1.1 Ohm's Law & DC Circuit Calculations

Every electrical calculation an examinee performs on the Technical Subjects portion of the board exam — from sizing a branch-circuit conductor to troubleshooting a tripped breaker — traces back to one relationship. Ohm's Law, named after German physicist Georg Simon Ohm, states that the current flowing through a conductor between two points is directly proportional to the voltage across those two points and inversely proportional to the resistance between them.

The Three Forms of Ohm's Law

Direct current (DC) is current that flows in one direction only, unlike alternating current (AC), which reverses direction periodically. In a DC circuit, three quantities are always related:

• Voltage (V), measured in volts, is the electrical "pressure" that pushes current through a circuit.
• Current (I), measured in amperes (A), is the rate of flow of electric charge.
• Resistance (R), measured in ohms (Ω), is the opposition to that flow.

Ohm's Law expresses this relationship as:

V = I × R

Because any one of the three variables can be the unknown, the formula rearranges into three usable forms:

Exam tip: Many reviewees memorize the "Ohm's Law wheel" — a circle divided into V, I, and R. Cover the unknown quantity and the remaining two show whether to multiply or divide.

Power Formulas

Power (P), measured in watts (W), is the rate at which electrical energy is converted into another form (heat, light, mechanical motion). Because power is the product of voltage and current, and because V and I are themselves related through resistance, power has three equivalent formulas:

These three power formulas, combined with the three Ohm's Law forms, form a complete formula "wheel" — every combination of two known quantities lets you solve for either of the other two (V, I, R, or P).

Worked Example 1 — Solving for Resistance

A DC-powered control relay is rated to draw 2.5 A when connected to a 24 V DC source. What is the resistance of the relay coil?

Using R = V / I:

R = 24 V / 2.5 A = 9.6 Ω

Worked Example 2 — Solving for Current from Power

A 120 V DC heating element is rated at 1,500 W. What current does it draw?

Using I = P / V (derived from P = V × I):

I = 1,500 W / 120 V = 12.5 A

Once current is known, resistance can also be found: R = V / I = 120 / 12.5 = 9.6 Ω. Cross-checking a result with a second formula — computing R two different ways — is a fast way to catch arithmetic mistakes on the exam.

Series Circuits

In a series circuit, components are connected end-to-end so there is only one path for current. Three rules govern series circuits:

1. Current is the same at every point in the circuit: I_T = I_1 = I_2 = I_3
2. Voltage drops add up to equal the source voltage: V_T = V_1 + V_2 + V_3
3. Resistances add directly: R_T = R_1 + R_2 + R_3

Worked Example 3 — Series Resistors

Two resistors, R1 = 20 Ω and R2 = 30 Ω, are connected in series across a 50 V DC source.

• Total resistance: R_T = 20 + 30 = 50 Ω
• Total (and branch) current: I = V / R_T = 50 / 50 = 1 A
• Voltage drop across R1: V1 = I × R1 = 1 × 20 = 20 V
• Voltage drop across R2: V2 = I × R2 = 1 × 30 = 30 V

Check: 20 V + 30 V = 50 V, matching the source — confirming Kirchhoff's Voltage Law (covered fully in Section 1.2).

Parallel Circuits

In a parallel circuit, components are connected across the same two nodes, so current has multiple paths available. The rules invert compared to series:

1. Voltage is the same across every branch: V_T = V_1 = V_2 = V_3
2. Currents add up to equal the total current: I_T = I_1 + I_2 + I_3
3. Resistances combine reciprocally: 1/R_T = 1/R_1 + 1/R_2 + 1/R_3

For exactly two resistors, this simplifies to the convenient "product over sum" shortcut: R_T = (R1 × R2) / (R1 + R2).

Worked Example 4 — Parallel Resistors

R1 = 20 Ω and R2 = 30 Ω are connected in parallel across the same 50 V source.

• Total resistance: R_T = (20 × 30) / (20 + 30) = 600 / 50 = 12 Ω
• Branch current through R1: I1 = V / R1 = 50 / 20 = 2.5 A
• Branch current through R2: I2 = V / R2 = 50 / 30 = 1.67 A
• Total current: I_T = 2.5 + 1.67 = 4.17 A

Notice that total resistance in parallel (12 Ω) is always less than the smallest individual resistor (20 Ω) — a useful sanity check on the exam.

Combined Series-Parallel Circuits

Real branch-circuit wiring is rarely purely series or purely parallel. A combined (series-parallel) circuit is solved by reducing it in stages: collapse parallel branches into a single equivalent resistance first, then treat that equivalent as one series element.

Worked Example 5 — Combined Circuit

A 100 V DC source feeds R1 = 10 Ω in series with a parallel combination of R2 = 20 Ω and R3 = 20 Ω.

1. Reduce the parallel pair: R_23 = (20 × 20)/(20+20) = 400/40 = 10 Ω
2. Add the series resistor: R_T = R1 + R_23 = 10 + 10 = 20 Ω
3. Total current from the source: I_T = V / R_T = 100 / 20 = 5 A
4. Voltage across the parallel section: V_23 = I_T × R_23 = 5 × 10 = 50 V
5. Each parallel branch carries: I2 = I3 = 50 V / 20 Ω = 2.5 A (equal because R2 = R3)

Formula Summary Table

Key Takeaways

• Ohm's Law (V = IR) and its two rearrangements (I = V/R, R = V/I) are the base of nearly every numeric problem in this exam group.
• Power has three equivalent formulas — pick the one that matches the two quantities you already know.
• Series circuits share one current path; parallel circuits share one voltage across all branches.
• Reduce combined circuits by collapsing parallel sections into an equivalent resistance before treating the circuit as series.