2.5 DC Circuit Analysis: Ohm's Law, Series and Parallel Networks, and Power

DC Circuit Analysis: Ohm's Law, Series and Parallel Networks, and Power

Quick Answer: NICET Level II domain 2.3 Troubleshooting and Analysis contains exactly one task — 2.3.1, "Calculate basic electrical quantities" — with the skills stated as "calculate voltage, current, and resistance in series and parallel circuits" and "calculate power (watts) in a circuit." The 2026 outlines re-state the same material at Level I (1.1.3 Describe basic electrical equations) and Level II (2.4.1 Apply fundamental electrical laws to simple circuits). This is guaranteed exam content and it is worth points that require no field experience at all.

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1. Ohm's law and the three rearrangements

$V = I \times R \qquad I = \frac{V}{R} \qquad R = \frac{V}{I}$

Where V is in volts, I in amperes, and R in ohms. The single most common arithmetic error on level exams is a unit-prefix slip — mixing milliamperes with amperes, or megohms with ohms. Convert everything to base units before you compute.

Worked example. A DLRO reports a breaker contact resistance of 45 µΩ. The breaker carries 1,200 A continuous. What is the voltage drop across the contacts?

$V = I \times R = 1{,}200 \times 45 \times 10^{-6} = 0.054\ \text{V} = 54\ \text{mV}$

2. Kirchhoff's laws

• Kirchhoff's Current Law (KCL): the algebraic sum of currents entering and leaving any node is zero. In a three-phase four-wire system the neutral carries the phasor sum of the three phase currents — which is why a balanced linear load produces near-zero neutral current, and why triplen harmonics from non-linear loads produce a neutral current that can exceed a phase current.
• Kirchhoff's Voltage Law (KVL): the algebraic sum of voltage rises and drops around any closed loop is zero. This is the law behind every "the readings do not add up" troubleshooting question.

3. Series circuits

Elements in series share one current path.

Voltage divider:

$V_x = V_S \times \frac{R_x}{R_T}$

Worked example. A 125 V DC station battery feeds a trip circuit consisting of a 20 Ω trip coil in series with 5 Ω of cable resistance.

• R_T = 20 + 5 = 25 Ω
• I = 125 / 25 = 5.0 A
• Voltage across the coil: V = 5.0 × 20 = 100 V

The coil only sees 100 V of the 125 V available. If cable resistance rises to 15 Ω through a corroded connection, current falls to 3.57 A and coil voltage falls to 71.4 V — potentially below the coil's minimum pickup, producing an intermittent failure to trip. This is the classic control-circuit troubleshooting scenario.

4. Parallel circuits

Elements in parallel share one pair of nodes.

For exactly two resistors the product-over-sum shortcut applies:

$R_T = \frac{R_1 \times R_2}{R_1 + R_2}$

For n equal resistors, R_T = R / n.

The rule that catches errors: total parallel resistance is always smaller than the smallest individual branch. If your arithmetic gives a larger number, you inverted something.

Worked example. Three parallel grounding electrodes measure 30 Ω, 25 Ω, and 20 Ω.

$\frac{1}{R_T} = \frac{1}{30} + \frac{1}{25} + \frac{1}{20} = 0.03333 + 0.04 + 0.05 = 0.12333$ $R_T = 8.11\ \Omega$

Note this is a theoretical figure. Real electrodes suffer mutual resistance interference when spaced closer than roughly twice their driven depth, so the measured combined resistance is always higher than the parallel calculation predicts. Recognizing that gap is a Level III analysis skill.

5. Series-parallel reduction

Reduce from the inside out.

Worked example. A 48 V DC control supply feeds a 4 Ω series resistance, then splits into two parallel branches of 12 Ω and 6 Ω.

1. Parallel section: (12 × 6)/(12 + 6) = 72/18 = 4 Ω
2. Total: 4 + 4 = 8 Ω
3. Total current: 48 / 8 = 6 A
4. Drop across the series resistor: 6 × 4 = 24 V
5. Voltage across the parallel section: 48 - 24 = 24 V
6. Branch currents: 24/12 = 2 A and 24/6 = 4 A — which sum to 6 A, confirming the result.

Always run that final check. If branch currents do not sum to the total, the reduction is wrong.

6. Power

$P = V \times I = I^2 R = \frac{V^2}{R}$

The I²R form is the one that matters in field testing, because it shows heating rises with the square of current.

Worked example. A bolted busway joint should measure 20 µΩ but measures 200 µΩ. The busway carries 2,000 A.

• Good joint: P = I² R = 2,000² × 20 × 10⁻⁶ = 4,000,000 × 0.00002 = 80 W
• Degraded joint: P = 4,000,000 × 0.0002 = 800 W

Eight hundred watts concentrated in a joint the size of a fist is a thermal runaway in progress: heat raises resistance, higher resistance produces more heat, and the joint eventually fails. That feedback loop is why contact resistance is trended rather than merely compared to a pass/fail limit, and why an infrared survey finds these joints before a DLRO campaign does.

7. Resistance, temperature, and conductor material

Metallic conductor resistance rises with temperature:

$R_2 = R_1 \times \frac{T + t_2}{T + t_1}$

where T is 234.5 for copper and 228 for aluminum, and t₁ and t₂ are the initial and final temperatures in °C.

Note the direction and contrast it with insulation: conductor resistance increases with temperature, insulation resistance decreases with temperature. Candidates routinely apply the insulation correction factor in the conductor direction. The two behave oppositely because conduction in metal is impeded by lattice vibration while conduction in a dielectric is aided by thermal activation of charge carriers.

Exam trap: A question gives three parallel branches and offers an answer equal to the sum of the branch resistances. Parallel resistance can never exceed the smallest branch — that answer is checkable on sight without doing the arithmetic.