3.1 Ohm's Law, Power, and Series/Parallel Circuits

Why This Matters for the Texas Journeyman Exam

The Calculations part of the TDLR/PSI journeyman exam (26 questions, 110 minutes) is built on two algebraic relationships: Ohm's Law and Watt's Law. Whether the question asks you to size a branch-circuit conductor, compute the voltage drop on a 250-foot feeder, or verify a motor's full-load current, every formula reduces to one of these two laws. The Knowledge part (59 questions, 130 min) also leans on series and parallel circuit behavior when it tests your understanding of equipment grounding paths, tap rules, and multi-wire branch circuits.

This section defines the variables, presents the formula set, and works through circuit examples you will see on exam day.

The Four Quantities

Every electrical calculation on the exam uses four variables:

The NEC uses E for voltage in its informational notes and examples; some textbooks use V. The formulas are identical either way.

Ohm's Law (E = I × R)

Ohm's Law states that the voltage dropped across a resistance equals the current through it multiplied by the resistance. Rearranged, it gives three working forms:

Worked Example 1 — Solving for Current

A 120-volt incandescent lighting circuit has a total resistance of 24 Ω. What current flows?

I = E ÷ R = 120 ÷ 24 = 5 A

Worked Example 2 — Solving for Resistance

A 277-volt lighting load draws 11 A. What is the equivalent resistance?

R = E ÷ I = 277 ÷ 11 = 25.2 Ω

Watt's Law (Power in a Circuit)

Watt's Law gives the real power dissipated by a resistive load. It has three forms, each useful depending on which two quantities you know:

Worked Example 3 — Power from Voltage and Current

A 240-volt electric water heater draws 18.3 A. What is the power consumption?

P = E × I = 240 × 18.3 = 4,392 W (≈ 4.4 kW)

Worked Example 4 — Power from Current and Resistance

A heating element carries 12 A through 10 Ω. Power?

P = I² × R = 12² × 10 = 144 × 10 = 1,440 W

Worked Example 5 — Power from Voltage and Resistance

A 208-volt resistive load has 16.6 Ω resistance. Power?

P = E² ÷ R = 208² ÷ 16.6 = 43,264 ÷ 16.6 = 2,604 W (≈ 2.6 kW)

Series Circuits

In a series circuit, all current flows through a single path — every load is connected end-to-end. Three rules govern series circuits:

1. Current is the same at every point: I_T = I1 = I2 = I3
2. Voltage adds across the loads: E_T = E1 + E2 + E3
3. Resistance adds directly: R_T = R1 + R2 + R3

Worked Example 6 — Series Circuit

A 120-volt source supplies three resistors in series: 4 Ω, 6 Ω, and 10 Ω.

• Total resistance: R_T = 4 + 6 + 10 = 20 Ω
• Total current: I = 120 ÷ 20 = 6 A (same through every resistor)
• Voltage across each: E(4Ω) = 6 × 4 = 24 V; E(6Ω) = 6 × 6 = 36 V; E(10Ω) = 6 × 10 = 60 V
• Check: 24 + 36 + 60 = 120 V ✓
• Total power: P = 120 × 6 = 720 W

A series circuit is rare in power wiring (an open on one load kills the whole string), but the arithmetic appears in voltage-drop and tap-rule questions.

Parallel Circuits

In a parallel circuit, each load is connected across the same two supply conductors. This is how every branch circuit and feeder in a building is wired. Three rules govern parallel circuits:

1. Voltage is the same across every branch: E_T = E1 = E2 = E3
2. Current adds at each junction: I_T = I1 + I2 + I3
3. Reciprocal resistance: 1/R_T = 1/R1 + 1/R2 + 1/R3

For two resistors only, the product-over-sum shortcut is faster:

R_T = (R1 × R2) ÷ (R1 + R2)

Worked Example 7 — Two-Resistor Parallel

A 120-volt circuit feeds two parallel loads: 10 Ω and 15 Ω.

• Branch currents: I(10Ω) = 120 ÷ 10 = 12 A; I(15Ω) = 120 ÷ 15 = 8 A
• Total current: I_T = 12 + 8 = 20 A
• Equivalent resistance (shortcut): R_T = (10 × 15) ÷ (10 + 15) = 150 ÷ 25 = 6 Ω
• Check: I = 120 ÷ 6 = 20 A ✓
• Total power: P = 120 × 20 = 2,400 W

Worked Example 8 — Three or More Resistors in Parallel

Three 30 Ω resistors are paralleled across a 240-volt supply.

1/R_T = 1/30 + 1/30 + 1/30 = 3/30 = 1/10 → R_T = 10 Ω

I_T = 240 ÷ 10 = 24 A (8 A per branch). P = 240 × 24 = 5,760 W.

How These Laws Connect to the NEC

The exam will rarely ask a bare theory question. Instead it wraps Ohm's and Watt's Laws inside NEC scenarios:

• Branch-circuit sizing (NEC 210.19, 210.20): Continuous load × 125% uses the current value you computed with I = P ÷ E.
• Motor full-load current (NEC 430.22, Table 430.250): Conductor sized at 125% of FLC; the FLC itself comes from the NEC table, but verifying it against a nameplate uses I = P ÷ (E × 1.732 × PF × η) for three-phase motors.
• Voltage drop (NEC 210.19(A) Informational Note No. 4, 215.2(A) Informational Note No. 2): VD = I × R, scaled by circuit length and the 1.732 or 2 factor — all built on Ohm's Law.
• Service calculations (NEC Article 220): Demand factors reduce the connected load in watts; the resulting current determines conductor and overcurrent sizes via I = P ÷ E.

Key Formula Reference Table