1.2 Electrical Fundamentals, Ohm's Law, Joule's Law & Power Formulas

1.2 Electrical Fundamentals, Ohm's Law, Joule's Law & Power Formulas

Core Electrical Quantities and Physical Definitions

Every electrical system operates on fundamental physical principles governing charge, force, and energy. Four primary quantities form the foundation of electrical trade analysis:

• Electromotive Force ($E$ or $V$): Measured in volts ($\text{V}$), electromotive force is the electrical potential difference driving electrons through a conductor ($1\text{ V} = 1\text{ J/C}$).
• Electric Current ($I$): Measured in amperes ($\text{A}$), current represents the rate of charge displacement. One ampere equals one coulomb flowing per second ($1\text{ A} = 1\text{ C/s} = 6.242 \times 10^{18}\text{ electrons/second}$).
• Electrical Resistance ($R$): Measured in ohms ($\Omega$), resistance is the opposition offered by a material to electron flow, converting electrical energy into heat.
• Electrical Power ($P$): Measured in watts ($\text{W}$), power represents the rate of performing work ($1\text{ W} = 1\text{ J/s} = 1\text{ V} \times 1\text{ A}$).

Ohm's Law and Algebraic Derivations

Formulated by Georg Simon Ohm, Ohm's Law establishes that current through a linear conductor is directly proportional to applied voltage and inversely proportional to circuit resistance:

$E = I \times R$ $I = \frac{E}{R}$ $R = \frac{E}{I}$

Joule's Law and Power Equations

James Prescott Joule demonstrated that heat generation is proportional to resistance and the square of the current. Combining Ohm's Law with Joule's findings yields the primary power expressions:

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

Solving for voltage and current from power and resistance:

$E = \sqrt{P \times R}$ $I = \sqrt{\frac{P}{R}}$

Alternating Current Fundamentals: Peak and RMS Values

Direct current maintains unidirectional flow, whereas alternating current periodically reverses direction following a sinusoidal waveform. In North American systems, generation operates at 60 Hertz ($60\text{ Hz}$), yielding a cycle duration of $16.67\text{ milliseconds}$.

• Peak Voltage ($V_{\text{peak}}$): Maximum instantaneous amplitude reached during the sinusoidal crest.
• Peak-to-Peak Voltage ($V_{\text{p-p}}$): Total voltage span between positive and negative alternating peaks ($V_{\text{p-p}} = 2 \times V_{\text{peak}}$).
• Root-Mean-Square Voltage ($V_{\text{RMS}}$): Because AC voltage varies continuously, peak values do not represent working thermal capacity. The RMS value provides the equivalent heating value of a DC system across an identical resistive load: $V_{\text{RMS}} = \frac{V_{\text{peak}}}{\sqrt{2}} = 0.7071 \times V_{\text{peak}}$ $V_{\text{peak}} = \sqrt{2} \times V_{\text{RMS}} = 1.4142 \times V_{\text{RMS}}$

A nominal 120-volt AC circuit represents an RMS value whose waveform reaches peaks of $120 \times 1.4142 = 169.7\text{ volts}$.

The AC Power Triangle: True, Reactive, and Apparent Power

In resistive AC circuits, current and voltage remain in phase ($\theta = 0^\circ$). In inductive circuits (motors, transformers), magnetic fields cause current to lag voltage. In capacitive circuits, current leads voltage. This phase angle divides AC power into three vector components:

• True Power ($P$): Expressed in watts ($\text{W}$) or kilowatts ($\text{kW}$), true power represents work converted to heat or mechanical output ($P = E \times I \times \cos\theta$).
• Reactive Power ($Q$): Expressed in volt-amperes reactive ($\text{VAR}$) or kilovars ($\text{kVAR}$), reactive power sustains electromagnetic fields without performing mechanical work.
• Apparent Power ($S$): Expressed in volt-amperes ($\text{VA}$) or kilovolt-amperes ($\text{kVA}$), apparent power is the vector sum of true and reactive power ($S = \sqrt{P^2 + Q^2} = E \times I$).
• Power Factor ($\text{PF}$): Ratio of true power to apparent power ($\text{PF} = \frac{P}{S} = \frac{\text{kW}}{\text{kVA}} = \cos\theta$). Low power factor requires larger conductors to deliver equivalent working wattage.

Three-Phase Calculations: Wye and Delta Systems

Three-phase systems deliver three sinusoidal voltages displaced by 120 electrical degrees, introducing the factor $\sqrt{3} \approx 1.732$:

• Wye ($Y$) Systems: $V_{\text{line}} = \sqrt{3} \times V_{\text{phase}} = 1.732 \times V_{\text{phase}}$ $I_{\text{line}} = I_{\text{phase}}$ Typical configurations include $120/208\text{V}$ and $277/480\text{V}$.
• Delta ($\Delta$) Systems: $V_{\text{line}} = V_{\text{phase}}$ $I_{\text{line}} = \sqrt{3} \times I_{\text{phase}} = 1.732 \times I_{\text{phase}}$
• Three-Phase Power Equations:
  • Apparent Power: $S = \sqrt{3} \times V_{\text{line}} \times I_{\text{line}}$
  • True Power: $P = \sqrt{3} \times V_{\text{line}} \times I_{\text{line}} \times \text{PF}$
  • Line Current: $I_{\text{line}} = \frac{P}{\sqrt{3} \times V_{\text{line}} \times \text{PF}}$

Motor Horsepower, Efficiency, and Calculations

Electric motors convert electrical power to mechanical shaft output rated in horsepower:

• $1\text{ Horsepower (HP)} = 746\text{ Watts}$
• Efficiency ($\eta$): Ratio of mechanical output power to electrical input power ($\eta = \frac{P_{\text{out}}}{P_{\text{in}}}$).
• Input Power: $P_{\text{in}} = \frac{\text{HP} \times 746}{\eta}$.

Worked Example: Three-Phase Motor Line Current

Calculate full-load line current for a 15-HP, 460-volt, three-phase motor operating at 89% efficiency with an 0.84 lagging power factor.

1. Determine input power in watts: $P_{\text{in}} = \frac{15\text{ HP} \times 746\text{ W/HP}}{0.89} = \frac{11,190}{0.89} = 12,573.03\text{ Watts}$
2. Determine input apparent power: $S = \frac{12,573.03\text{ W}}{0.84} = 14,967.89\text{ VA}$
3. Calculate full-load line current: $I_{\text{line}} = \frac{14,967.89\text{ VA}}{1.73205 \times 460\text{ V}} = \frac{14,967.89}{796.74} = 18.79\text{ Amperes}$