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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
These laws are not approximations. They are consequences of charge conservation and energy conservation. They work for DC, AC, and everything in between.

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: simple loop

12V + − R₁ = 6Ω R₂ = 6Ω −6V −6V loop direction KVL: +12V − 6V − 6V = 0
A simple series circuit: the 12V source drives current through two 6Ω resistors. Applying KVL: \(+12\text{V} - 6\text{V} - 6\text{V} = 0\).

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. Electrons don't accumulate at a junction (in steady state), and they don't vanish. Whatever flows in must flow out.

Visual example: parallel branches

12V + − A R₁ = 12Ω R₂ = 6Ω R₃ = 4Ω B I_total I₁ I₂ I₃ KCL at node A: I_total = I₁ + I₂ + I₃
Parallel resistors: current splits at node A, with each branch carrying current based on its resistance. KCL: \(I_{\text{total}} = I_1 + I_2 + I_3\).

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 \(V = I \times R\). 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?

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

Example: finding voltage drop

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

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

Ohm's Law works for individual components and entire circuits. Combined with KVL and KCL, you can solve for all voltages and currents in any resistive network.

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

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.

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 in parallel on 24V—what is \(R_\text{total}\) and the supply current?
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

By substituting Ohm's Law, we get three equivalent forms:

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

Example: power dissipation

In our earlier example with 2A through a 6Ω resistor:

\[ P = I^2 R = (2\text{A})^2 \times 6\Omega = 24\text{W} \]

This resistor dissipates 24 watts as heat. If it's only rated for 10W, it will overheat and fail.

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. The control circuit is just low-current coils and switches arranged in series and parallel, exactly like the resistor combinations above.

  • 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 in parallel with the start button so the coil holds itself energized after a momentary press.
  • Timer auto-off: add a time-delay relay contact in series to drop the coil after a set time, 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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