Electricity Fundamentals - Free Electrical Engineering Board Exam Reviewer

LessonAbout 1 h

Electricity Basic Principles

Electricity is the physical phenomenon associated with the presence and movement of electric charge.

1. Electric Charge, Potential Difference, and Current

1.1 Electric Charge

Electric charge is a fundamental property of subatomic particles that causes them to experience a force in an electric field.

Unit: coulomb (\(\text{C}\))
Electron charge: \(-1.602 \times 10^{-19}\ \text{C}\)
Proton charge: \(+1.602 \times 10^{-19}\ \text{C}\)

Electric Charge

\[ Q = n e \]

EE Board Exam October 1998

Example. Determine the total charge in coulombs of \(3.12 \times 10^{19}\) electrons.

Solution.

\[ Q = n e = (3.12 \times 10^{19})(1.602 \times 10^{-19}\ \text{C}) = 5\ \text{C} \]

Answer: \(5\ \text{C}\)

1.2 Potential Difference

Potential difference is the work done per unit charge in moving a charge between two points.

Potential Difference

\[ V = \dfrac{W}{Q} \]

Example. A total charge of \(80\ \text{C}\) passes through a potential difference of \(110\ \text{V}\). Find the energy expended in joules.

Solution.

\[ W = Q V = (80\ \text{C})(110\ \text{V}) = 8800\ \text{J} \]

Answer: \(8800\ \text{J}\)

1.3 Electric Current

Electric current is the rate of flow of electric charge through a conductor cross-section.

Electric Current

\[ I = \dfrac{Q}{t} \]

EE Board Exam October 1991

Example. A conductor carries a current of \(4\ \text{A}\). Find the charge in coulombs passing through a section of the conductor in \(2\ \text{minutes}\).

Solution.

\[ Q = I t = (4\ \text{A})(2 \times 60\ \text{s}) = 480\ \text{C} \]

Answer: \(480\ \text{C}\)

2. Resistance and Resistivity

2.1 Specific Resistance and Resistivity

Resistance is opposition to electric current flow, while resistivity measures intrinsic material opposition.

Conductor Resistance

\[ R = \rho \dfrac{L}{A} \]

Example. A copper conductor \(500\ \text{m}\) long has a resistivity of \(1.72 \times 10^{-8}\ \Omega\cdot\text{m}\) and a cross-sectional area of \(2\ \text{mm}^2\). Calculate its resistance.

Solution.

\[ R = \rho \dfrac{L}{A} = \dfrac{(1.72 \times 10^{-8}\ \Omega\cdot\text{m})(500\ \text{m})}{2 \times 10^{-6}\ \text{m}^2} = 4.3\ \Omega \]

Answer: \(4.3\ \Omega\)

2.2 Circular Mils

A circular mil is the area of a circle having a diameter of 1 mil (\(0.001\ \text{inch}\)).

Circular Mil Area

\[ \text{CM} = d^2 \]

\[ A = \dfrac{\pi}{4} d^2 \qquad\qquad \text{sq mil} \]

Where: \(d\) = diameter in mils (\(1\ \text{inch} = 1000\ \text{mils}\))

EE Board Exam April 1992

Example. Determine the cross-sectional area in circular mils of a wire with a diameter of \(0.0808\ \text{inch}\).

Solution.

\[ d = 0.0808 \times 1000 = 80.8\ \text{mils} \]

\[ \text{CM} = d^2 = (80.8)^2 = 6528.64\ \text{CM} \]

Answer: \(6528.64\ \text{CM}\)

2.3 Resistance in Terms of Volume

Volume resistance relations express conductor resistance using total volume rather than cross-sectional area alone.

Resistance in Terms of Volume

\[ R = \rho \dfrac{L^2}{V} \]

\[ R = \rho \dfrac{V}{A^2} \]

Example. A copper wire has a resistance of \(0.5\ \Omega\) and a volume of \(100\ \text{cm}^3\). If the resistivity of copper is \(1.72 \times 10^{-6}\ \Omega\cdot\text{cm}\), find the length of the wire.

Solution.

\[ L = \sqrt{\dfrac{R V}{\rho}} = \sqrt{\dfrac{(0.5\ \Omega)(100\ \text{cm}^3)}{1.72 \times 10^{-6}\ \Omega\cdot\text{cm}}} = 5393.6\ \text{cm} = 53.94\ \text{m} \]

Answer: \(53.94\ \text{m}\)

3. Wire Drawing and Conductor Resizing

3.1 Rewinding Coils

Coil rewinding preserves total winding volume while changing wire gauge and total turn count.

Rewound Coil Resistance

\[ \dfrac{R_2}{R_1} = \left(\dfrac{N_2}{N_1}\right)^2 = \left(\dfrac{A_1}{A_2}\right)^2 = \left(\dfrac{d_1}{d_2}\right)^4 \]

Where: \(N\) = number of turns

EE Board Exam October 1997

Example. A coil wound with No. 22 AWG wire has \(2000\ \text{turns}\) and a resistance of \(120\ \Omega\). If rewound using No. 25 AWG wire having half the cross-sectional area in the same slot space, find the new resistance.

Solution.

\[ R_2 = R_1 \left(\dfrac{A_1}{A_2}\right)^2 = 120\ \Omega \left(\dfrac{2}{1}\right)^2 = 480\ \Omega \]

Answer: \(480\ \Omega\)

3.2 Wire Drawing and Volume Conservation

Wire drawing stretches a conductor to increase length while conserving total metal volume.

Drawn Wire Resistance

\[ R_2 = R_1 \left(\dfrac{L_2}{L_1}\right)^2 = R_1 \left(\dfrac{A_1}{A_2}\right)^2 = R_1 \left(\dfrac{d_1}{d_2}\right)^4 \]

EE Board Exam October 1992

Example. A wire with a resistance of \(10\ \Omega\) is drawn out so that its length is tripled. Find the new resistance of the wire.

Solution.

\[ R_2 = R_1 \left(\dfrac{L_2}{L_1}\right)^2 = 10\ \Omega (3)^2 = 90\ \Omega \]

Answer: \(90\ \Omega\)

3.3 Conductor Material Comparison and Replacement

Conductor replacement determines the size of an alternate material giving equal length and resistance.

Equal Resistance Replacement

\[ \dfrac{\rho_1}{A_1} = \dfrac{\rho_2}{A_2} \]

\[ \dfrac{d_2}{d_1} = \sqrt{\dfrac{\rho_2}{\rho_1}} \]

EE Board Exam April 1995

Example. An aluminum wire is to replace a copper wire of \(0.5\ \text{cm}\) diameter, keeping length and resistance unchanged. If resistivity of aluminum is \(2.83 \times 10^{-8}\ \Omega\cdot\text{m}\) and copper is \(1.72 \times 10^{-8}\ \Omega\cdot\text{m}\), find the aluminum wire diameter.

Solution.

\[ d_2 = d_1 \sqrt{\dfrac{\rho_2}{\rho_1}} = 0.5\ \text{cm} \sqrt{\dfrac{2.83 \times 10^{-8}}{1.72 \times 10^{-8}}} = 0.641\ \text{cm} \]

Answer: \(0.641\ \text{cm}\)

4. Temperature Effects on Resistance

4.1 Resistance Change with Temperature

Temperature variation changes conductor resistance linearly above inferred absolute zero temperature.

Inferred Absolute Temperature Relation

\[ \dfrac{R_2}{R_1} = \dfrac{|T| + t_2}{|T| + t_1} \]

Where: \(|T|\) = inferred absolute zero temperature (\(234.5^\circ\text{C}\) for annealed copper)

EE Board Exam October 1993

Example. A copper coil has a resistance of \(20\ \Omega\) at \(20^\circ\text{C}\). Find its resistance at \(80^\circ\text{C}\) if inferred absolute zero for copper is \(-234.5^\circ\text{C}\).

Solution.

\[ R_2 = R_1 \dfrac{234.5 + t_2}{234.5 + t_1} = 20\ \Omega \dfrac{234.5 + 80}{234.5 + 20} = 24.715\ \Omega \]

Answer: \(24.715\ \Omega\)

4.2 Temperature Coefficient of Resistance

The temperature coefficient of resistance is the fractional change in resistance per degree temperature change.

Temperature Coefficient

\[ \alpha_1 = \dfrac{1}{|T| + t_1} \]

\[ \alpha_2 = \dfrac{\alpha_1}{1 + \alpha_1 (t_2 - t_1)} \]

\[ R_2 = R_1 [1 + \alpha_1 (t_2 - t_1)] \]

Where: \(\alpha_1\) = temperature coefficient at reference temperature \(t_1\)

EE Board Exam April 1990

Example. The resistance of a wire is \(125\ \Omega\) at \(20^\circ\text{C}\) and \(150\ \Omega\) at \(70^\circ\text{C}\). Find the temperature coefficient of resistance at \(20^\circ\text{C}\).

Solution.

\[ \alpha_1 = \dfrac{R_2 - R_1}{R_1 (t_2 - t_1)} = \dfrac{150 - 125}{125 (70 - 20)} = 0.004\ \text{per }^\circ\text{C} \]

Answer: \(0.004\ \text{per }^\circ\text{C}\)

5. Insulation Resistance of Cables

5.1 Radial Insulation Resistance of Cables

Insulation resistance of a single-core cable measures opposition to radial leakage current through the dielectric sheath.

Cable Insulation Resistance

\[ R_i = \dfrac{\rho}{2 \pi L} \ln\left(\dfrac{r_2}{r_1}\right) \]

Where: \(r_1\) = conductor core radius, \(r_2\) = outer insulation radius

EE Board Exam October 1998

Example. A single-core cable \(1\ \text{km}\) long has a conductor radius of \(1\ \text{cm}\) and insulation thickness of \(0.5\ \text{cm}\). If insulation resistivity is \(5 \times 10^{12}\ \Omega\cdot\text{m}\), calculate the cable insulation resistance.

Solution.

\[ r_1 = 1\ \text{cm} \qquad\qquad r_2 = 1 + 0.5 = 1.5\ \text{cm} \]

\[ R_i = \dfrac{\rho}{2 \pi L} \ln\left(\dfrac{r_2}{r_1}\right) = \dfrac{5 \times 10^{12}}{2 \pi (1000)} \ln\left(\dfrac{1.5}{1}\right) = 322.74 \times 10^6\ \Omega = 322.74\ \text{M}\Omega \]

Answer: \(322.74\ \text{M}\Omega\)

6. Conductance

6.1 Electrical Conductance

Conductance is the reciprocal of electrical resistance, measuring how easily electric current flows.

Conductance

\[ G = \dfrac{1}{R} = \sigma \dfrac{A}{L} \]

Where: \(\sigma\) = electrical conductivity (\(\sigma = \dfrac{1}{\rho}\))

Example. A conductor has a resistance of \(0.025\ \Omega\). Determine its conductance in siemens.

Solution.

\[ G = \dfrac{1}{R} = \dfrac{1}{0.025\ \Omega} = 40\ \text{S} \]

Answer: \(40\ \text{S}\)

Formula Summary

Formula Name	Formula
Electric Charge	\(Q = n e\)
Potential Difference	\(V = \dfrac{W}{Q}\)
Electric Current	\(I = \dfrac{Q}{t}\)
Conductor Resistance	\(R = \rho \dfrac{L}{A}\)
Circular Mil Area	\(\text{CM} = d^2 \quad A = \dfrac{\pi}{4} d^2\)
Resistance in Terms of Volume	\(R = \rho \dfrac{L^2}{V} \quad R = \rho \dfrac{V}{A^2}\)
Rewound Coil Resistance	\(\dfrac{R_2}{R_1} = \left(\dfrac{N_2}{N_1}\right)^2 = \left(\dfrac{A_1}{A_2}\right)^2 = \left(\dfrac{d_1}{d_2}\right)^4\)
Drawn Wire Resistance	\(R_2 = R_1 \left(\dfrac{L_2}{L_1}\right)^2 = R_1 \left(\dfrac{A_1}{A_2}\right)^2 = R_1 \left(\dfrac{d_1}{d_2}\right)^4\)
Equal Resistance Replacement	\(\dfrac{\rho_1}{A_1} = \dfrac{\rho_2}{A_2} \quad \dfrac{d_2}{d_1} = \sqrt{\dfrac{\rho_2}{\rho_1}}\)
Inferred Absolute Temperature Relation	\(\dfrac{R_2}{R_1} = \dfrac{\|T\| + t_2}{\|T\| + t_1}\)
Temperature Coefficient	\(\alpha_1 = \dfrac{1}{\|T\| + t_1} \quad R_2 = R_1 [1 + \alpha_1 (t_2 - t_1)]\)
Cable Insulation Resistance	\(R_i = \dfrac{\rho}{2 \pi L} \ln\left(\dfrac{r_2}{r_1}\right)\)
Conductance	\(G = \dfrac{1}{R} = \sigma \dfrac{A}{L}\)