Comprehensive Electronics Study Notes: Voltage, Current, Resistance, Ohm's Law, Power, and Series Circuit Analysis

Fundamentals of Electrical Quantities and Atomic Structure

Hydro-Electrical Analogy (Water/Hose Model)

• Charge ($Q$): Equivalent to the volume of water.

• Voltage ($V$): Equivalent to water pressure. Increasing the height of a water reservoir increases the potential energy of the water, which directly corresponds to increasing voltage.

• Current ($I$): Equivalent to the flow rate of the water through a pipe.

• Power ($P$): Equivalent to the total amount of water flowing in a given time period. A small pipe under high pressure or a large pipe under low pressure can deliver the exact same power.

• Resistance ($R$): Equivalent to the internal roughness or narrow width of the pipe restricting flow.

Atomic Structure and the Bohr Model

• Nucleus: The central core of an atom containing positively charged protons and electrically neutral neutrons.

• Electrons: Negatively charged subatomic particles orbiting the nucleus in discrete energy shells.

• Atomic Number: Equal to the number of protons in the nucleus. In a neutral, electrically balanced atom, the number of electrons equals the number of protons.

• Fundamental Atom Structures:

  • Hydrogen ($\text{H}$): Atomic number $1$ (contains $1$ proton in the nucleus and $1$ orbiting electron).

  • Helium ($\text{He}$): Atomic number $2$ (contains $2$ protons and $2$ neutrons in the nucleus, and $2$ orbiting electrons).

Electron Shells, Energy Levels, and Photons

• Discrete Energy Levels: Electrons can only orbit within specific, quantized energy shells labeled $n = 1, 2, 3, \text{etc.}$ Ground state represents the lowest energy level ($n = 1$).

• Binding Energy: Electrons in orbits farther from the nucleus possess higher energy levels and are less tightly bound to the positive attraction of the nucleus.

• Photon Absorption and Emission:

  • When an electron absorbs a discrete packet of light energy (a photon) equal to the gap between energy levels, it transitions to a higher energy shell.

  • When returning to its lower ground state, the electron emits a photon of identical energy.

Classification of Electrical Materials

• Valence Shell & Valence Electrons: The outermost electron shell of an atom is its valence shell. The electrons residing in this shell determine the material's chemical and electrical conductivity properties.

• Conductors:

  • Materials that readily permit the flow of electric current due to a high density of free electrons.

  • Characterized atomic structure: $1$ to $3$ valence electrons.

  • Examples: Silver ($1$ valence electron, highest conductivity), Copper ($1$ valence electron, most widely used due to cost/availability), Gold ($1$ valence electron), Mercury ($2$ valence electrons), Aluminum ($3$ valence electrons).

  • Copper Core Model: A copper atom has $29$ protons and $29$ electrons arranged in four shells ($2, 8, 18, 1$). The core consists of the nucleus and inner electrons with a net positive charge of $+1$, weakly holding its single valence electron.

• Insulators:

  • Nonmetallic materials that severely impede electric current because valence electrons are tightly bound to the nucleus, leaving virtually no free electrons.

  • Characterized atomic structure: Greater than $4$ valence electrons.

  • Examples: Glass, porcelain, teflon, polyethylene.

• Semiconductors:

  • Materials whose conductive properties lie between conductors and insulators.

  • Characterized atomic structure: Exactly $4$ valence electrons.

  • Examples: Silicon ($\text{Si}$) and Germanium ($\text{Ge}$), which form the foundation of solid-state electronic devices like diodes, transistors, and integrated circuits (ICs).

Electric Charge, Coulomb's Law, and Electric Fields

• Electric Charge ($Q$): An inherent property of matter resulting from an excess or deficiency of electrons. Charge is measured in coulombs ($\text{C}$).

  • Charge of an electron ($e^-$): $-1.6 \times 10^{-19}\text{\,C}$.

  • One coulomb ($1.0\text{\,C}$) equals the total charge of $6.25 \times 10^{18}$ electrons.

• Coulomb's Law of Electrostatic Forces:

  • Uncharged bodies exert no electrostatic force.

  • Opposite charges ($+$ and $-$) attract each other with force $F$.

  • Like positive charges ($+$ and $+$) or like negative charges ($-$ and $-$) repel each other.

• Electric Fields: Represented by lines of force extending outward from positive charges toward negative charges.

Core Electrical Quantities

• Voltage ($V$):

  • Definition: The amount of potential energy ($W$) available per unit charge ($Q$) to move electrons between two points in a circuit.

  • Formula: $V = \frac{W}{Q}$

  • Unit: Volt ($\text{V}$), where $1\text{\,V} = 1\text{\,J/C}$.

  • Example A: Moving a charge of $10\text{\,C}$ using $50\text{\,J}$ of energy yields a voltage of $V = \frac{50\text{\,J}}{10\text{\,C}} = 5\text{\,V}$.

  • Example B: Moving $50\text{\,C}$ across a potential difference of $12\text{\,V}$ requires energy of $W = V \times Q = 12\text{\,V} \times 50\text{\,C} = 600\text{\,J}$.

• Current ($I$):

  • Definition: The time rate at which electric charge flows past a specific cross-sectional point in a conductor.

  • Formula: $I = \frac{Q}{t}$

  • Unit: Ampere ($\text{A}$), where $1\text{\,A} = 1\text{\,C/s}$.

  • Example: If $2.0\text{\,C}$ of charge pass a point in $5.0\text{\,s}$, the resulting current is $I = \frac{2.0\text{\,C}}{5.0\text{\,s}} = 0.40\text{\,A}$.

• Resistance ($R$):

  • Definition: The opposition to the flow of electric current.

  • Unit: Ohm ($\Omega$). One ohm ($1\text{\,}\Omega$) of resistance exists when a potential difference of $1\text{\,V}$ produces a current of $1\text{\,A}$.

• Conductance ($G$):

  • Definition: The ability of a circuit or element to conduct electric current; mathematically defined as the reciprocal of resistance.

  • Formula: $G = \frac{1}{R}$

  • Unit: Siemens ($\text{S}$).

Resistor Identification and Components

Resistor Construction

Resistors are fixed or variable components constructed to provide a calibrated opposition to electric current. Carbon-composition resistors consist of a resistive carbon mixture, an insulating protective coating, solid metal leads, and color-coded bands.

4-Band Resistor Color Code System

• Band 1: First significant digit of the resistance value.

• Band 2: Second significant digit of the resistance value.

• Band 3: Multiplier (power of 10 or number of zeros following the second digit).

• Band 4: Percent tolerance.

Color Code Values

Mnemonic: "Big Beautiful Roses Occupy Your Garden But Violets Grow Wild"

4-Band Examples

• Gray-Red-Black-Gold: $8$, $2$, $\times 1$ ($0$ zeros), $\pm 5\% \implies 82\text{\,}\Omega \pm 5\%$.

• *330\text{\,k}\Omega \pm 5\%$:* 3$(Orange),$3$(Orange),$\times 10^4$/$4$zeros (Yellow),$\pm 5\%$(Gold)$\implies Orange-Orange-Yellow-Gold.\n* *Green-Blue-Brown-Gold:* 5$,$6$,$\times 10$($1$zero),$\pm 5\% \implies 560\text{\,}\Omega \pm 5\%.\n\n## 5-Band Precision Resistor Color Code System\n\n\n\nUsed for precision resistors with tight tolerances (\le 2\%).\n* **Band 1:** First significant digit.\n* **Band 2:** Second significant digit.\n* **Band 3:** Third significant digit.\n* **Band 4:** Multiplier (Gold = 0.1$, Silver =$0.01).\n* **Band 5:** Tolerance: Red = \pm 2\%$, Brown =$\pm 1\%$, Green =$\pm 0.5\%$, Blue =$\pm 0.25\%$, Violet =$\pm 0.1\%.\n\n## Alphanumeric Resistor Labeling\n\nUses digits and letters (`R`, `K`, `M`) where the letter signifies both the multiplier and the decimal point location:\n* `22R` = 22\text{\,}\Omega\n* `2M2` = 2.2\text{\,M}\Omega\n* `220K` = 220\text{\,k}\Omega\n* `0R1` = 0.1\text{\,}\Omega\n\n## Variable Resistors\n\n* **Potentiometer:** A three-terminal variable resistor used primarily as an adjustable voltage divider.\n* **Rheostat:** A two-terminal variable resistor used primarily to adjust current.\n\n# Basic Electric Circuits and Measurements\n\n## Circuit Structure\n\nAll functional electric circuits require three essential components:\n1. **Voltage Source:** Provides potential difference (e.g., battery, generator, solar cell).\n2. **Load:** Consumes electrical energy (e.g., lamp, resistor, motor).\n3. **Complete Current Path:** Conductive wiring enabling loop closure.\n4. *(Optional) Control Element:* Regulates current (e.g., switch).\n\n## Circuit Control Elements (Switches)\n\n* **Pole:** The movable contact arm in a switch. Represents the number of independent circuits controlled.\n* **Throw:** The number of fixed contact positions connected by a single pole movement.\n* **Single-Pole Single-Throw (SPST):** A basic two-terminal on/off switch.\n* **Closed Circuit:** A complete, unbroken path allowing current (I) to flow freely.\n* **Open Circuit:** A broken path created by an open switch, blown fuse, or severed conductor. Current drops to exactly zero (I = 0\text{\,A}).\n\n## Electrical Measurements and Safety\n\n* **Digital Multimeter (DMM):** Multi-function meter used to measure Voltage, Current, and Resistance.\n* **Safety Standards:**\n * *Electric Shock:* Know the exact location of lab emergency shutoff controls.\n * *Fire Safety:* Use Class-C rated fire extinguishers for electrical fires.\n * *Thermal Hazards:* Remove rings, watches, and conductive jewelry to prevent severe burn injuries.\n\n# Ohm's Law and Linear Circuit Relationships\n\n## Ohm's Law Equations\n\nFormulated by Georg Simon Ohm, stating that current in a circuit is directly proportional to voltage and inversely proportional to resistance.\n\n\begin{aligned} I &= \frac{V}{R} \[1em] V &= I \times R \[1em] R &= \frac{V}{I} \end{aligned}\n\n## Graphical Representations\n\n### Current versus Voltage (I$vs.$V)\n\n* Plotting current (I$) on the vertical axis against voltage ($V) on the horizontal axis for a fixed resistor yields a straight line passing through the origin (linear relationship).\n* **Slope:** Represents conductance (G).\n\nG = \frac{\Delta I}{\Delta V}\n\n* *Data Fit Calculation:* For a straight line through (0\text{\,V}, 0\text{\,mA})$and$(10.0\text{\,V}, 14.8\text{\,mA}):\n\nG = \frac{14.8\text{\,mA} - 0\text{\,mA}}{10.0\text{\,V} - 0\text{\,V}} = 1.48\text{\,mS}\n\nR = \frac{1}{G} = \frac{1}{1.48\text{\,mS}} = 676\text{\,}\Omega\n\n* *Graph Slope Example:* A graph indicating R = 2.7\text{\,k}\Omega yields a conductance of:\n\nG = \frac{1}{2.7\text{\,k}\Omega} = 0.37\text{\,mS}\n\n### Current versus Resistance (I$vs.$R)\n\nPlotting current against resistance at a constant voltage yields a non-linear hyperbola (I = \frac{V}{R}).\n\n## Worked Ohm's Law Calculations\n\n* **Finding Current:**\n * For V = 2.5\text{\,V}$across$R = 5.6\text{\,k}\Omega:\n\nI = \frac{2.5\text{\,V}}{5.6\text{\,k}\Omega} = 0.4464\text{\,mA} \approx 0.45\text{\,mA}\n\n * A 75.0\text{\,W}$light bulb with a hot resistance of$192\text{\,}\Omega$connected to$120\text{\,V}:\n\nI = \frac{120\text{\,V}}{192\text{\,}\Omega} = 0.625\text{\,A} = 625\text{\,mA}\n\n* **Finding Voltage:**\n * Voltage across a 680\text{\,}\Omega$resistor carrying$26.5\text{\,mA}:\n\nV = (26.5\text{\,mA})(680\text{\,}\Omega) = 18.02\text{\,V} \approx 18\text{\,V}\n\n * Voltage developed across a 5.0\text{\,m}\Omega$sense resistor carrying$20\text{\,A}:\n\nV = (20\text{\,A})(0.005\text{\,}\Omega) = 0.1\text{\,V} = 100\text{\,mV}\n\n * Voltage across a 270\text{\,k}\Omega$resistor with$11.1\text{\,}\mu\text{A} current:\n\nV = (11.1\text{\,}\mu\text{A})(270\text{\,k}\Omega) = 2.997\text{\,V} \approx 3.0\text{\,V}\n\n* **Finding Resistance:**\n * Resistance if 5.0\text{\,V}$causes$333\text{\,}\mu\text{A} current:\n\nR = \frac{5.0\text{\,V}}{333\text{\,}\mu\text{A}} = 15.015\text{\,k}\Omega \approx 15\text{\,k}\Omega\n\n * Resistance if 66\text{\,mV}$causes$200\text{\,}\mu\text{A} current:\n\nR = \frac{66\text{\,mV}}{200\text{\,}\mu\text{A}} = 330\text{\,}\Omega\n\n * Resistance if 15.0\text{\,V}$causes$165\text{\,}\mu\text{A} current:\n\nR = \frac{15.0\text{\,V}}{165\text{\,}\mu\text{A}} = 90.9\text{\,k}\Omega\n\n# Energy, Power, and Efficiency\n\n## Energy vs. Power\n\n* **Energy (W$):** The capacity to do work, measured in Joules ($\text{J}$). One Joule is the work performed when a force of$1\text{\,N}$moves an object through$1\text{\,m}$($1\text{\,J} = 1\text{\,N}\cdot\text{m} = 1\text{\,W}\cdot\text{s}).\n* **Power (P$):** The time rate of doing work or transferring electrical energy. Measured in Watts ($\text{W}$), where$1\text{\,W} = 1\text{\,J/s}.\n\n\n\n* *Lifting Analogy:* Lifting a box requires a fixed amount of **energy** regardless of speed. Lifting the box faster increases the **power** required, but energy remains constant.\n\n## Units of Measurement and Conversions\n\n* **Kilowatt-Hour (\text{kWh}):** Standard commercial utility unit for electrical energy.\n\n1\text{\,kWh} = (1000\text{\,W})(3600\text{\,s}) = 3.6 \times 10^6\text{\,J}\n\n * *Heater Example:* Operating a 1200\text{\,W}$heater for$20\text{\,minutes}$($\frac{1}{3}\text{\,hour}):\n\nW = (1.2\text{\,kW}) \times \left(\frac{1}{3}\text{\,h}\right) = 0.4\text{\,kWh}\n\n* **Horsepower (\text{hp}):** Industrial rating for mechanical/electrical motor output.\n\n1\text{\,hp} = 746\text{\,W}\n\n * *Motor Conversion:* Power of a 5.0\text{\,hp} motor in kilowatts:\n\nP = 5.0\text{\,hp} \times \left(\frac{746\text{\,W}}{1\text{\,hp}}\right) \times \left(\frac{1\text{\,kW}}{1000\text{\,W}}\right) = 3.73\text{\,kW}\n\n * *Energy consumed over 24 hours:* W = (3.73\text{\,kW})(24\text{\,h}) = 89.5\text{\,kWh}.\n\n## Watt's Law Equations\n\n\begin{aligned} P &= V \times I \[1em] P &= I^2 \times R \[1em] P &= \frac{V^2}{R} \end{aligned}\n\n### Worked Watt's Law Examples\n\n* *Example 1:* Power in a 27\text{\,}\Omega$resistor carrying$0.135\text{\,A}:\n\nP = I^2 R = (0.135\text{\,A})^2(27\text{\,}\Omega) = 0.492\text{\,W} \approx 0.49\text{\,W}\n\n* *Example 2:* Power dissipated by a heater drawing 12\text{\,A}$from$120\text{\,V}:\n\nP = VI = (120\text{\,V})(12\text{\,A}) = 1440\text{\,W}\n\n* *Example 3:* Power dissipated in a 100\text{\,}\Omega$resistor across$5.0\text{\,V}:\n\nP = \frac{V^2}{R} = \frac{(5.0\text{\,V})^2}{100\text{\,}\Omega} = 0.25\text{\,W}\n\n* *Example 4:* A heater with R = 25.0\text{\,}\Omega$operated at$120.0\text{\,V}:\n    a. Current supplied: I = \frac{120.0\text{\,V}}{25.0\text{\,}\Omega} = 4.8\text{\,A}.\n    b. Energy in Joules in 10.0\text{\,s}$:$E = I^2 R t = (4.8\text{\,A})^2(25.0\text{\,}\Omega)(10.0\text{\,s}) = 5760\text{\,J}.\n* *Example 5 (Television Energy Cost):* A color TV draws 2.0\text{\,A}$on$120\text{\,V}$, operated$8.00\text{\,hours/day}.\n    a. Power: P = VI = (120\text{\,V})(2.0\text{\,A}) = 240\text{\,W}.\n    b. Daily energy consumption: E = \frac{(240\text{\,J/s})(8\text{\,h})(3600\text{\,s/h})}{3.6 \times 10^6\text{\,J/kWh}} = 1.92\text{\,kWh/day}.\n    c. Cost for 30 days at \$0.15/\text{kWh}$:$\text{Cost} = (1.92\text{\,kWh/day})(30\text{\,days})(\0.15/\text{kWh}) = \8.64.\n\n## Resistor Power Ratings and Thermal Failure\n\n* **Power Rating:** The maximum heat power a resistor can safely dissipate without changing its resistance value or damaging itself.\n* *Physical Factors:* Determined by body surface area, material composition, shape, and physical size—not ohmic resistance value.\n* *Engineering Rule of Thumb:* Select a power rating at least **twice** the maximum anticipated power dissipation.\n* *Overheating Consequences:* Burning, charring of color bands, resistance value permanent drift, open-circuit failure, and potential fire hazards.\n\n### Overheating Evaluations\n\n* *Circuit (a):* 9\text{\,V}$battery,$100\text{\,}\Omega$resistor rated$\frac{1}{4}\text{\,W}$($0.25\text{\,W}).\n\nP = \frac{(9\text{\,V})^2}{100\text{\,}\Omega} = 0.81\text{\,W}\n\n    *Since 0.81\text{\,W} > 0.25\text{\,W}, it will overheat and fail.*\n\n* *Circuit (b):* 24\text{\,V}$battery,$1.5\text{\,k}\Omega$resistor rated$\frac{1}{2}\text{\,W}$($0.50\text{\,W}).\n\nP = \frac{(24\text{\,V})^2}{1500\text{\,}\Omega} = 0.384\text{\,W}\n\n    *Since 0.384\text{\,W} \le 0.50\text{\,W}, the power rating is adequate.*\n\n* *Circuit (c):* 5\text{\,V}$battery,$10\text{\,}\Omega$resistor rated$1\text{\,W}.\n\nP = \frac{(5\text{\,V})^2}{10\text{\,}\Omega} = 2.5\text{\,W}\n\n    *Since 2.5\text{\,W} > 1\text{\,W}, it will overheat and fail.*\n\n## Power Grid and High-Voltage Transmission\n\nPower line loss is governed by P_{\text{loss}} = I^2 R_{\text{line}}.\n* For a power line with R = 1.00\text{\,}\Omega$carrying$I = 10.0\text{\,A}$:$P_{\text{loss}} = (10.0\text{\,A})^2(1.00\text{\,}\Omega) = 100\text{\,W}.\n* For the same line carrying I = 1.00\text{\,A}$:$P_{\text{loss}} = (1.00\text{\,A})^2(1.00\text{\,}\Omega) = 1.00\text{\,W}.\n* *Key Insight:* Stepping up voltage for power transmission drops current proportionally, drastically reducing I^2 R heat losses.\n\n## Power Supply Efficiency and Battery Ratings\n\n\n\n* **Efficiency Formulas:**\n\n\text{Efficiency} = \left(\frac{P_{\text{OUT}}}{P_{\text{IN}}}\right) \times 100\%\n\nP_{\text{LOSS}} = P_{\text{IN}} - P_{\text{OUT}}\n\n* *Efficiency Example:* Power supply requiring 25\text{\,W}$input power delivering$20\text{\,W} output power:\n\n\text{Efficiency} = \frac{20\text{\,W}}{25\text{\,W}} = 0.80 = 80\%\n\nP_{\text{LOSS}} = 25\text{\,W} - 20\text{\,W} = 5\text{\,W}\n\n* *Related Calculation:* Efficiency = 92\%$,$P_{\text{IN}} = 50\text{\,W}$, find$P_{\text{OUT}}:\n\nP_{\text{OUT}} = (0.92)(50\text{\,W}) = 46\text{\,W}\n\n* **Ampere-Hour (Ah) Battery Capacity:**\n\n\text{Rating (Ah)} = I \times t\n\n * *Example A:* Battery rated 10\text{\,Ah}$delivering$0.5\text{\,A}$lasts$t = \frac{10\text{\,Ah}}{0.5\text{\,A}} = 20\text{\,hours}.\n * *Example B:* Battery rated 70\text{\,Ah}$delivering$2\text{\,A}$lasts$t = \frac{70\text{\,Ah}}{2\text{\,A}} = 35\text{\,hours}.\n * *Example C:* Battery delivering 10\text{\,A}$for$6\text{\,hours}$has capacity$\text{Rating} = (10\text{\,A})(6\text{\,h}) = 60\text{\,Ah}.\n\n# Series Circuits and Circuit Analysis\n\n## Characteristics of Series Circuits\n\n\n\nA series circuit provides only **one single current path** connecting components end-to-end.\n* **Current Rule:** The current at every point in a series circuit is identical.\n\nI_T = I_1 = I_2 = I_3 = \dots = I_n\n\n* **Total Resistance Rule:** Total series resistance equals the sum of all individual series resistors.\n\nR_T = R_1 + R_2 + R_3 + \dots + R_n\n\n### Series Resistance Calculations\n\n* *Example 1:* Five resistors in series: 56\text{\,}\Omega, 100\text{\,}\Omega, 27\text{\,}\Omega, 10\text{\,}\Omega, 47\text{\,}\Omega.\n\nR_T = 56\text{\,}\Omega + 100\text{\,}\Omega + 27\text{\,}\Omega + 10\text{\,}\Omega + 47\text{\,}\Omega = 240\text{\,}\Omega\n\n* *Example 2:* Six resistors in series: 39\text{\,}\Omega, 100\text{\,}\Omega, 47\text{\,}\Omega, 100\text{\,}\Omega, 180\text{\,}\Omega, 68\text{\,}\Omega.\n\nR_T = 39\text{\,}\Omega + 100\text{\,}\Omega + 47\text{\,}\Omega + 100\text{\,}\Omega + 180\text{\,}\Omega + 68\text{\,}\Omega = 534\text{\,}\Omega\n\n* *Example 3 (Finding unknown resistor R_4$):* Ohmmeter reads total resistance$R_T = 17.9\text{\,k}\Omega$. Known resistors are$R_1 = 1.0\text{\,k}\Omega$,$R_2 = 2.2\text{\,k}\Omega$,$R_3 = 4.7\text{\,k}\Omega.\n\nR_4 = R_T - (R_1 + R_2 + R_3) = 17.9\text{\,k}\Omega - (1.0\text{\,k}\Omega + 2.2\text{\,k}\Omega + 4.7\text{\,k}\Omega) = 10.0\text{\,k}\Omega\n\n## Series Voltage Sources\n\nVoltage sources connected in series combine algebraically based on their polarities.\n* *Series-Aiding:* Polarities orient in the same direction (-$to$+$,$-$to$+, etc.). Voltages sum together.\n * Three 9\text{\,V}$batteries aiding:$V_T = 9\text{\,V} + 9\text{\,V} + 9\text{\,V} = 27\text{\,V}.\n* *Reversed Battery:* If one battery is connected in reverse polarity, its voltage subtracts.\n * V_T = 9\text{\,V} + 9\text{\,V} - 9\text{\,V} = 9\text{\,V}.\n\n## Kirchhoff's Voltage Law (KVL)\n\n\n\nKVL states that the sum of all voltage drops around a single closed path equals the total source voltage, or algebraically, the sum of all voltages around a closed loop is zero.\n\nV_S = V_1 + V_2 + V_3 + \dots + V_n\n\n\sum_{i=1}^n V_i = 0\n\n### KVL Worked Calculations\n\n* *Circuit with Fuse (Fig 5-33):* Drop across fuse is 0\text{\,V}$. Drops across resistors are$V_1 = 5\text{\,V}$and$V_2 = 10\text{\,V}.\n\nV_S = 5\text{\,V} + 10\text{\,V} = 15\text{\,V}\n\n* *Finding Unknown Drop V_4$and Resistor$R_4$(Fig 5-34):*$V_S = 50\text{\,V}$,$I = 10\text{\,mA}$,$R_1 = 100\text{\,}\Omega$,$R_2 = 470\text{\,}\Omega$,$R_3 = 1.0\text{\,k}\Omega.\n\nV_1 = (10\text{\,mA})(100\text{\,}\Omega) = 1.0\text{\,V}\n\nV_2 = (10\text{\,mA})(470\text{\,}\Omega) = 4.7\text{\,V}\n\nV_3 = (10\text{\,mA})(1.0\text{\,k}\Omega) = 10.0\text{\,V}\n\nV_4 = V_S - (V_1 + V_2 + V_3) = 50\text{\,V} - (1.0\text{\,V} + 4.7\text{\,V} + 10.0\text{\,V}) = 34.3\text{\,V}\n\nR_4 = \frac{V_4}{I} = \frac{34.3\text{\,V}}{10\text{\,mA}} = 3.43\text{\,k}\Omega\n\n## Voltage Divider Rule\n\nThe voltage drop across any resistor in a series circuit equals the ratio of its resistance to the total resistance, multiplied by the source voltage.\n\nV_x = \left(\frac{R_x}{R_T}\right) V_S\n\n### Voltage Divider Calculations\n\n* *Two-Resistor Ratio:* If R_1 = 2 R_2$and$V_S = 12\text{\,V}$, then$R_1$drops$\frac{2}{3}$of$V_S$($V_1 = 8.0\text{\,V}$) and$R_2$drops$\frac{1}{3}$of$V_S$($V_2 = 4.0\text{\,V}).\n* *Example:* V_S = 20\text{\,V}$,$R_1 = 15\text{\,k}\Omega$,$R_2 = 10\text{\,k}\Omega$.$R_T = 25\text{\,k}\Omega.\n\nV_2 = \left(\frac{10\text{\,k}\Omega}{25\text{\,k}\Omega}\right)(20\text{\,V}) = (0.40)(20\text{\,V}) = 8.0\text{\,V}\n\n## Power in Series Circuits\n\nTotal power dissipated in a series circuit equals the sum of powers dissipated by each individual resistor.\n\nP_T = P_1 + P_2 + P_3 + \dots + P_n\n\n* *Example:* V_S = 20\text{\,V}$,$R_1 = 470\text{\,}\Omega$,$R_2 = 330\text{\,}\Omega$.$R_T = 800\text{\,}\Omega$.

$V_1 = \left(\frac{470\text{\,}\Omega}{800\text{\,}\Omega}\right)(20\text{\,V}) = 11.75\text{\,V}$

$V_2 = \left(\frac{330\text{\,}\Omega}{800\text{\,}\Omega}\right)(20\text{\,V}) = 8.25\text{\,V}$

$P_1 = \frac{(11.75\text{\,V})^2}{470\text{\,}\Omega} = 0.29\text{\,W}$

$P_2 = \frac{(8.25\text{\,V})^2}{330\text{\,}\Omega} = 0.21\text{\,W}$

$P_T = P_1 + P_2 = 0.29\text{\,W} + 0.21\text{\,W} = 0.50\text{\,W}$

Circuit Ground and Subscript Notation

• Reference Ground: A common reference potential point ($0\text{\,V}$).

• Single Subscript ($V_A$): Voltage at point A measured relative to ground.

• Double Subscript ($V_{AB}$): Potential difference between points A and B ($V_{AB} = V_A - V_B$).

Ground Shift Comparison

• Configuration 1 (Ground at C): $V_S = +12\text{\,V}$, $R_1 = 5.0\text{\,k}\Omega$, $R_2 = 10\text{\,k}\Omega$. $R_T = 15\text{\,k}\Omega$.

  • $I = \frac{12\text{\,V}}{15\text{\,k}\Omega} = 0.80\text{\,mA}$.

  • $V_C = 0\text{\,V}$.

  • $V_B = (0.80\text{\,mA})(10\text{\,k}\Omega) = +8.0\text{\,V}$.

  • $V_A = +12\text{\,V}$.

  • $V_{AB} = V_A - V_B = 12\text{\,V} - 8.0\text{\,V} = 4.0\text{\,V}$.

• Configuration 2 (Ground moved to B):

  • $V_B = 0\text{\,V}$.

  • $V_A = +4.0\text{\,V}$.

  • $V_C = -8.0\text{\,V}$.

  • $V_{AB} = V_A - V_B = 4.0\text{\,V} - 0\text{\,V} = 4.0\text{\,V}$.

Systematic Troubleshooting Techniques

Initial Diagnostic Questions

1. Has the circuit ever worked properly?

2. If it once worked, under what specific conditions did it fail?

3. What are the observed failure symptoms?

4. What are the likely physical or electrical causes?

Planning and Analysis

• Review schematics, wiring diagrams, and equipment manuals.

• Analyze failure history and symptom patterns.

Half-Splitting Method

An isolation strategy where testing begins at the midpoint of a circuit path to establish whether the fault lies upstream or downstream, halving the remaining search area with each subsequent measurement.

Self-Check Questions and Practice Problems

Section 5-4 Checkup Problems

1. A $6\text{\,V}$ battery is connected across three resistors in series. What is the current through each resistor?

   • Solution: Current is identical through each resistor: $I = \frac{6\text{\,V}}{R_T}$.

2. Voltage required to produce $50\text{\,mA}$ through $R_1 = 10\text{\,}\Omega$, $R_2 = 5.6\text{\,}\Omega$, $R_3 = 5.6\text{\,}\Omega$ in series:

   • Solution: $R_T = 10 + 5.6 + 5.6 = 21.2\text{\,}\Omega$. $V_S = (50\text{\,mA})(21.2\text{\,}\Omega) = 1.06\text{\,V}$.

3. Voltage dropped across each resistor when current is $50\text{\,mA}$:

   • Solution: $V_1 = (50\text{\,mA})(10\text{\,}\Omega) = 0.50\text{\,V}$; $V_2 = (50\text{\,mA})(5.6\text{\,}\Omega) = 0.28\text{\,V}$; $V_3 = (50\text{\,mA})(5.6\text{\,}\Omega) = 0.28\text{\,V}$.

4. Four equal-value resistors in series with $5\text{\,V}$ source produce $4.63\text{\,mA}$. Value of each resistor:

   • Solution: $R_T = \frac{5\text{\,V}}{4.63\text{\,mA}} = 1079.9\text{\,}\Omega \approx 1080\text{\,}\Omega$. Each resistor $R = \frac{1080\text{\,}\Omega}{4} = 270\text{\,}\Omega$.

5. Series current-limiting resistor for red LED at $10\text{\,mA}$ with $3\text{\,V}$ source and $1.7\text{\,V}$ LED drop:

   • Solution: $V_R = 3\text{\,V} - 1.7\text{\,V} = 1.3\text{\,V}$. $R = \frac{1.3\text{\,V}}{10\text{\,mA}} = 130\text{\,}\Omega$.

Chapter Quiz Answer Key

Chapter 2 Review Quiz Answers

1. Atomic number is the number of protons in the nucleus. (Option a)

2. Valence electrons are in the outer shell, involved in chemical reactions, and loosely bound. (Option d)

3. Particle responsible for current in solid metallic conductors is the electron. (Option b)

4. Symbol for charge is $Q$. (Option c)

5. Definition for voltage is $V = \frac{W}{Q}$. (Option c)

6. Gray-Red-Black-Gold resistor value is $82\text{\,}\Omega$. (Option b)

7. $330\text{\,k}\Omega \pm 5\%$ resistor color bands are Orange-Orange-Yellow-Gold. (Option b)

8. Solar cell converts light energy to electrical energy. (Option b)

Chapter 3 (Energy & Power) Review Quiz Answers

1. Unit of power is Watt. (Option d - none of the above)

2. SI unit of energy is Joule. (Option b)

3. If voltage for a resistive circuit doubles, power quadruples. (Option d)

4. Smallest power rating for $330\text{\,}\Omega$ with $12\text{\,V}$ ($0.436\text{\,W}$) is $\frac{1}{2}\text{\,W}$. (Option b)

5. Power dissipated by light operating on $12\text{\,V}$ with $3.0\text{\,A}$ is $36\text{\,W}$. (Option c)

6. Power rating of a resistor is determined mainly by surface area. (Option a)

7. Circuit with largest power dissipation among $10\text{\,V}/100\text{\,}\Omega$ ($1\text{\,W}$), $15\text{\,V}/200\text{\,}\Omega$ ($1.125\text{\,W}$), $20\text{\,V}/300\text{\,}\Omega$ ($1.33\text{\,W}$), $25\text{\,V}/400\text{\,}\Omega$ ($1.56\text{\,W}$) is Circuit (d). (Option d)

8. Circuit with smallest power dissipation is Circuit (a). (Option a)

9. Battery rated $20\text{\,Ah}$ supplying $2.0\text{\,A}$ lasts $10\text{\,hours}$. (Option c)

10. Efficiency is determined by dividing output power by input power. (Option a)

Chapter 4 (Series Circuits) Review Quiz Answers

1. In a series circuit with more than one resistor, current is always the same in all resistors. (Option c)

2. Voltage is larger across larger resistors. (Option a)

3. Three equal resistors in series produce total resistance three times the value of one resistor. (Option c)

4. A series circuit cannot have more than one path. (Option c)

5. In a closed loop, the algebraic sum of all voltages is $0$. (Option a)

6. Current in $10\text{\,k}\Omega$ resistor with $24\text{\,V}$ source and $R_2 = 2.0\text{\,k}\Omega$ ($R_T = 12\text{\,k}\Omega$) is $2.0\text{\,mA}$. (Option b)

7. Output voltage from voltage divider across $2.0\text{\,k}\Omega$ resistor is $4.0\text{\,V}$. (Option b)

8. Smallest output voltage available from potentiometer voltage divider is $0\text{\,V}$. (Option a)

9. Total power dissipated in a series circuit equals the sum of power in all resistors. (Option d)

10. Voltage $V_{AB}$ means the voltage difference between points A and B. (Option d)