Basic concept

Every electrical and electronic system, from a house wiring circuit to a microprocessor, obeys a few basic relationships between voltage, current, resistance, power and energy. This topic sets out those relationships: Ohm's law, how resistors combine in series and parallel, how to convert star (Y) and delta (Δ) networks, Kirchhoff's laws, and the vocabulary used to classify circuits (linear/non-linear, bilateral/unilateral, active/passive). MCQs on this topic are usually quick calculations or definition checks, so learn the formulas and the exact meaning of each term.

Key concepts

• Electric charge and current: Current is the rate at which charge flows, $I = dQ/dt$, measured in amperes (1 A = 1 C/s). Conventional current flows from + to −, opposite to electron flow.
• Voltage (potential difference): The work done per unit charge in moving a charge between two points, $V = W/Q$ (1 V = 1 J/C). EMF is the voltage produced by a source; potential difference is the voltage across an element.
• Ohm's law: At constant temperature, the current through a metallic conductor is proportional to the voltage across it: $V = IR$. It does not hold for non-linear (non-ohmic) devices such as diodes, transistors and arc lamps.
• Resistance: $R = \rho L / A$. It increases with length, decreases with area, and for metals rises with temperature ($R_t = R_0(1+\alpha t)$). Carbon, semiconductors and insulators have a negative temperature coefficient.
• Power and energy: Power is the rate of doing work, $P = VI = I^2R = V^2/R$ (watts). Energy is $W = Pt$; the commercial unit is the kWh ("unit"), $1\ \text{kWh} = 3.6\times10^6$ J.
• Conducting and insulating materials:

Class	Resistivity (approx.)	Examples	Use
Conductors	about $10^{-8}$ Ω·m	silver, copper, aluminium	wires, busbars
Semiconductors	between conductors and insulators	Si, Ge	devices
Insulators	very high (about $10^{8}$ Ω·m and above)	mica, PVC, porcelain, glass	cable insulation, supports

Silver is the best conductor; copper is the most used; aluminium is used in overhead lines for its low weight and cost. Insulators are rated by dielectric strength (kV/mm).

• Series circuit: Same current through all elements; voltages add; $R_s = R_1 + R_2 + \dots$. Voltage divider: $V_1 = V\,R_1/(R_1+R_2)$.
• Parallel circuit: Same voltage across all branches; currents add; $1/R_p = 1/R_1 + 1/R_2 + \dots$. Current divider: $I_1 = I\,R_2/(R_1+R_2)$. $R_p$ is always smaller than the smallest branch resistance.
• Star–delta conversion: Lets you simplify bridge-type networks that are neither series nor parallel.
• 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 any closed loop is zero (conservation of energy). Basis of mesh analysis.
• Linear vs non-linear: A linear element has parameters (R, L, C) that do not change with voltage or current; it obeys superposition and homogeneity. Diodes and iron-cored inductors are non-linear.
• Bilateral vs unilateral: A bilateral element behaves the same for current in either direction (resistor, inductor, capacitor). A unilateral element does not (diode, transistor).
• Active vs passive: Active elements can supply energy (voltage/current sources, op-amps, transistors when biased). Passive elements only absorb or store energy (R absorbs; L and C store).

High-yield points

• $1\ \text{kWh} = 3.6\times10^6$ J (one commercial "unit").
• Equal resistors: $R_Y = R_\Delta/3$, $R_\Delta = 3R_Y$.
• KCL follows from conservation of charge; KVL from conservation of energy.
• Stretching a wire to $n$ times its length (same volume) makes its resistance $n^2R$.
• Parallel resistance is always less than the smallest branch resistance.
• Diode: non-linear and unilateral. Resistor: linear, bilateral, passive.

Formulas

$R = \frac{\rho L}{A}$ $\rho$ = resistivity (Ω·m), $L$ = length (m), $A$ = cross-sectional area (m²).

$P = VI = I^2 R = \frac{V^2}{R}$ $P$ in W, $V$ in V, $I$ in A, $R$ in Ω.

Delta to star (delta resistors $R_{ab}, R_{bc}, R_{ca}$): $R_a = \frac{R_{ab}R_{ca}}{R_{ab}+R_{bc}+R_{ca}}$ and similarly for $R_b$, $R_c$ (product of the two adjacent delta resistors over the sum).

Star to delta: $R_{ab} = R_a + R_b + \frac{R_aR_b}{R_c}$

For equal resistors: $R_Y = R_\Delta/3$ and $R_\Delta = 3R_Y$.

Worked example

Three 9 Ω resistors are connected in delta. What is the equivalent star resistance per arm, and if two arms of that star are in series across a 12 V supply, what current flows?

1. $R_Y = R_\Delta/3 = 9/3 = 3$ Ω.
2. Two arms in series: $3 + 3 = 6$ Ω.
3. $I = V/R = 12/6 = 2$ A.
4. Check with delta directly: between two terminals, 9 Ω is in parallel with (9+9) = 18 Ω, giving $9\times18/27 = 6$ Ω. Same answer.

Answer: 3 Ω per arm, 2 A.

Exam tips

• Parallel resistance of $n$ equal resistors is $R/n$; series is $nR$. Two equal resistors in parallel give half.
• If a wire is stretched to $n$ times its length (volume constant), its resistance becomes $n^2$ times.
• A 100 W bulb has lower resistance than a 40 W bulb at the same rated voltage ($R = V^2/P$); in series the 40 W bulb glows brighter.
• KCL comes from conservation of charge, KVL from conservation of energy — a frequent one-mark question.
• Diode: non-linear and unilateral. Resistor: linear, bilateral, passive.

Exam-style questions

From Calmon's question bank, which has 25 questions on this topic.

1. A circuit element whose voltage–current relationship is different for the two directions of current, such as a semiconductor diode, is called:

• A. Bilateral
• B. Unilateral
• C. Linear
• D. Passive

Answer: B. A unilateral element behaves differently for opposite directions of current; a diode conducts readily only in the forward direction.

2. Node N is connected to the positive terminal of a 12 V source through 2 Ω, to the positive terminal of a 6 V source through 3 Ω, and to the reference node through 6 Ω. Both source negative terminals are at the reference node. The voltage of node N is:

• A. 9 V
• B. 8 V
• C. 7.2 V
• D. 6 V

Answer: B. KCL at N: $\frac{12-V}{2} + \frac{6-V}{3} = \frac{V}{6}$; multiplying by 6: $36 - 3V + 12 - 2V = V$, so $V = 8$ V.

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