How Electric Circuits Work

Three Variables That Explain Every Electric Circuit

Every electric circuit analysis reduces to three variables and one equation:

• Voltage (V) — electric potential difference, measured in volts (V). The "pressure" that drives electrons through a circuit.
• Current (I) — flow of electric charge, measured in amperes (A). The rate at which electrons pass a point.
• Resistance (R) — opposition to current flow, measured in ohms (Ω).

Ohm's Law: V = I × R

If you know any two of these three values, you can calculate the third. From this single equation, all of basic circuit analysis follows.

Ohm's Law: The Foundation

Rearrangements of V = IR:

• Solve for current: I = V/R
• Solve for resistance: R = V/I

Simple example: A resistor connected to a 9V battery draws 30 mA (0.030 A) of current. What is its resistance? R = V/I = 9 / 0.030 = 300 Ω

Power in a resistor: P = V × I = I² × R = V²/R

The same 9V battery and 300Ω resistor: P = V × I = 9 × 0.030 = 0.27 W (270 milliwatts)

This power is dissipated as heat. A resistor rated for 0.25 W would overheat in this circuit; you'd need a 0.5 W or 1 W rated resistor.

Series Circuits: Resistors End-to-End

In a series circuit, components are connected in a single path. The same current flows through every component.

Rules for series circuits:

• Total resistance: R_total = R₁ + R₂ + R₃ + ...
• Current is the same throughout: I_total = I₁ = I₂ = I₃
• Voltages add up: V_total = V₁ + V₂ + V₃

Worked example: Three resistors in series R₁ = 100 Ω, R₂ = 220 Ω, R₃ = 470 Ω, connected to a 12V source.

R_total = 100 + 220 + 470 = 790 Ω I = V/R = 12/790 = 0.0152 A (15.2 mA)

Voltage across each resistor:

• V₁ = I × R₁ = 0.0152 × 100 = 1.52 V
• V₂ = I × R₂ = 0.0152 × 220 = 3.34 V
• V₃ = I × R₃ = 0.0152 × 470 = 7.14 V

Check: 1.52 + 3.34 + 7.14 = 12.0 V ✓ (voltages must sum to source voltage)

Key implication: In a series circuit, if any component fails (opens), the entire circuit stops working — there's only one path for current. This is why old Christmas lights would go dark when one bulb burned out.

Parallel Circuits: Multiple Paths

In a parallel circuit, components are connected side-by-side, giving current multiple paths to flow through.

Rules for parallel circuits:

• Voltage is the same across all branches: V₁ = V₂ = V₃ = V_total
• Currents add: I_total = I₁ + I₂ + I₃
• Total resistance (reciprocal formula): 1/R_total = 1/R₁ + 1/R₂ + 1/R₃

Simplified for equal resistors: For n identical resistors of value R: R_total = R/n

Worked example: Three 100 Ω resistors in parallel with a 12V source

1/R_total = 1/100 + 1/100 + 1/100 = 3/100 R_total = 100/3 = 33.3 Ω

Total current: I_total = V/R_total = 12/33.3 = 0.360 A (360 mA)

Current in each branch (all equal since all resistors are equal): I_branch = V/R = 12/100 = 0.12 A (120 mA per branch)

Check: 0.12 + 0.12 + 0.12 = 0.36 A ✓

Key implication: In a parallel circuit, removing one branch doesn't affect the others. Household wiring is parallel — when you turn off one light, the others stay on. Each outlet maintains 120V (or 230V) regardless of other loads on the circuit.

Why parallel resistance is always less than the smallest individual resistance: Adding a parallel path always gives current another route, reducing the total opposition to flow. Three 100Ω resistors in parallel = 33.3Ω — less than any individual resistor.

Kirchhoff's Laws: Analyzing Complex Circuits

For circuits that are combinations of series and parallel (mixed circuits), or for circuits with multiple voltage sources, Kirchhoff's two laws provide a systematic framework.

Kirchhoff's Current Law (KCL): Current Conservation

At any node (junction) in a circuit, the sum of currents entering equals the sum of currents leaving.

This is conservation of charge — charge can't accumulate at a node or appear from nowhere.

Example: Three wires meet at a node. Wire A carries 2A in, wire B carries 1A in. Wire C carries current out. I_C = I_A + I_B = 2 + 1 = 3A out

Kirchhoff's Voltage Law (KVL): Voltage Conservation

Around any closed loop in a circuit, the sum of all voltage rises and drops equals zero.

This is conservation of energy — energy gained from voltage sources must equal energy dissipated in resistors.

Example: A simple loop with one battery (9V) and two resistors (R₁ = 100Ω, R₂ = 200Ω)

Going around the loop: +9V (battery rise) - V_R1 - V_R2 = 0 V_R1 + V_R2 = 9V I × 100 + I × 200 = 9 300I = 9 I = 0.03A = 30mA

Then: V_R1 = 0.03 × 100 = 3V, V_R2 = 0.03 × 200 = 6V Check: 3 + 6 = 9V ✓

Series-Parallel Combination Circuits

Worked example: R₁ = 100Ω is in series with a parallel combination of R₂ = 200Ω and R₃ = 200Ω. Source voltage = 12V.

Step 1: Find the parallel combination of R₂ and R₃: R_parallel = 200/2 = 100Ω (two equal resistors in parallel)

Step 2: Add the series resistor: R_total = R₁ + R_parallel = 100 + 100 = 200Ω

Step 3: Find total current: I_total = 12/200 = 0.06A (60mA)

Step 4: Find voltage across each section:

• V_R1 = 0.06 × 100 = 6V (across R₁)
• V_parallel = 0.06 × 100 = 6V (across the R₂/R₃ combination)

Check: 6 + 6 = 12V ✓

Step 5: Find current in each parallel branch:

• I_R2 = 6/200 = 0.03A (30mA)
• I_R3 = 6/200 = 0.03A (30mA)

Check: 30 + 30 = 60mA = I_total ✓

Common Real-World Applications

LED circuits: LEDs require a specific voltage and can only handle a limited current. A current-limiting resistor in series with an LED prevents it from burning out. For a red LED (forward voltage 2V) powered by 5V, with desired 20mA: R = (V_supply - V_LED) / I = (5 - 2) / 0.020 = 150Ω

Home wiring: A typical 15-amp circuit at 120V has a maximum power capacity of P = IV = 15 × 120 = 1,800 W. Connecting too many appliances causes the current to exceed 15A, tripping the breaker.

Battery packs: Connecting batteries in series adds their voltages (two 1.5V AA batteries in series = 3V). Connecting in parallel maintains the same voltage but doubles current capacity.