Basic Electrical Engineering Formulas with Examples

Electrical engineering relies on fundamental formulas to analyze circuits, size components, and predict system behavior. Whether you’re studying dc circuits or designing three phase power systems, having the right equations at your fingertips saves time and prevents costly errors.

This guide compiles the most important basic electrical engineering formulas with clear parameter definitions and solved examples you can apply immediately.

Electric Field, Coulomb’s Law and Gauss’s Law

The electric field concept is critical in electrical engineering for insulation design, transmission line calculations, and understanding force on charged particles.

Coulomb’s Law: It gives the force between two point charges. It is measured in Newton (N).

$\boxed{F = \frac{1}{4\pi \varepsilon} \· \frac{Q_1 Q_2}{r^2}}$

For a material, the permittivity is given by,

$\boxed{\varepsilon = \varepsilon_0 \varepsilon_r}$

where ε_r is the relative permittivity (dielectric constant) of the material and ε₀ is the permittivity of the free space or air and is given by,

$\boxed{\varepsilon_0 = 8.854 \times 10^{-12} F/m}$

Electric field intensity: Electric field intensity at a point is defined as the force experienced by a unit positive test charge placed at that point in an electric field.

$\boxed{E = \frac{F}{Q_0} }$

For a point charge Q, the electric field intensity at a distance r is given by,

$\boxed{E = \frac{1}{4\pi \varepsilon} \· \frac{Q}{r^2}}$

In free space or air, approximately,

$\boxed{E = \frac{9 \times 10^9 \, Q}{r^2}}$

Electric flux: It is obtained from the surface integral of the electric field over a closed surface.

$\boxed{\Phi_e = \iint_S E.dA}$

Gauss’s Law: The total electric flux through any closed surface is equal to the net electric charge enclosed by the surface divided by the permittivity (ε) of the medium.

$\boxed{\Phi_e = \frac{Q_{\text{enclosed}}}{\varepsilon}}$

Example:

Find the electric field strength at a distance of 0.1 m from a point charge of 1 μC in air. (ε_₀ = 8.854 × 10⁻¹² F/m):

The electric field due to a point charge is given by

$E = \frac{9 \times 10^9 \, Q}{r^2}$

$E=\frac{(9\times10^9)(1\times10^{-6})}{(0.1)^2}$

$E=9\times10^5\ \text{N/C} = 900\ \text{kN/C}}$

The field is directed away from the charge because the charge is positive.

Basic Electrical Quantities and Ohm’s Law

Electric charge, electric current, potential difference, and electrical resistance are the building blocks of basic electrical engineering. Every circuit analysis problem starts with these quantities.

Charge (Q): Electric charge is measured in coulombs (C).

$\boxed{Q = ne}$

where an electron (e) carries a charge of −1.6 × 10⁻¹⁹ C. Electric charge is quantized as integral multiples of elementary charge, and electric charge is carried by electrons and protons.

Current (I): It is the rate of flow of electric charge, given by the below given formula. current is measured in amperes (A).

$\boxed {I = \frac{Q}{t} }$

Voltage (V): It is the work done in moving a unit charge, measured in volts (V).

$\boxed{V = \frac{\text{Work done}}{\text{Charge}} =\frac{W}{Q} }$

Voltage is directly proportional to current in a conductor (for ohmic material).

Resistance (R): It is the opposition for the flow of electric current through a conductor. Resistance is denoted by the symbol R and is measured in Ohms (Ω). Resistance is calculated using,

$\boxed {R = \frac{\rho \ell }{A}}$

where ρ is the resistivity is a material property opposing current flow.

Conductance: Conductance is the reciprocal of resistance, and is measured in siemens. It is given by the formula

$\boxed{G = \frac{1}{R}}$

Ohm’s Law gives the relation between the voltage, current and resistance, given by the formula,

$\boxed{V = IR}$

It applies to ohmic conductors under constant temperature. The formula can be rearranged to find the value of current and resistance, as shown below

$\boxed{\text{Current, I} = \frac{V}{R}}$

$\boxed{\text{Resistance, R} = \frac{V}{I}}$

Electric Power:

Power is the rate at which energy is consumed or supplied. It is given by the physical formula,

$\boxed{P=\frac{dW}{dt}}$

For DC Supply, the electrical power is given by the below formula. It is measured in watts (W)

$\boxed{P = VI}$

By using ohm’s law, the power can be modified as below,

$\boxed{ P = I^2R}$

$\boxed{ P = \frac{V^2}{R} }$

Example

Consider a copper wire which has the resistivity (ρ) of 1.68 × 10⁻⁸ Ω·m. It is 50 m long with cross-section 2.5 mm². Assume the applied voltage as 5 V. Calculate its resistance and current flowing through it.

Using the resistance formula,

$R=\frac{\rho\ell}{A}$

$R=\frac{(1.68\times10^{-8})(50)}{2.5\times10^{-6}}$

$R=0.336 \, \Omega$

Using Ohm’s law,

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

$I=\frac{5}{0.336} = 14.88\ \text{A}$

DC Circuits formulas

Understanding series and parallel connections plus Kirchhoff’s laws is essential for analyzing any electrical circuit in dc circuits.

In a series circuit, electrical elements are connected one after another, so the same current flows through all elements.

$\boxed{R_{\text{total}}=R_1+R_2+\cdots+R_n}$

The voltage is divided among the resistors according to their resistance. The voltage across resistor (R₁) is:

$\boxed{V_1=V_{\text{total}} \, \frac{R_1}{R_{\text{total}}}}$

In a parallel circuit, electrical elements are connected across the same two nodes. Therefore, the voltage is the same across each branch, while the current divides among the branches.

$\boxed{\frac{1}{R_{\text{total}}}= \frac{1}{R_1}+\frac{1}{R_2}+\cdots+\frac{1}{R_n}}$

For two resistors in parallel:

$\boxed{R_{\text{total}}=\frac{R_1R_2}{R_1+R_2}}$

Kirchhoff’s Current Law states that the total current entering a junction is equal to the total current leaving the junction.

$\boxed{\sum I_{\text{in}}=\sum I_{\text{out}}}$

KCL is based on the principle of conservation of electric charge.

Kirchhoff’s Voltage Law states that the algebraic sum of all voltages around any closed loop is zero.

$\boxed{\sum V=0}$

KVL is based on the principle of conservation of energy.

Example

Consider a 12 V battery connected to two resistors, R1 = 4 Ω and R2 = 6 Ω, in series. This series combination is connected in parallel with R3 = 10 Ω. Find the total resistance, total current and current in each parallel branch.

Since 4 Ω and 6 Ω are connected in series, the total series resistance is given by,

$R_s=R_1+R_2=4+6=10 \,\Omega$

This series combination is in parallel with R3 = 10 Ω, the total resistance is obtained as,

$R_{\text{total}}= \frac{R_3\times R_s}{R_3+R_s}$

$R_{\text{total}}= \frac{10\times10}{10+10} =5 \,\Omega$

The total current is calculated as,

$I_{\text{total}}= \frac{V}{R_{\text{total}}} =\frac{12}{5} = 2.4 \,A$

Since both parallel branches have the same resistance of 10 Ω,

$I_1=I_2=\frac{V}{R}=\frac{12}{10}=1.2 \, A$

AC Fundamental formulas

Alternating current (AC) is characterized by its frequency (f), measured in hertz (Hz). Frequency represents the number of complete cycles occurring per second. The relationship between frequency and time period is,

$\boxed{f=\frac{1}{T}}$

where (T) is the time period, or the time required to complete one cycle.

The angular frequency is given by,

$\boxed{\omega=2\pi f}$

where ω is measured in radians per second (rad/s).

Inductive reactance (X_L) is the opposition offered by an inductor to the flow of alternating current. It is given by the formula,

$\boxed{X_L=2\pi fL=\omega L}$

The inductive reactance increases with frequency. Inductance (L) is measured in henrys (H).

Capacitive reactance (X_C) is the opposition offered by a capacitor to the flow of alternating current. It is given by

$\boxed{X_C=\frac{1}{2\pi fC}=\frac{1}{\omega C}}$

Unlike inductive reactance, capacitive reactance decreases as frequency increases. Capacitance (C) is measured in farads (F).

Impedance (Z) is the total opposition offered by an AC circuit to the flow of current. It combines the effects of resistance and reactance and is measured in ohms (Ω).

For a series RLC circuit, the magnitude of impedance is,

$\boxed{Z=\sqrt{R^2+X^2}}$

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

For a purely resistive AC circuit, the reactance is zero, so Z = R.

In a series RLC circuit, resonance occurs when the inductive reactance equals the capacitive reactance,

$X_L=X_C$

The resonant frequency is,

$\boxed{f_0=\frac{1}{2\pi\sqrt{LC}}}$

At resonance, the net reactance is zero and the circuit has only resistance (Z = R). So the circuit impedance becomes minimum. Therefore, for a given supply voltage, the current reaches its maximum value.

AC Ohm’s Law : Ohm’s law for AC circuits is expressed using impedance,

$\boxed{V=IZ}$

In phasor form:

$\boxed{\overline{V}=\overline{I}Z}$

Example:

A 230 V, 50 Hz single-phase AC supply feeds a load that draws 10 A at a power factor of 0.8 lagging. Determine the active power, reactive power, and apparent power.

Given:

$V=230\text{ V},\qquad I=10\text{ A},\qquad \cos\phi=0.8$

$\text{Active Power, P}=VI\cos\phi$

$P=230\times10\times0.8 =1840\text{ W}$

$\text{Reactive Power, Q}=VI\sin\phi$

$Q=230\times10\times0.6 =1380\text{ VAR}$

Because the power factor is lagging, the reactive power is inductive.

$\text{Apparent Power, S}=VI =230\times10 =2300\text{ VA}$

Therefore, the load consumes 1.84 kW of active power, 1.38 kVAR of reactive power, and has an apparent power of 2.3 kVA.

Single-Phase and Three-Phase AC Power Formulas

Single-phase AC systems are commonly used for residential applications and small electrical loads, whereas three-phase AC systems are widely used for power generation, transmission, distribution, and industrial loads such as large motors. Three-phase systems provide more efficient and balanced power delivery for high-power applications.

Single-phase AC power

In a single-phase AC circuit, electrical power is classified into active power, reactive power, and apparent power.

Active power is the actual power consumed or converted into useful work by the load. It is measured in watts (W).

$\boxed{P=VI\cos\phi}$

Reactive power is the power that alternately flows between the source and reactive elements such as inductors and capacitors. It is measured in volt-amperes reactive (VAR).

$\boxed{Q=VI\sin\phi}$

Apparent power is the product of the RMS voltage and RMS current. It is measured in volt-amperes (VA).

$\boxed{S=VI}$

The relationship among active, reactive, and apparent power is:

$\boxed{S^2=P^2+Q^2}$

Power factor (pf) is the ratio of active power to apparent power.

$\boxed{\text{pf}=\cos\phi=\frac{P}{S}}$

For an inductive load, the power factor is lagging, while for a capacitive load, it is leading.

These AC circuit formulas are fundamental for analyzing single-phase and three-phase AC circuits, including circuits containing resistors, inductors, and capacitors.

These impedance formulas and concepts apply across all phase ac circuits.

Three Phase System

In a balanced three-phase system with a star (Y) connection, the relationship between line and phase quantities is:

$\boxed{V_L=\sqrt{3} \,V_{Ph}}$

$\boxed{I_L=I_{Ph}}$

where V_L and I_L are the line voltage and line current, while V_Ph and I_Ph are the phase voltage and phase current.

For a balanced three-phase system with a delta (Δ) connection:

$\boxed{V_L=V_{Ph}}$

$\boxed{I_L=\sqrt{3} \,I_{Ph}}$

Three-Phase AC Power

The total active power in a balanced three-phase AC circuit is,

$\boxed{P=\sqrt{3} \,V_LI_L\cos\phi}$

Alternatively, using phase quantities:

$\boxed{P=3V_{Ph}I_{Ph}\cos\phi}$

Similarly, the total reactive and apparent powers are:

$\boxed{Q=\sqrt{3} \, V_LI_L\sin\phi}$

$\boxed{S=\sqrt{3}V_LI_L}$

These formulas are applicable to balanced three-phase systems, regardless of whether the load is connected in star or delta, provided the corresponding line and phase relationships are used correctly.

AC Electrical Energy : Electrical energy consumed by an AC load is calculated from,

$\boxed{E=Pt}$

If power is expressed in kilowatts and time in hours, the resulting energy is in kilowatt-hours:

$\boxed{E(\text{kWh})=P(\text{kW})\times t(\text{h})}$

The kilowatt-hour (kWh) is the commonly used unit for measuring electrical energy consumption.

Example:

A three-phase motor is supplied from a 400 V line-to-line supply and draws 20 A at a power factor of 0.85. Determine the active power consumed by the motor.

Given:

$V_L=400\text{ V}\,\qquad I_L=20\text{ A}\,\qquad \cos\phi=0.85$

Using the three-phase power formula:

$P=\sqrt{3}V_LI_L\cos\phi$

$P=\sqrt{3}\times400\times20\times0.85$

$P = 11,778\text{ W} = 11.78\text{ kW}$

Energy Storage Elements

A capacitor stores electric charge in an electrostatic field, while inductors store energy in a magnetic field. Both are critical electrical components in every electrical system.

Capacitor formulas:

Capacitance is defined as

$\boxed{C = \frac{Q}{V}}$

The unit of capacitance is microfarad (μF). Charge Q is calculated as

$\boxed{Q = C \times V}$

For a parallel-plate capacitor, capacitance can be expressed as

$\boxed{C = \frac{\varepsilon A}{d}}$

Energy stored in a capacitor is calculated from the formula,

$\boxed{W_C = \frac{1}{2}CV^2}$

Inductor formulas:

The formula for inductance is

$\boxed{L = \frac{N\phi}{I}}$

Inductance is the property that stores energy in a magnetic field. Inductance relates magnetic flux linkage to current.

Induced EMF is given by the formula,

$\boxed{ e_L = −L \frac{di}{dt}}$

Energy stored in the inductor is given by,

$\boxed{ W_L = \frac{1}{2} LI^2 }$

Magnetic circuit formulas:

Reluctance is expressed as,

$\boxed{ \mathcal{R} = \frac{\ell}{\mu A} }$

Magnetic flux is calculated from the formula,

$\boxed{\phi = \frac{NI}{\mathcal{R}} }$

Magnetic field intensity is determined from the formula,

$\boxed{ H = \frac{NI}{\ell} }$

Flux density is obtained from the formula,

$\boxed{ B = \frac{\phi}{A} = \mu H }$

Example

Calculate the energy in a 100 μF capacitor at 50 V and calculate the energy in a 10 mH inductor at 5 A.

Energy stored in a capacitor is given by,

$W_C = \frac{1}{2}CV^2$

$W_C = \frac{1}{2}\times 100 \times 50^2 = 0.125 \, J$

Energy stored in a inductor is given by,

$W_L = \frac{1}{2} LI^2$

$W_L = \frac{1}{2} \times 10 \times 10^{-3} \times 5^2 = 0.125 \, J$

Electrical Machine Formulas

Faraday’s law of electromagnetic induction is the fundamental principle behind the operation of many electrical machines, including generators, transformers, and induction motors.

Faraday’s Law: It states that an electromotive force (EMF) is induced in a coil (N turns) whenever the magnetic flux (ϕ) linking the coil changes with time (t).

$\boxed{e = -N \frac{d\phi}{dt} }$

Transformer essential formulas:

EMF equation of a transformer is given by the formula,

$\boxed {E = 4.44 f \phi_m N }$

Transformation ratio relates the primary and secondary voltages and currents by the formula,

$\boxed{ k = \frac{N_2}{N_1} = \frac{V_2}{V_1} = \frac{I_1}{I_2} }$

Electrical efficiency can be calculated as the ratio of output power to input power.

$\boxed{ \eta = \frac{P_{\text{out}}}{P_{\text{in}}}}$

Voltage regulation of a transformer is calculated from the formula,

$\boxed {\text{Voltage Regulation} = \frac{V_{\text{no-load}} - V_{\text{full-load}}}{V_{\text{full-load}}} \times 100 \% }$

DC machine formulas:

EMF generated by the DC generator is given by,

$\boxed{E = \frac{P\phi ZN}{60A} }$

where P = poles, ϕ = flux/pole, Z = armature conductors, N = speed (rpm), A = parallel paths

Torque of a DC Motor is given by the formula,

$\boxed{T = \frac{P\phi ZI_a}{2\pi A}}$

It can be simplified as

$\boxed{ T \propto \phi I_a }$

Speed-voltage relation is defined by the formula,

$\boxed{ N \propto \frac{(V - I_a R_a)}{\phi} }$

Hysteresis Loss is the loss that occurs due to continuous magnetic reversal. It is expressed as,

$\boxed{P_{hy} = \eta \,B_{\text{max}}^{1.6}\, f\, v }$

Eddy current loss is the circulating current in the electric machine, expressed as,

$\boxed{P_e =K_e B_{\text{max}}^2 \,f^2\, v\,t^2 }$

Induction motor formulas:

Synchronous speed is determined from the formula,

$\boxed{N_s = \frac{120f}{P} }$

Slip of a induction motor is obtained from,

$\boxed{s = \frac{N_s - N}{N_s} }$

Mechanical power of the induction motor is given by,

$\boxed{ P_m = \frac{2\phi NT}{60} }$

Example

A 4-pole, 50 Hz induction motor operates at a rotor speed of 1440 rpm. Determine the synchronous speed and slip.

The synchronous speed is given by the formula,

$N_s=\frac{120f}{P}$

$N_s =\frac{120\times 50}{4} = 1500 \text{ rpm}$

The slip is given by,

$s=\frac{N_s-N}{N_s} =\frac{1500-1440}{1500} =0.04$

$s=4 \%$

Power factor correction formulas

Required reactive power:

$\boxed{Q_c = P(tan \phi_1 - tan\phi_2)}$

For a single-phase capacitor connected across an AC supply, the capacitor reactive power is determined from the formula,

$\boxed{ Q_c = V^2 \omega C }$

The required capacitance is calculated from,

$\boxed{ C = \frac{Q_c}{2\pi fV^2} }$

Example

A 10 kW load operates at a power factor of 0.7 lagging. The power factor is to be improved to 0.95 lagging. Determine the required capacitor rating.

The initial and final power-factor angles are:

$\phi_1=\cos^{-1}(0.7)=45.57^\circ$

$\phi_2=\cos^{-1}(0.95)=18.19^\circ$

The required reactive power compensation is:

$Q_c=P(\tan\phi_1-\tan\phi_2)$

$Q_c=10[\tan(45.57^\circ)-\tan(18.19^\circ)]$

$Q_c=10(1.02-0.329)$

$Q_c = 6.91\text{ kVAR}$

Therefore, a capacitor bank of approximately 6.91 kVAR is required to improve the power factor from 0.7 lagging to 0.95 lagging.

Solved Examples

This section brings together some of the essential electrical engineering formulas through worked examples covering DC circuits, AC series RLC circuits, and three-phase systems. These examples are useful for practicing the application of basic electrical formulas.

Example 1 : A 24 V DC source supplies a load that draws 4 A. Determine the resistance, power, and energy consumed in 3 hours.

$R=\frac{V}{I}=\frac{24}{4}=6\,\Omega$

$P=VI=24\times4=96\,W$

$E=Pt=96\times3=288 \,Wh=0.288 \, kWh$

Therefore, the load has a resistance of 6 Ω, consumes 96 W of power, and uses 0.288 kWh of energy in 3 hours.

Example 2: A series RLC circuit has R = 20 Ω, L = 50 mH, C = 100 μF. It is connected to a 230 V, 50 Hz AC supply. Determine the inductive reactance, capacitive reactance, impedance, and current.

$\text{Inductive reactance, }X_L=2\pi fL$

$X_L=2 \times\pi \times 50 \times 0.05$

$X_L =15.71 \,\Omega$

$\text{Capacitive reactance, }X_C=\frac{1}{2\pi fC}$

$X_C=\frac{1}{2 \times \pi \times 50 \times 100\times10^{-6}}$

$X_C=31.83 \,\Omega$

For a series RLC circuit , Impedance is determined from,

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

$Z=\sqrt{20^2+(15.71-31.83)^2}$

$Z=\sqrt{400+259.85}$

$Z = 25.69\,\Omega$

Using AC Ohm’s law,

$I=\frac{V}{Z} =\frac{230}{25.69}$

$I = 8.95\, A$

Since (X_C > X_L), the circuit behaves as a capacitive load, and the current leads the supply voltage.

Example 3: A balanced three-phase load is supplied by a 415 V line voltage and draws a line current of 30 A at a power factor of 0.9. Determine the active power.

Using the three-phase power formula,

$P=\sqrt{3}V_LI_L\cos\phi$

$P=\sqrt{3}\times415\times30\times0.9$

$P = 19,406 \,W = 19.41 \,kW}$

Basic Electrical Engineering Formulas with Examples

Electric Field, Coulomb’s Law and Gauss’s Law

Example:

Basic Electrical Quantities and Ohm’s Law

Example

DC Circuits formulas

Example

AC Fundamental formulas

Example:

Single-Phase and Three-Phase AC Power Formulas

Single-phase AC power

Three Phase System

Three-Phase AC Power

Example:

Energy Storage Elements

Example

Electrical Machine Formulas

Example

Power factor correction formulas

Example

Solved Examples

Basic Terms in Electric Circuits | Types of Electric Networks

Electrical Quantities and Units with Tables and Conversions

Parallel operation of Alternator

Testing of dc machine

Commutation in DC Machine

Three point starter – Construction & Working

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Electric Field, Coulomb’s Law and Gauss’s Law

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Single-phase AC power

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Energy Storage Elements

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Electrical Machine Formulas

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Power factor correction formulas

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