.068\ \Omega$

Trap: area must be in m². Leaving mm² unconverted inflates R by 10⁶.

Temperature note (FSc): metal resistance typically rises with temperature; $R = R_0(1 + \alpha\Delta T)$ with α the temperature coefficient. Semiconductors often show the opposite trend — know the qualitative distinction.

Series and Parallel Circuits

Series (one path):

• Same current through every resistor: $I$ common
• Voltages add: $V = V_1 + V_2 + \cdots$
• $R_{eq} = R_1 + R_2 + \cdots$
• Voltage divider: $V_i = I R_i = V\dfrac{R_i}{R_{eq}}$

Parallel (shared nodes):

• Same voltage across each branch: $V$ common
• Currents add: $I = I_1 + I_2 + \cdots$
• $\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \cdots$
• For two resistors: $R_{eq} = \dfrac{R_1 R_2}{R_1 + R_2}$

Worked example — mixed network. A 6 Ω and 3 Ω resistor in parallel are then placed in series with 4 Ω across a 12 V battery. Find R_eq, total current, and current through the 3 Ω branch.

Parallel part: $R_p = \dfrac{6\times 3}{6+3} = 2\ \Omega$.

$R_{eq} = 2 + 4 = 6\ \Omega,\quad I_{total} = \frac{12}{6} = 2\ \mathrm{A}$

Voltage across parallel pair: $V_p = I_{total} R_p = 2 \times 2 = 4\ \mathrm{V}$.

Current through 3 Ω: $I_3 = V_p/3 = 4/3 \approx 1.33\ \mathrm{A}$ (and through 6 Ω: $4/6 \approx 0.67\ \mathrm{A}$; sum = 2 A).

Electrical Power and Energy

Instantaneous power delivered to a device:

$P = VI = I^2 R = \frac{V^2}{R}$

Energy transferred in time t: $W = Pt = VIt = I^2 Rt$. Domestic energy is often quoted in kWh: $1\ \mathrm{kWh} = 3.6 \times 10^6\ \mathrm{J}$.

Worked example — heating. A 100 Ω heater runs from 220 V for 30 minutes. Find current, power, and energy in joules.

$I = \frac{220}{100} = 2.2\ \mathrm{A},\quad P = (2.2)^2(100) = 484\ \mathrm{W}$

$W = 484 \times (30 \times 60) = 8.712 \times 10^5\ \mathrm{J}$

(Alternatively $P = V^2/R = 48400/100 = 484\ \mathrm{W}$.)

Joule heating $H = I^2 Rt$ is why undersized wiring overheats: higher I or higher R (loose connection) dumps energy as heat.

EMF, Internal Resistance, and Terminal Voltage

A cell of EMF ε and internal resistance r driving external R:

$I = \frac{\varepsilon}{R + r},\quad V_{terminal} = \varepsilon - Ir = IR$

When current is drawn, terminal voltage falls below EMF. Short-circuit current (R → 0) is ε/r — conceptual upper bound, not a recommended experiment.

Worked example. A battery of EMF 12 V and r = 1 Ω feeds a 5 Ω load. Find I and terminal voltage.

$I = 12/(5+1) = 2\ \mathrm{A},\quad V = 12 - 2(1) = 10\ \mathrm{V}$

Kirchhoff's Rules (Introduction)

When a circuit cannot be collapsed by series–parallel alone (e.g., bridge networks, multiple batteries), use Kirchhoff:

1. Junction rule (KCL): Algebraic sum of currents at a node is zero — charge conservation. Currents in = currents out.
2. Loop rule (KVL): Algebraic sum of potential changes around any closed loop is zero — energy conservation. Walking with the current through a resistor: −IR; against the current: +IR; from − to + through a battery: +ε.

Worked example — simple two-loop sketch. Two EMFs and three resistors often appear as exam diagrams. Assign loop currents, write two KVL equations, solve. Even if arithmetic is messy, showing correct sign convention earns method marks.

Minimal illustration: single loop with ε₁ = 6 V, ε₂ = 3 V opposing, R = 3 Ω.

Net EMF = 6 − 3 = 3 V → $I = 3/3 = 1\ \mathrm{A}$ in the direction of the larger EMF. That is KVL in one line: $6 - 3 - IR = 0$.

Exam Strategy

• Redraw the circuit; mark series vs parallel clearly before calculating.
• Prefer $V^2/R$ when only voltage and resistance are given; prefer $I^2R$ when current is known.
• Internal resistance questions almost always need $I = \varepsilon/(R+r)$ first.
• Capacitor series/parallel memory sometimes contaminates resistor problems — keep the tables distinct.