1.1 Electrical Theory, Units & Basic Formulas

1.1 Electrical Theory, Units & Basic Formulas

Quick Reference:

• Ohm's Law: $E = I \times R$ (Electromotive Force = Current $\times$ Resistance)
• Watt's Law: $P = E \times I = I^2 R = \frac{E^2}{R}$ (Power in Watts)
• Series Circuits: $I_T = I_1 = I_2 = \dots$; $R_T = R_1 + R_2 + \dots$; $E_T = E_1 + E_2 + \dots$
• Parallel Circuits: $E_T = E_1 = E_2 = \dots$; $I_T = I_1 + I_2 + \dots$; $\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \dots$
• AC Effective Voltage (RMS): $V_{\text{RMS}} = 0.707 \times V_{\text{peak}}$; $V_{\text{peak}} = 1.414 \times V_{\text{RMS}}$
• Inductive Reactance: $X_L = 2\pi f L$ (Current lags voltage by up to 90°)
• Capacitive Reactance: $X_C = \frac{1}{2\pi f C}$ (Current leads voltage by up to 90°)
• Total Impedance: $Z = \sqrt{R^2 + (X_L - X_C)^2}$
• Three-Phase Real Power: $P = \sqrt{3} \times V_{\text{line}} \times I_{\text{line}} \times \text{PF}$ (where $\sqrt{3} \approx 1.73205$)

Every electrical installation governed by the National Electrical Code (NEC) operates on universal mathematical and physical laws. Sizing conductors, selecting overcurrent protection devices, evaluating circuit breaker interrupting capacity, and diagnosing voltage drop across commercial and industrial feeders all require mastery of electrical fundamentals. On the Connecticut E-2 Unlimited Journeyperson exam, fundamental theory questions test not merely abstract physics, but the practical mathematics required to keep electrical systems safe, functional, and code-compliant.

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1. Fundamental Units & Atomic Electrical Theory

Electricity is the movement of free electrons through conductive materials. In electrical trade work, four primary physical quantities govern every calculation:

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2. Ohm's Law and Watt's Law (The PIER Formulas)

In pure direct current (DC) circuits and purely resistive alternating current (AC) circuits, voltage, current, resistance, and power are inextricably linked through Ohm's Law and Watt's Law. In field examinations, these relationships are collectively visualized as the PIER Wheel ($P$, $I$, $E$, $R$).

                  P (Watts)
               /             \
          E x I               I² x R
         /                         \
     E²/R                           P/I
    /                                   \
E (Volts) -------- PIER WHEEL -------- I (Amperes)
    \                                   /
     I x R                          E/R
         \                         /
          √(P x R)            √(P/R)
               \             /
                  R (Ohms)

The 12 Fundamental Formulas

Practical Worked Example: Voltage Drop on Heating Elements

Problem: An electric water heater element is rated at 240 volts and 4,500 watts. If this heating element is connected to a 208-volt commercial supply network, what is the new wattage dissipated by the element?

Step 1: Calculate the internal resistance of the heater element.
The physical resistance of the element remains constant (ignoring minor temperature coefficients): $R = \frac{E_{\text{rated}}^2}{P_{\text{rated}}} = \frac{240^2}{4500} = \frac{57600}{4500} = 12.8\ \Omega$

Step 2: Calculate the real power dissipated at 208 volts.
$P_{\text{actual}} = \frac{E_{\text{actual}}^2}{R} = \frac{208^2}{12.8} = \frac{43264}{12.8} = 3,379.75\text{ Watts} \approx 3,380\text{ W}$

Exam Insight: Power varies with the square of the voltage ratio: $(208 / 240)^2 = (0.8667)^2 = 0.7511$ (75.1% of rated capacity). Never assume power drops linearly with voltage!

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3. Series, Parallel, and Combination Circuits

Understanding current distribution, voltage drops, and equivalent resistance across different circuit topologies is essential for load calculations and fault diagnosis.

Series Circuits

A series circuit provides only a single continuous path for electron flow.

1. Current is constant: The current through every component is identical:
   $I_T = I_1 = I_2 = I_3 = \dots = I_n$
2. Total resistance is additive: Sum the resistance of each component:
   $R_T = R_1 + R_2 + R_3 + \dots + R_n$
3. Voltage drops divide: Total source voltage equals the sum of component voltage drops (Kirchhoff's Voltage Law):
   $E_T = E_1 + E_2 + E_3 + \dots + E_n$
4. Voltage Divider Rule: The voltage drop across any resistor $R_x$ in series is proportional to its resistance:
   $E_x = E_T \times \left(\frac{R_x}{R_T}\right)$

Parallel Circuits

A parallel circuit provides two or more independent paths (branches) connected to the same voltage potential.

1. Voltage is constant: Every branch experiences the full line voltage:
   $E_T = E_1 = E_2 = E_3 = \dots = E_n$
2. Current is additive: Total circuit current equals the sum of branch currents (Kirchhoff's Current Law):
   $I_T = I_1 + I_2 + I_3 + \dots + I_n$
3. Equivalent resistance decreases: Adding parallel branches always decreases total resistance. Total equivalent resistance is always less than the smallest individual branch resistance.
   • General Reciprocal Formula:
     $\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n}$
   • Two Resistors in Parallel (Product-over-Sum Shortcut):
     $R_T = \frac{R_1 \times R_2}{R_1 + R_2}$
   • $N$ Equal Resistors in Parallel:
     $R_T = \frac{R}{N}$

Circuit Comparison Matrix

Combination (Series-Parallel) Network Analysis

In field control circuits and branch feeds, combination circuits frequently appear. To solve:

1. Identify parallel branch blocks and simplify them into single equivalent resistances using product-over-sum.
2. Re-draw the circuit schematic with the simplified equivalent blocks.
3. Add all remaining series resistances to determine total equivalent circuit resistance ($R_T$).
4. Apply Ohm's Law to find total current ($I_T = E_T / R_T$), then trace branch currents backwards through the network.

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4. Alternating Current (AC) Fundamentals

Unlike direct current, which flows continuously in a single direction, alternating current periodically reverses direction and changes magnitude continuously following a sinusoidal waveform.

   + Peak Voltage (V_peak = 1.414 x V_RMS)
      |   ***
      |  *   *
      | *     *
   ---+*-------*-------*-------*---> Time
      |         *     *
      |          *   *
      |           ***
   - Peak Voltage
      |<-- Period (T = 1/60s) -->|

Key AC Waveform Terms

• Cycle: One complete sequence of positive and negative alternating values (360 electrical degrees).
• Frequency ($f$): Number of complete cycles per second, measured in Hertz (Hz). North American electrical utility distribution operates at a standardized frequency of 60 Hz.
• Period ($T$): The time required to complete one full cycle:
  $T = \frac{1}{f} = \frac{1}{60\text{ Hz}} \approx 0.01667\text{ seconds} = 16.67\text{ ms}$
• Angular Velocity ($\omega$): The rate of electrical rotation: $\omega = 2\pi f = 2 \times 3.14159 \times 60 \approx 377\text{ rad/s}$.

Peak, Peak-to-Peak, and Root-Mean-Square (RMS)

Because an AC wave is constantly changing, instantaneous voltage fluctuates from zero to maximum positive and negative peaks. To establish an equivalent working value compared to DC:

$V_{\text{RMS}} = 0.7071 \times V_{\text{peak}} \quad \text{and} \quad V_{\text{peak}} = 1.4142 \times V_{\text{RMS}}$

$V_{\text{peak-to-peak}} = 2 \times V_{\text{peak}} = 2.828 \times V_{\text{RMS}}$

• Root-Mean-Square (RMS) / Effective Value: The effective value of an alternating current is the value that produces the exact same heating effect in a pure resistor as an equivalent direct current. Standard test instruments (multimeters) read RMS voltage. When an electrician measures 120 volts at a duplex receptacle, 120V is the RMS value; $V_{\text{peak}} = 120 \times 1.4142 \approx 169.7\text{ Volts}$ $V_{\text{p-p}} = 169.7 \times 2 \approx 339.4\text{ Volts}$

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5. Inductance, Capacitance, Reactance & Impedance

In AC circuits containing inductive or capacitive loads, current encounters opposition beyond simple DC ohmic resistance.

Inductive Reactance ($X_L$)

Inductors (coils, motor windings, transformers, ballasts) generate a counter-electromotive force (CEMF) opposing any change in current. This opposition is called inductive reactance ($X_L$), measured in ohms:

$X_L = 2\pi f L$

• $f$ = frequency in Hertz (Hz)
• $L$ = inductance in Henrys (H)
• In a purely inductive circuit, current lags voltage by 90 electrical degrees.

Capacitive Reactance ($X_C$)

Capacitors (power factor correction banks, motor run capacitors) store electrical energy in an electrostatic field. Capacitors oppose changes in voltage. This opposition is capacitive reactance ($X_C$), measured in ohms:

$X_C = \frac{1}{2\pi f C}$

• $f$ = frequency in Hertz (Hz)
• $C$ = capacitance in Farads (F)
• In a purely capacitive circuit, current leads voltage by 90 electrical degrees.

Memory Aid: "ELI the ICE man"

• ELI: Voltage ($E$) leads Current ($I$) in an Inductive circuit ($L$).
• ICE: Current ($I$) leads Voltage ($E$) in a Capacitive circuit ($C$).

Total Impedance ($Z$)

Resistance and reactance operate 90° out of phase with each other. Therefore, they cannot be added arithmetically; they must be combined using vector (Pythagorean) addition:

$Z = \sqrt{R^2 + (X_L - X_C)^2}$

                  ^ +j (Inductive Reactance XL)
                  |
                  |         /| Total Impedance Z = √(R² + X_net²)
                  |        / |
                  |       /  | Net Reactance (XL - XC)
                  |  θ   /   |
  ----------------+-----+----+-----------------> +R (Resistance)
                  |      \
                  |       \
                  v -j (Capacitive Reactance XC)

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6. Single-Phase vs. Three-Phase Power & Power Factor

The Power Triangle

In AC circuits, power consists of three components:

1. Apparent Power ($S$): The total power delivered to the circuit, measured in volt-amperes (VA) or kilovolt-amperes (kVA):
   $S = V \times I$
2. Real (True) Power ($P$): The actual power converted into work or heat, measured in watts (W) or kilowatts (kW):
   $P = V \times I \times \cos(\theta)$
3. Reactive Power ($Q$): The non-working power oscillating between magnetic/electrostatic fields and the source, measured in volt-amperes reactive (VAR) or kVAR:
   $Q = V \times I \times \sin(\theta)$

$S = \sqrt{P^2 + Q^2} \quad \text{and} \quad \text{Power Factor (PF)} = \frac{\text{Real Power (kW)}}{\text{Apparent Power (kVA)}} = \cos(\theta)$

Single-Phase Power Formulas

• Apparent Power: $S = E \times I$
• Real Power: $P = E \times I \times \text{PF}$
• Full-Load Current: $I = \frac{P}{E \times \text{PF}}$

Three-Phase Power Formulas

Three-phase electrical systems deliver continuous, smooth power using three sine waves displaced by 120 electrical degrees. In balanced three-phase calculations, line-to-line voltage ($V_{\text{L-L}}$) and line current ($I_{\text{line}}$) are related by the constant factor $\sqrt{3} \approx 1.73205$:

$S_{\text{3-phase}} = \sqrt{3} \times V_{\text{L-L}} \times I_{\text{line}}$

$P_{\text{3-phase}} = \sqrt{3} \times V_{\text{L-L}} \times I_{\text{line}} \times \text{PF}$

$I_{\text{line}} = \frac{P}{\sqrt{3} \times V_{\text{L-L}} \times \text{PF}} = \frac{S}{\sqrt{3} \times V_{\text{L-L}}}$

Wye vs. Delta System Relationships

Practical Worked Example: Three-Phase Commercial Feeder Sizing

Problem: A 480-volt, three-phase commercial heating and air-conditioning equipment package has an apparent power rating of 75 kVA. Calculate the full-load line current that the feeder conductors must carry.

Step 1: Identify the given values.

• Line Voltage ($V_{\text{L-L}}$) = 480 V
• Apparent Power ($S$) = 75 kVA = 75,000 VA
• Configuration = 3-Phase balanced

Step 2: Apply the 3-phase line current formula. $I_{\text{line}} = \frac{S}{\sqrt{3} \times V_{\text{L-L}}}$ $I_{\text{line}} = \frac{75,000}{1.73205 \times 480} = \frac{75,000}{831.38} = 90.21\text{ Amperes}$

Field Note: Sizing a 3-phase feeder using single-phase math ($75,000 / 480 = 156.25\text{ A}$) results in a catastrophic error, severely oversizing conductors and raceways. Always remember the $\sqrt{3}$ multiplier for three-phase systems!