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Electronics Fundamentals — Building from the ground up

Before diving into complex AC systems, we start with the foundation. This chapter covers Kirchhoff's Voltage Law (KVL), Kirchhoff's Current Law (KCL), and Ohm's Law—the three fundamental principles that make circuit analysis possible.

DC circuits KVL • KCL • Ohm's Law visual examples

1) The building blocks of circuit analysis

Every circuit, no matter how complex, follows three fundamental rules. Master these, and you have the tools to analyze virtually any electrical network:

  • KVL (Kirchhoff's Voltage Law): voltages around any closed loop sum to zero
  • KCL (Kirchhoff's Current Law): currents into any node equal currents out
  • Ohm's Law: the relationship between voltage, current, and resistance
KVL and KCL are not approximations—they're consequences of charge conservation and energy conservation, and they hold for DC and AC alike. Ohm's Law is different: it describes how a resistor behaves, not a conservation law, and in AC circuits it generalizes to \(V = I \cdot Z\) (impedance). See the Phasors chapter.
How to read this chapter. Each law comes with a Predict step before the answer. Commit to a number first, then open the reveal — a guess you have to correct sticks far better than a result you only read.

Entry check: open return path

A 12 V battery is wired to one side of a small lamp, but the lamp's other terminal is not connected back to the battery. Predict the current through the lamp.

Predict the lamp current, then reveal

0 A flows through the lamp. Current is loop flow: charge must have a complete path out of the source, through the load, and back to the source. One wire can leave a voltage difference at the open terminal, but it cannot sustain lamp current.

2) Kirchhoff's Voltage Law (KVL)

KVL states: The algebraic sum of all voltages around any closed loop in a circuit is zero.

\[ \sum V = 0 \]

Why it works

Voltage is potential difference—energy per unit charge. If you trace a complete loop starting and ending at the same point, you've traveled through zero net potential difference. Any energy gained from sources must equal energy dropped across loads.

Visual example: loop energy balance

Predict first. A 12 V source drives two equal 6 Ω resistors in series. Before you open the reveal, commit to two numbers: (a) how many volts does each resistor drop, and (b) what do all the voltages around the loop add up to?
KVL example circuit with animated loop trace A 12 volt source feeds two 6 ohm resistors in series. The animated trace follows the loop, while the ledger shows plus 12 volts from the source and two minus 6 volt drops returning to zero. KVL: trace the loop — predict the drops + - 12V R1 = 6Ω R2 = 6Ω ? ? chosen loop direction 0V +12V source R1: ? R2: ? 0V
A complete loop starts and ends at the same potential: the source's rise is balanced by the drops across the two equal resistors.
Reveal: did your loop sum to zero?

Each resistor drops 6 V (equal resistors split the 12 V equally), and around the loop +12 − 6 − 6 = 0. The source raises each charge by 12 V; the two resistors give all of it back. A complete loop always returns to the same potential.

Myth: voltage flows around the loop and gets used up. Voltage is a difference between two points, not a substance that travels. A “6 V drop” means the resistor's two ends differ by 6 V. What the resistors consume is energy — the voltage just bookkeeps where it goes.

Now make the resistors unequal

Same 12 V source, but the two series resistors are now different: 4 Ω then 8 Ω. Equal resistors split the voltage equally — these aren't equal. Predict the voltage across the 8 Ω resistor.

Predict the 8 Ω drop, then reveal

I = 12 / (4 + 8) = 1 A, so the 8 Ω drops V = I × 8 = 8 V (and 4 V + 8 V = 12 V, KVL). The bigger series resistor drops more, not less.

Practical application

KVL lets you write equations for any loop. In a series circuit with a voltage source \(V_s\) and resistors \(R_1, R_2, \ldots, R_n\):

\[ V_s - V_{R_1} - V_{R_2} - \cdots - V_{R_n} = 0 \]

This is how you solve for unknown voltages when you know the current (or vice versa with Ohm's Law).

3) Kirchhoff's Current Law (KCL)

KCL states: The algebraic sum of currents entering a node (junction) equals zero. Equivalently: current in = current out.

\[ \sum I_{\text{in}} = \sum I_{\text{out}} \]

Why it works

Charge is conserved, and an ideal node stores none of it. Electrons don't pile up at a junction and they don't vanish—so whatever flows in must flow out, at every instant.

Visual example: current splits, then rejoins

12 V is applied across three parallel resistors: R1 = 12 Ω, R2 = 6 Ω, R3 = 4 Ω. Each branch shares the same two nodes, so each branch sees the same 12 V.

Predict the total source current, then reveal

Each branch sees the full 12 V (parallel = shared voltage), so I1 = 12/12 = 1 A, I2 = 12/6 = 2 A, and I3 = 12/4 = 3 A. KCL says the source current is the sum leaving node A: 1 + 2 + 3 = 6 A. Current splits and rejoins; it is not used up.

KCL example circuit with three parallel branches A 12 volt source feeds three resistors connected between node A and node B. The supply return is routed below the branch network so it does not short across the resistors. KCL at A: Itotal = I1 + I2 + I3 + - 12V A B Itotal I1 I2 I3 R1 = 12Ω R2 = 6Ω R3 = 4Ω The return rail runs below the branches, so every path from A to B must pass through a resistor.
Parallel resistors share the same two nodes: current splits at node A, passes through the branches, and rejoins at node B.
Myth: current is “used up” as it flows. A node stores no charge — whatever flows in flows out, every instant. A lamp doesn't swallow current; the same current that enters leaves the other side. What gets consumed is energy, not current.

Practical application

At any node with multiple branches, you can write:

\[ I_{\text{in}} = I_{\text{out,1}} + I_{\text{out,2}} + \cdots + I_{\text{out,n}} \]

KCL is essential for analyzing parallel circuits and complex networks with multiple loops.

4) Ohm's Law — The workhorse equation

Ohm's Law describes the linear relationship between voltage, current, and resistance:

\[ V = I \cdot R \]

Where:

  • \(V\) = voltage (volts, V)
  • \(I\) = current (amperes, A)
  • \(R\) = resistance (ohms, Ω)

Three forms, one relationship

You can rearrange Ohm's Law to solve for any variable:

V I R Ohm's Law triangle
Cover the unknown: need V? Use \(V = I \cdot R\). Need I? Use \(I = V / R\). Need R? Use \(R = V / I\).

\[V = I \cdot R\]

\[I = \frac{V}{R}\]

\[R = \frac{V}{I}\]

Three forms of the same law. All are equivalent, just rearranged for convenience.

Example: finding current

Given a 12V battery connected to a 6Ω resistor, what is the current?

Predict the current, then reveal

\[ I = \frac{V}{R} = \frac{12\text{V}}{6\Omega} = 2\text{A} \]

Example: finding voltage drop

If 0.5A flows through a 100Ω resistor, what is the voltage drop?

Predict the voltage drop, then reveal

\[ V = I \cdot R = 0.5\text{A} \times 100\Omega = 50\text{V} \]

Ohm's Law applies to resistive (ohmic) components, and to a whole network through its equivalent resistance. Combined with KVL and KCL, you can solve for all voltages and currents in any resistive network.
Myth: current is a fixed amount the supply “pushes out”. Current isn't set by the source alone — it responds to both voltage and resistance: \(I = V/R\). Keep the 12 V and double the resistance to 12 Ω and the current halves to 1 A. The resistor, not the battery, sets how much flows.

5) Combining the laws — Putting it all together

The real power comes when you use all three laws together. Let's work a complete example.

Example: Series-parallel circuit

Series-parallel example circuit A 24 volt source drives R1 (4 ohms) in series, then R2 (12 ohms) in parallel with R3 (6 ohms), then R4 (2 ohms) in series, before returning to the source. Goal: find the total current. 24V R₁ = 4Ω R₂ = 12Ω R₃ = 6Ω R₄ = 2Ω I Find total current I
A mixed circuit: R₁ and R₄ are in series, while R₂ and R₃ are in parallel.
Predict first. Before working the steps: will the total resistance be more or less than \(R_1 + R_4 = 6\,\Omega\)? Roughly what total current do you expect from the 24 V source — closer to 1 A, 2 A, or 5 A? Then check yourself against the worked steps below.

Solution steps

Step 1: Simplify parallel combination

\(R_2\) and \(R_3\) in parallel:

\[ R_{23} = \frac{R_2 \cdot R_3}{R_2 + R_3} = \frac{12\Omega \cdot 6\Omega}{12\Omega + 6\Omega} = \frac{72}{18} = 4\Omega \]

Step 2: Find total resistance

Now the circuit is just three resistors in series:

\[ R_{\text{total}} = R_1 + R_{23} + R_4 = 4\Omega + 4\Omega + 2\Omega = 10\Omega \]

Step 3: Apply Ohm's Law

\[ I = \frac{V}{R_{\text{total}}} = \frac{24\text{V}}{10\Omega} = 2.4\text{A} \]

Step 4: Find voltage drops (KVL)

\[ V_{R_1} = I \cdot R_1 = 2.4\text{A} \times 4\Omega = 9.6\text{V} \]

\[ V_{R_{23}} = I \cdot R_{23} = 2.4\text{A} \times 4\Omega = 9.6\text{V} \]

\[ V_{R_4} = I \cdot R_4 = 2.4\text{A} \times 2\Omega = 4.8\text{V} \]

Check with KVL: \(9.6\text{V} + 9.6\text{V} + 4.8\text{V} = 24\text{V}\) ✓

Step 5: Find branch currents (KCL)

Since \(R_2\) and \(R_3\) have 9.6V across them:

\[ I_2 = \frac{9.6\text{V}}{12\Omega} = 0.8\text{A} \]

\[ I_3 = \frac{9.6\text{V}}{6\Omega} = 1.6\text{A} \]

Check with KCL: \(0.8\text{A} + 1.6\text{A} = 2.4\text{A}\) ✓

Quick questions: series, parallel, mixed

  • Series: Two 10Ω resistors on 12V—what is the current and how does each drop voltage?
  • Parallel: Three 12Ω branches on 9V—what is the branch current in each and the total?
  • Mixed: A 4Ω resistor in series with two 8Ω resistors in parallel on 24V—what is \(R_\text{total}\) and the supply current?
Predict all three, then reveal the answer key
  • Series: \(R = 20\Omega\), so \(I = 12\text{V}/20\Omega = 0.6\text{A}\); each resistor drops \(0.6\text{A} \times 10\Omega = 6\text{V}\).
  • Parallel: each branch carries \(9\text{V}/12\Omega = 0.75\text{A}\); total \(= 3 \times 0.75\text{A} = 2.25\text{A}\).
  • Mixed: \(8\Omega \parallel 8\Omega = 4\Omega\), so \(R_\text{total} = 4\Omega + 4\Omega = 8\Omega\) and \(I = 24\text{V}/8\Omega = 3\text{A}\).
This systematic approach—simplify, apply Ohm's Law, verify with KVL and KCL—works for any resistive network, no matter how complex.

6) Power — Where energy meets reality

Electrical power is the rate of energy transfer:

\[ P = V \cdot I \]

Power equations from Ohm's Law

\(P = V \cdot I\) is always true. For a resistor, substituting Ohm's Law gives two more equivalent forms:

\[ P = V \cdot I = I^2 R = \frac{V^2}{R} \]

Which way is the power flowing? (the passive sign convention)

\(P = V \cdot I\) gives a number, but its sign tells you the direction of energy flow — and that depends on a reference choice. Under the passive sign convention you draw the current reference arrow into the terminal marked \(+\); then \(P = V\cdot I\) is the power the element absorbs:

  • \(P > 0\): the element absorbs power — it acts as a load (a resistor always does, \(P = I^2 R \ge 0\)).
  • \(P < 0\): the element delivers power — it acts as a source.

A battery being discharged delivers power (conventional current leaves its \(+\) terminal, so the referenced \(I\) is negative and \(P<0\)); the same battery being charged absorbs power (\(P>0\)). Energy is conserved: with every element referenced this way the powers sum to zero — what the sources deliver, the loads absorb.

The sign is bookkeeping, not physics: flip the assumed current arrow (or which terminal you call \(+\)) and \(P\) flips sign, while the real energy flow is unchanged. This same "load reference" is what the complex-power formula \(\underline{S}=\underline{V}\,\underline{I}^*\) uses — it is why an inductive load reads \(Q>0\) (absorbing reactive power).

Example: power dissipation and rating

A 12 V source is connected across a 100 Ω resistor. Common parts on the bench are rated 0.25 W, 0.5 W, 1 W, and 2 W. What is the smallest safe resistor rating you should choose?

Predict the resistor rating, then reveal

\(I = V/R = 12/100 = 0.12\,\text{A}\). The heat load is \(P = V \cdot I = 12 \times 0.12 = 1.44\,\text{W}\) (the same as \(V^2/R = 144/100\)).

A 1 W resistor is below that heat load, and an exact 1.44 W part would have no margin. From the listed choices, pick the next higher rating: 2 W. Voltage alone does not decide whether the part survives; power does.

Power ratings matter. Always check that components can handle the power they'll dissipate. Heat is the enemy of reliability.

7) Contactors and relays — simple motor control

An asynchronous motor is usually switched with a contactor (or power relay) so the heavy current stays away from operator buttons. We only analyze the low-current control circuit here—coils and switches arranged in series and parallel, exactly like the resistor combinations above—not the motor's AC power path. (An asynchronous motor is an AC machine; its power wiring is a separate topic.)

  • Start/stop with safety: place the stop and emergency-stop buttons in series with the coil so any open switch kills power (safety first).
  • Start from two places: wire multiple start buttons in parallel; any closed button sends a “1” to the coil while stops stay in series to enforce “0” on fault.
  • Seal-in/hold: use an auxiliary contact (an extra contact that moves with the contactor) in parallel with the start button so the coil holds itself energized after a momentary press.
  • Timer auto-off: add a time-delay relay—whose contact switches a set time after it's triggered—in series to drop the coil after that delay, implementing an automatic “0”.
Each contact is a binary element: closed = 1, open = 0. Series wiring implements logical AND (all must be 1), parallel implements OR (any 1 is enough). The same logic you applied to resistors maps directly to safe motor control.

8) Where to go from here

You now have the foundation for DC circuit analysis. These same principles extend to AC circuits, but with some important additions:

  • Impedance replaces resistance (includes capacitors and inductors)
  • Phase relationships between voltage and current
  • Complex power and power factor
  • Phasors for steady-state AC analysis
Ready for the next level? Check out the Phasors chapter to see how these fundamentals extend to AC systems and three-phase power.

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