Chapters/Ch 2· beginner

Basic Laws

Everything for Basic Laws in one place — 8 topics, 10 key formulas, a live simulator and a 10-question quiz.

01Ohm's Law

Ohm's law states that the voltage across a resistive element is directly proportional to the current through it:

$v = i \cdot R$

where $R$ is the resistance $in o hm s (Ω) . O n eo hmi s 1 V / A .$

Resistance of a uniform conductor depends on geometry and material:

$R = \frac{ρ L}{A}$

$ρ$ is the resistivity of the material (Ω·m), $L$ is length, $A$ is cross-sectional area. Copper: $ρ \approx 1.72 \times 1 0^{- 8}$ Ω·m; aluminium: $2.82 \times 1 0^{- 8}$ ; nichrome (heating element): $1.10 \times 1 0^{- 6}$ .

Conductance $G = \frac{1}{R}$ has units of siemens (S). High conductance = low resistance.

Limitations. Ohm's law applies only to ohmic (linear) materials within their rated range. It fails for:

Diodes (exponential $i$ - $v$ curve)
Transistors (depend on gate voltage)
Resistors at very high temperature (resistance rises with $T$ )
Discharge tubes, neon lamps (nonlinear conduction)

Solved examples

A 220 - Ω res i s t or ha s 11 Va cross i t . F in d t h ec u rre n t .

$i = v / R = 11/220$

Answer

i = 50

Common pitfalls

Applying Ohm's law to a non-linear element (diode, transistor) — won't give correct results.
$F or g e tt in g t oco n v er t u ni t s . R ink Ω an d iinm A g i v es v in v o lt s — b u t o n l y i f yo u^{'} reco n s i s t e n t .$

02Nodes, Branches, and Loops

These are the three structural pieces of any circuit. Every analysis method (KVL, KCL, mesh, nodal) starts by identifying them.

Branch — a single two-terminal element (one resistor, one capacitor, one source, one wire segment).
Node — a junction where *two or more* branches meet. Wires connecting nodes without an intervening element are part of the same node.
Loop — any closed path through the circuit. An independent loop is one that contains at least one branch not in any other loop. Mesh is the term for a loop with no loops inside it (think 'window pane').

Counting rule — for a connected circuit with $b$ branches, $n$ nodes, and $ℓ$ independent loops:

$ℓ = b - n + 1$

This tells you exactly how many independent KVL equations you need to write. KCL gives you $n - 1$ independent equations (one node is redundant since charge in = charge out globally).

Solved examples

A circuit has 8 branches and 5 nodes. How many independent loops does it have?

$ℓ = b - n + 1 = 8 - 5 + 1$

Answer4 independent loops

Common pitfalls

Counting nodes as 'every wire bend' — only true junctions count, plus the points where elements terminate.
Confusing mesh (smallest loop) with loop (any closed path). Mesh analysis uses meshes; KVL uses any loop.

03Kirchhoff's Current Law (KCL)

Kirchhoff's Current Law is the statement of conservation of charge at every node:

$n = 1 \sum N i_{n} = 0$

The algebraic sum of all currents entering a node is zero. Equivalently: the sum of currents *in* equals the sum of currents *out*.

Sign convention: pick one direction (in or out) and call it positive. Currents flowing that direction are +; currents flowing the other way are −. Be consistent within each node equation.

Why it works: a node has no capacity to store charge. Whatever flows in must flow out, instant by instant. (Real wires can have tiny capacitance, but in ideal circuit analysis we ignore it.)

KCL applies at every node — but only $n - 1$ of the $n$ node equations are independent. Skipping one (usually the ground/reference) gives you a non-redundant system.

Solved examples

At a node,

i_{1} = 3

A and

i_{2} = 4

A flow in;

i_{3} = 2

A flows out. Find the fourth branch current

i_{4}

(define + as out of node).

KCL: in = out → $3 + 4 = 2 + i_{4}$
$i_{4} = 7 - 2$

Answer

i_{4} = 5

A (flowing out)

Common pitfalls

Mixing 'in' and 'out' signs within the same equation. Pick one convention per node and stick to it.
Forgetting that a current source forces a specific current — you don't compute it, you just use its value.

04Kirchhoff's Voltage Law (KVL)

Kirchhoff's Voltage Law is the statement of conservation of energy around every closed loop:

$n = 1 \sum N v_{n} = 0$

The algebraic sum of all voltages around any closed path is zero — energy gained equals energy spent.

Sign convention: walk around the loop in a chosen direction (clockwise or counter-clockwise). When you enter an element at its + terminal, write *+v*; when you enter at the − terminal, write *−v*. Sources work the same way: + to − across a source means an EMF rise.

Why it works: a charge taken once around any closed path returns to its starting potential. The net energy gained = net energy spent.

KVL with $ℓ$ independent loops gives $ℓ$ equations. Combined with $n - 1$ KCL equations and $b$ Ohm's law element equations, you have enough to solve for every branch current and node voltage in any resistive network.

Solved examples

A loop has a 24-V source, then voltage drops of 9 V across

R_{1}

, then

V_{2}

across

R_{2}

, then 4 V across

R_{3}

, back to the source. Find

V_{2}

KVL: source rise = sum of drops
$24 = 9 + V_{2} + 4$
$V_{2} = 24 - 13$

Answer

V_{2} = 11

Common pitfalls

Reversing the sign of a source when traversing it from − to +. Track polarity carefully — most KVL errors come from this.
Writing KVL for a path that isn't closed. A loop must end where it started.

05Series Resistors and Voltage Division

Two or more resistors are in series when they share a single connection and the same current flows through all of them. The equivalent resistance is the sum:

$R_{e q} = R_{1} + R_{2} + \dots + R_{n}$

Voltage Divider Rule (VDR). When $n$ resistors in series share a source voltage $v$ , each resistor takes a fraction of $v$ proportional to its own value:

$v_{k} = \frac{R _{k}}{R _{1} + R _{2} + \dots + R _{n}} \cdot v$

For two resistors specifically:

$v_{1} = \frac{R _{1}}{R _{1} + R _{2}} \cdot v, v_{2} = \frac{R _{2}}{R _{1} + R _{2}} \cdot v$

Use VDR when: you need a voltage across one resistor in a series chain and don't want to compute the current first.

Solved examples

A 12-V source feeds

R_{1} = 4

kΩ and

R_{2} = 8

kΩ in series. Find the voltage across

R_{2}

$v_{2} = \frac{R _{2}}{R _{1} + R _{2}} \cdot V = \frac{8}{12} \times 12$

Answer

v_{2} = 8

Common pitfalls

Applying VDR when current is being drawn through a third path. VDR assumes only the series chain — if a load is connected across one of the resistors, you need a Thévenin analysis instead.
Forgetting that the largest resistor in a series chain drops the most voltage.

06Parallel Resistors and Current Division

Two or more resistors are in parallel when their terminals share both nodes and the same voltage appears across all of them. Equivalent resistance:

$\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \dots + \frac{1}{R _{n}}$

For exactly two resistors, the *product-over-sum* form is faster:

$R_{e q} = \frac{R _{1} R _{2}}{R _{1} + R _{2}}$

Notice $R_{e q}$ is always *smaller* than the smallest of the parallel resistors.

Current Divider Rule (CDR). For two resistors in parallel sharing a total current $i$ , the current through one of them is given by the *other's* resistance divided by the sum:

$i_{1} = \frac{R _{2}}{R _{1} + R _{2}} \cdot i, i_{2} = \frac{R _{1}}{R _{1} + R _{2}} \cdot i$

Current divides inversely with resistance — the smaller resistor takes more current.

Solved examples

R_{1} = 6

Ω and

R_{2} = 12

Ω are in parallel. Find

R_{e q}

, and the current through

R_{1}

if the total current is 9 A.

$R_{e q} = (6 \times 12) / (6 + 12) = 72/18 = 4$ Ω
$i_{1} = \frac{R _{2}}{R _{1} + R _{2}} \cdot i = \frac{12}{18} \times 9$

Answer

R_{e q} = 4

Ω;

i_{1} = 6

Common pitfalls

Confusing the CDR sign — for the current through $R_{1}$ , the numerator is $R_{2}$ (the OTHER resistor), not $R_{1}$ .
Applying the product/sum form to three or more resistors. That shortcut only works for exactly two.

07Wye-Delta (Y-Δ) Transformations

$S o m e n e tw or k s a re n^{'} tp u reser i es / p a r a ll e l — t h reeres i s t ors a rr an g e d ina t r ian g l e (Δ) or ina s t a r (Y, a l so w r i tt e n T) c a n^{'} t b es im pl i f i e dd i rec tl y . T h e$ wye-delta transformation converts between the two forms so the rest of the circuit can be reduced.

Label the three Δ resistors $R_{a}, R_{b}, R_{c}$ between three nodes 1, 2, 3 and the three Y resistors $R_{1}, R_{2}, R_{3}$ from the same three nodes meeting at a common centre node.

$Δ \to Y$ (use when the network has a triangle you want to simplify):

$R_{1} = \frac{R _{b} R _{c}}{R _{a} + R _{b} + R _{c}}, R_{2} = \frac{R _{a} R _{c}}{R _{a} + R _{b} + R _{c}}, R_{3} = \frac{R _{a} R _{b}}{R _{a} + R _{b} + R _{c}}$

$Y \to Δ$ $(u se w h e n t h e n e tw or kha s aY / T yo u w an tt o mak e in t o a Δ) :$

$R_{a} = \frac{R _{1} R _{2} + R _{2} R _{3} + R _{3} R _{1}}{R _{1}}$

and similarly for $R_{b}$ (denominator $R_{2}$ ) and $R_{c}$ (denominator $R_{3}$ ).

Symmetric special case: when all three resistors are equal,

$R_{Y} = \frac{R _{Δ}}{3}, R_{Δ} = 3 R_{Y}$

Solved examples

C o n v er t aba l an ce d Δ w i t h e a c ha r m = 30Ω t o i t s Y e q u i v a l e n t .

Symmetric case: $R_{Y} = R_{Δ} /3 = 30/3$

Answer

E a c hYa r m = 10Ω

Common pitfalls

Mixing up which Y resistor corresponds to which Δ pair. Each Y arm is opposite the Δ arm with the matching subscript — $R_{1}$ is opposite $R_{a}$ .
$F or g e tt in g t ha tt h ece n t re n o d eo f t h e Y d i d n^{'} t e x i s t in t h e Δ— i t^{'} s a v i r t u a l n o d e in t ro d u ce d b y t h e t r an s f or ma t i o n .$

08Applications: Lighting, DC Meters

Lighting circuits. Series Christmas-tree lights: if one bulb fails open, the whole string goes dark (no current path). Parallel household lighting: each bulb has its own path; one failing leaves the others lit. Trade-off: series uses less wire but is fragile; parallel costs more wire but is reliable.

DC ammeter measures current. It must be inserted in series with the branch you want to measure (so all the current flows through it). The ammeter's internal resistance must be very low so it doesn't disturb the circuit. A typical multimeter ammeter has $R_{shunt} \approx 0.1$ Ω.

DC voltmeter measures voltage. It must be connected in parallel with the element you want to measure (so it sees the same voltage). Its internal resistance must be very high so almost no current flows through it. A digital multimeter voltmeter has $R_{in} \approx 10$ MΩ.

The cardinal sin: connecting an ammeter directly across a voltage source. The ammeter's near-zero resistance becomes a short circuit and either blows a fuse or destroys the meter.

Common pitfalls

Inserting a voltmeter in series (you'll read tiny current and break the circuit) or an ammeter in parallel (instant short).
$I g n or in g t h e m e t e r^{'} s in t er na l res i s t an ce in p rec i s i o nm e a s u re m e n t s . A 1 - M Ω res i s t or m e a s u re d b y a 10 - M Ω v o lt m e t er i s 10$

§Key formulas

v = i R

Ohm's Law · V

R = ρ L / A

Resistance · Ohms (Ω)

\sum i_{in} = \sum i_{o u t}

KCL · A

\sum v = 0 (a ro u n d an yc l ose d l oo p)

KVL · V

R_{e q} = R_{1} + R_{2} + \dots + R_{n}

Series Resistance · Ω

\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \dots + \frac{1}{R _{n}}

Parallel Resistance · Ω

v_{1} = \frac{R _{1}}{R _{1} + R _{2}} v

Voltage Divider · V

i_{1} = \frac{R _{2}}{R _{1} + R _{2}} i

Current Divider · A

R_{a} = \frac{R _{1} R _{2} + R _{2} R _{3} + R _{3} R _{1}}{R _{1}}

Y to Δ · Ω

R_{1} = \frac{R _{a} R _{b}}{R _{a} + R _{b} + R _{c}}

Δ to Y · Ω

Open in cheat sheet

Basic Laws

01Ohm's Law

02Nodes, Branches, and Loops

03Kirchhoff's Current Law (KCL)

04Kirchhoff's Voltage Law (KVL)

05Series Resistors and Voltage Division

06Parallel Resistors and Current Division

07Wye-Delta (Y-Δ) Transformations

08Applications: Lighting, DC Meters

§Key formulas

Simulator & graphs

Chapter quiz