13.1 Electrical Fundamentals, Ohm's Law, and DC Circuits

13.1 Electrical Fundamentals, Ohm's Law, and DC Circuits

Core FE Exam Principle: Direct current (DC) circuit analysis relies on charge conservation (Kirchhoff's Current Law) and energy conservation (Kirchhoff's Voltage Law). In steady-state DC conditions, capacitors behave as open circuits ((I_C = 0)) and ideal inductors behave as short circuits ((V_L = 0)).

Fundamental Electrical Quantities and Ohm's Law

Electric circuit analysis evaluates the transfer of energy via charge movement through circuit elements. The primary electrical quantities tested on the NCEES FE exam are:

• Electric Charge ((q)): Quantized property of matter measured in Coulombs (C).
• Electric Current ((I)): Time rate of net charge flow through a cross-sectional boundary: $I = \frac{dq}{dt} \quad [\text{Amperes, A} = \text{C/s}]$
• Voltage / Potential Difference ((V)): Energy required per unit charge to move charge between two points: $V = \frac{dW}{dq} \quad [\text{Volts, V} = \text{J/C}]$
• Resistance ((R)): Opposition to charge flow offered by a material, governed by physical dimensions and intrinsic resistivity ((\rho)): $R = \frac{\rho L}{A} \quad [\text{Ohms, } \Omega]$ where (L) is length (m) and (A) is cross-sectional area ((\text{m}^2)). Conductance is the inverse of resistance: (G = \frac{1}{R}) (measured in Siemens, S, or mhos).

Ohm's Law and Electric Power

Ohm's Law establishes that the voltage drop across a linear resistor is directly proportional to the current flowing through it:

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

Electric power (P) is the rate at which electrical energy is converted into heat or work:

$P = \frac{dW}{dt} = V I = I^2 R = \frac{V^2}{R} \quad [\text{Watts, W} = \text{J/s}]$

Total energy dissipated or absorbed over a time interval (t) is given by:

$W = \int_{0}^{t} P(\tau) d\tau = P t \quad [\text{Joules, J} = \text{W}\cdot\text{s}]$

Kirchhoff's Laws and Resistor Network Reduction

Kirchhoff's Current Law (KCL)

Based on the principle of conservation of charge, KCL states that the algebraic sum of all electric currents entering any circuit junction (node) is identically zero:

$\sum_{k=1}^{n} I_k = 0 \implies \sum I_{\text{entering}} = \sum I_{\text{leaving}}$

Kirchhoff's Voltage Law (KVL)

Based on the principle of conservation of energy, KVL states that the algebraic sum of all potential differences (voltages) around any closed loop in a circuit is zero:

$\sum_{k=1}^{m} V_k = 0$

Series and Parallel Resistor Networks

For two resistors in parallel, the equivalent resistance simplifies to the product-over-sum formula:

$R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2}$

Thevenin and Norton Equivalent Circuits

Any linear active circuit containing independent sources, dependent sources, and resistors can be reduced to an equivalent two-terminal representation.

 Thevenin Equivalent Circuit               Norton Equivalent Circuit
     +---[ R_th ]---+ A                         +-------+-------+ A
     |              |                           |       |       |
 ( + V_th - )       o                        ( I_n )  [R_n]     o
     |              |                           |       |       |
     +--------------+ B                         +-------+-------+ B

Determining Thevenin and Norton Parameters

1. Thevenin Voltage ((V_{\text{th}})): The open-circuit voltage measured across terminals A and B: $V_{\text{th}} = V_{AB,\text{open}}$
2. Norton Current ((I_{\text{n}})): The short-circuit current flowing from terminal A to B when A-B are shorted: $I_{\text{n}} = I_{AB,\text{short}}$
3. Equivalent Resistance ((R_{\text{th}} = R_{\text{n}})): The internal resistance looking into terminals A-B with all independent sources deactivated (voltage sources replaced by short circuits, current sources replaced by open circuits): $R_{\text{th}} = R_{\text{n}} = \frac{V_{\text{th}}}{I_{\text{n}}}$

Maximum Power Transfer Theorem

Maximum power is delivered from a linear source network to a load resistor (R_L) when the load resistance equals the internal Thevenin resistance:

$R_L = R_{\text{th}}$

Under maximum power transfer conditions, the power absorbed by (R_L) is:

$P_{\text{max}} = \frac{V_{\text{th}}^2}{4 R_{\text{th}}}$

Energy Storage Elements and First-Order Transients

Capacitors and Inductors

• Capacitors: Store potential energy in an electric field between conductive plates separated by a dielectric: $q = C V, \quad i(t) = C \frac{dv}{dt}, \quad W_C = \frac{1}{2} C V^2 \quad [\text{Joules, J}]$

  • Series Capacitance: (\frac{1}{C_{\text{eq}}} = \sum \frac{1}{C_i})
  • Parallel Capacitance: (C_{\text{eq}} = \sum C_i)

• Inductors: Store kinetic energy in a magnetic field established by current flow through conductive coils: $v(t) = L \frac{di}{dt}, \quad W_L = \frac{1}{2} L I^2 \quad [\text{Joules, J}]$

  • Series Inductance: (L_{\text{eq}} = \sum L_i)
  • Parallel Inductance: (\frac{1}{L_{\text{eq}}} = \sum \frac{1}{L_i})

First-Order RC and RL Transients

When a switch changes position in an RC or RL circuit, state variables (capacitor voltage (v_C) or inductor current (i_L)) change exponentially according to the general step response equation:

$x(t) = x(\infty) + \left[ x(0^+) - x(\infty) \right] e^{-t / \tau}$

where:

• (x(0^+)) is the initial value immediately following switching (noting (v_C(0^+) = v_C(0^-)) and (i_L(0^+) = i_L(0^-)))
• (x(\infty)) is the final DC steady-state value
• (\tau) is the circuit time constant: $\tau_{RC} = R_{\text{th}} C, \quad \tau_{RL} = \frac{L}{R_{\text{th}}}$

Comprehensive Worked Engineering Example

Problem Statement

A DC power system shown in the schematic below consists of a source (V_S = 60 \text{ V}) connected to a resistor network where (R_1 = 10 \ \Omega), (R_2 = 30 \ \Omega), and (R_3 = 15 \ \Omega). An uncharged capacitor (C = 200 \ \mu\text{F}) is connected across terminals A-B at time (t = 0).

           R1 = 10 ohm         R3 = 15 ohm
  +-------[======]-------+----[======]-------o Terminal A
  |                      |                   |
(+ 60 V)             [=] R2 = 30 ohm        --- C = 200 uF
  |                      |                  --- 
  |                      |                   |
  +----------------------+-------------------o Terminal B (Ground)

Determine:

1. The Thevenin equivalent voltage (V_{\text{th}}) and Thevenin resistance (R_{\text{th}}) looking into terminals A-B.
2. The circuit time constant (\tau) when the capacitor is connected across A-B.
3. The capacitor voltage (v_C(t)) at time (t = 9.0 \text{ ms}) after the switch closes.
4. The total energy stored in the electric field of the capacitor at (t = 9.0 \text{ ms}).

Step-by-Step Solution

Step 1: Find Thevenin Equivalent Voltage (V_{\text{th}})

• Disconnect the capacitor to find open-circuit voltage (V_{AB,\text{open}}).
• With A-B open, no current flows through (R_3). Therefore, (V_{\text{th}}) equals the voltage across (R_2).
• Apply the voltage divider rule across (R_1) and (R_2): $V_{\text{th}} = V_S \left( \frac{R_2}{R_1 + R_2} \right) = 60 \text{ V} \times \left( \frac{30}{10 + 30} \right) = 60 \times \frac{30}{40} = 45.0 \text{ V}$

Step 2: Find Thevenin Equivalent Resistance (R_{\text{th}})

• Deactivate the voltage source (V_S) by replacing it with a short circuit to ground.
• Looking into terminals A-B, (R_1) and (R_2) are in parallel, and that combination is in series with (R_3): $R_{\text{th}} = R_3 + (R_1 \parallel R_2) = 15 + \frac{10 \times 30}{10 + 30} = 15 + \frac{300}{40} = 15 + 7.5 = 22.5 \\ \Omega$

Step 3: Compute Time Constant (\tau) and Voltage at (t = 9.0 \text{ ms})

• Time constant (\tau): $\tau = R_{\text{th}} C = 22.5 \\ \Omega \times (200 \times 10^{-6} \text{ F}) = 4.5 \times 10^{-3} \text{ s} = 4.5 \text{ ms}$
• Calculate (v_C(t)) using the charging equation with (v_C(0) = 0 \text{ V}) and (v_C(\infty) = V_{\text{th}} = 45.0 \text{ V}): $v_C(t) = V_{\text{th}} \left( 1 - e^{-t / \tau} \right)$
• At (t = 9.0 \text{ ms}), (\frac{t}{\tau} = \frac{9.0 \text{ ms}}{4.5 \text{ ms}} = 2.0): $v_C(9.0 \text{ ms}) = 45.0 \times \left( 1 - e^{-2.0} \right) = 45.0 \times (1 - 0.135335) = 45.0 \times 0.864665 = 38.91 \text{ V}$

Step 4: Compute Stored Energy at (t = 9.0 \text{ ms})

• Using the capacitor energy storage formula: $W_C = \frac{1}{2} C [v_C(9.0 \text{ ms})]^2 = \frac{1}{2} \times (200 \times 10^{-6} \text{ F}) \times (38.91 \text{ V})^2$ $W_C = 1.0 \times 10^{-4} \times 1513.99 = 0.1514 \text{ J} = 151.4 \text{ mJ}$

Final Answer: (V_{\text{th}} = 45.0 \text{ V}), (R_{\text{th}} = 22.5 \ \Omega), (\tau = 4.5 \text{ ms}), (v_C(9.0 \text{ ms}) = 38.91 \text{ V}), and stored energy (W_C = 151.4 \text{ mJ}).