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 provide useful context for 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. A useful two-loop exercise uses two EMFs and three resistors. 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: $6 - 3 - IR = 0$.

Practice 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.