Use the formulas: $$V = I \times R,\quad I = \frac{V}{R},\quad R = \frac{V}{I}$$
For DC: $$P = V \times I$$
For AC (real power): $$P = V \times I \times \cos\phi$$
$$E = P \times t$$ where P is in kW and t in hours.
$$\Delta V = I \times \left(\rho \times \frac{L}{A}\right)$$
$$I_{fault} = \frac{V}{Z_{total}}$$
Determine breaker rating based on load current and inrush multiplier.
Estimate required cable capacity with derating factors.
$$kVA = \frac{kW}{PF}$$ and Efficiency: $$\eta = \frac{Output}{Input} \times 100\%$$.
$$I_{FL} = \frac{P}{\sqrt{3} \, V \, \cos\phi \, \eta}$$ and Locked Rotor Current = Multiplier × I₍FL₎.
$$Q_c = kW \left(\tan(\arccos(PF_{initial})) - \tan(\arccos(PF_{target}))\right)$$
$$\text{Required Lumens} = \text{Area} \times \text{Desired Lux}$$
$$THD = \frac{\sqrt{V_2^2 + V_3^2 + \ldots}}{V_1}$$
For a single rod: $$R_e \approx \frac{\rho}{2\pi L} \ln\left(\frac{4L}{d}\right)$$
$$\Delta T = \frac{P}{h \times A}$$ where P is power loss.
$$\text{Capacity (Ah)} = \frac{\text{Load Current (A)} \times \text{Time (h)}}{\text{DoD}}$$
For a full-wave rectifier: $$V_{DC} \approx 0.9 \times V_{AC} - \text{Diode Drop}$$
$$\text{Fill Ratio} = \frac{\text{Total Cable Area}}{\text{Conduit Internal Area}}$$
Estimate current capacity using a rough guideline: capacity ≈ (Cross-sectional Area in mm²) × 1.6 (A/mm²).
$$A = \frac{MTBF}{MTBF + MTTR}$$
Life-Cycle Cost and Payback Period Analysis.