Fundamental of Electrical Engineering · 10 min read

Basic circuit concept

Every electrical circuit rests on a few basic quantities — voltage, current, resistance, power and energy — and a few rules that connect them: Ohm's law and Kirchhoff's laws. This topic also covers the materials that conduct or insulate, how resistors combine in series and parallel, the star-delta transformation used to simplify bridge-like networks, and the vocabulary used to classify circuits (linear/non-linear, bilateral/unilateral, active/passive). These definitions are favourites for short NEC MCQs.

Key concepts

  • Charge: coulomb (C); one electron carries 1.602×10−191.602\times10^{-19} C.
  • Current (I): rate of flow of charge, I=dQ/dtI = dQ/dt, in ampere. Conventional current flows from + to − through the external circuit, opposite to electron flow.
  • Voltage (V): work done per unit charge, V=W/QV = W/Q (J/C = volt). EMF is the source's driving voltage; potential difference is the drop across an element.
  • Ohm's law: at constant temperature, current through a conductor is proportional to the voltage across it, V=IRV = IR. Not valid for non-linear devices (diodes, transistors, arc lamps, electrolytes, semiconductors).
  • Resistance: R=ρl/AR = \rho l/A; ρ = resistivity (Ω·m). Temperature effect: Rt=R0(1+α0t)R_t = R_0(1 + \alpha_0 t); α is positive for metals, negative for carbon, semiconductors and insulators, almost zero for alloys like manganin and constantan (used in standard resistors).
  • Power and energy: P=VI=I2R=V2/RP = VI = I^2R = V^2/R (W); energy W=PtW = Pt (J). Commercial unit: 1 kWh = 1 "unit" = 3.6 MJ.
  • Conducting materials: low resistivity — silver (best), copper (most used), gold, aluminium (light, cheap, overhead lines). High-resistivity alloys: nichrome (heating elements), manganin and constantan (standard resistors). Tungsten is a pure metal used for lamp filaments because of its very high melting point.
  • Insulating materials: very high resistivity — PVC, XLPE, rubber, mica, porcelain, glass, paper, transformer oil, SF₆ gas. Key property: dielectric strength (kV/mm) — the field at which breakdown occurs; insulation classes (Y, A, E, B, F, H, C) are rated by maximum temperature.
  • Series circuit: same current; Req=R1+R2+…R_{eq} = R_1 + R_2 + …; voltage divides in proportion to resistance, V1=VR1R1+R2V_1 = V\frac{R_1}{R_1+R_2}.
  • Parallel circuit: same voltage; 1Req=∑1Ri\frac{1}{R_{eq}} = \sum\frac1{R_i}; current divides inversely, I1=IR2R1+R2I_1 = I\frac{R_2}{R_1+R_2}. ReqR_{eq} is less than the smallest resistance. Home wiring is parallel so each load gets full voltage.
  • Kirchhoff's current law (KCL): the algebraic sum of currents at a node is zero — conservation of charge. Basis of nodal analysis.
  • Kirchhoff's voltage law (KVL): the algebraic sum of voltages around a closed loop is zero — conservation of energy. Basis of mesh analysis.
  • Star-delta and delta-star conversion: equivalent three-terminal networks (see formulas). For equal resistors, RY=RΔ/3R_Y = R_\Delta/3.
  • Linear vs non-linear circuit: linear elements have parameters (R, L, C) independent of voltage and current — superposition applies; non-linear elements (diode, iron-cored coil near saturation, thermistor) do not.
  • Bilateral vs unilateral: bilateral elements behave the same for current in either direction (resistor, inductor, capacitor, transmission line); unilateral ones do not (diode, transistor, rectifier).
  • Active vs passive: active elements can deliver net energy to the circuit (voltage and current sources, generators, batteries, transistors/op-amps as amplifiers with power supply); passive elements only absorb (R) or store and return energy (L, C).
  • Lumped vs distributed: lumped elements are physically separate (resistor); distributed parameters are spread along length (transmission lines).

High-yield points

  • R=ρl/AR = \rho l/A; Rt=R0(1+α0t)R_t = R_0(1 + \alpha_0 t); α negative for carbon and semiconductors, near zero for manganin/constantan.
  • 1 kWh = 1 unit = 3.6 MJ.
  • Delta of three equal R → star of R/3; star R → delta 3R.
  • KCL: conservation of charge; KVL: conservation of energy.
  • Parallel equivalent resistance is less than the smallest branch resistance.
  • Superposition applies only to linear circuits; diodes are non-linear and unilateral.

Formulas

Delta → star: RA=RABRCARAB+RBC+RCA (similarly for RB,RC)R_A = \frac{R_{AB}R_{CA}}{R_{AB} + R_{BC} + R_{CA}}\ (\text{similarly for } R_B, R_C)

Star → delta: RAB=RA+RB+RARBRC (similarly for others)R_{AB} = R_A + R_B + \frac{R_AR_B}{R_C}\ (\text{similarly for others})

P=VI=I2R=V2R,1 kWh=3.6×106 JP = VI = I^2R = \frac{V^2}{R},\qquad 1\ \text{kWh} = 3.6\times10^6\ \text{J}

Worked example

A 230 V, 1000 W heater is connected to 200 V. Find its resistance and power drawn (assume R constant).

  1. R=V2/P=2302/1000=52 900/1000=52.9 ΩR = V^2/P = 230^2/1000 = 52\,900/1000 = 52.9\ \Omega.
  2. At 200 V: P=2002/52.9=40 000/52.9=756P = 200^2/52.9 = 40\,000/52.9 = 756 W.
  3. Energy in 5 h = 0.756 × 5 = 3.78 kWh (units).

Exam tips

  • For three equal resistors R in delta, the star equivalent is R/3 per arm; star R → delta 3R.
  • Ohm's law fails for non-linear and unilateral devices.
  • KCL ↔ conservation of charge; KVL ↔ conservation of energy.
  • Resistors in parallel: the equivalent is always less than the smallest one.
  • Silver is the best conductor; copper is most widely used; carbon has a negative temperature coefficient.

Exam-style questions

From Calmon's question bank, which has 23 questions on this topic.

1. A 6 Ω6\,\Omega resistor in parallel with a 3 Ω3\,\Omega resistor is connected in series with a 4 Ω4\,\Omega resistor across a 12 V source. The current drawn from the source is:

  • A. 3 A
  • B. 0.92 A
  • C. 2 A
  • D. 1 A

Answer: C. 6∥3=2 Ω6 \parallel 3 = 2\,\Omega; total =2+4=6 Ω= 2 + 4 = 6\,\Omega; I=12/6=2I = 12/6 = 2 A.

2. A wire 100 m long has a cross-sectional area of 1 mm21\,\text{mm}^2 and resistivity 1.7×10−8 Ω m1.7\times10^{-8}\,\Omega\,\text{m}. Its resistance is:

  • A. 1.7 Ω1.7\,\Omega
  • B. 0.017 Ω0.017\,\Omega
  • C. 0.17 Ω0.17\,\Omega
  • D. 17 Ω17\,\Omega

Answer: A. R=ρl/A=1.7×10−8×100/(1×10−6)=1.7 ΩR = \rho l/A = 1.7\times10^{-8}\times100/(1\times10^{-6}) = 1.7\,\Omega.

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