Comprehensive Study Notes on Current Electricity, Circuit Analysis, and Electrical Energy

Electrical Energy and Cost Calculation

• Definition of Kilowatt-Hour ($\text{kWh}$):

  • A kilowatt-hour ($\text{kWh}$) represents $1000\,\text{W}$ of electrical power consumed or generated continuously over a duration of one hour ($1\,\text{h}$).

  • It serves as a standard unit of measurement for total electrical energy consumed and total work performed.

  • The kilowatt-hour ($\text{kWh}$) is the commercial billing unit utilized by electrical utility providers to charge consumers.

• Step-by-Step Energy and Cost Calculation Procedure:

  • Step 1: Calculate Electrical Power ($P$):

  • Multiply electrical potential difference ($V$) by current ($I$):       $P = V \times I$

  • In this relationship, potential difference is measured in volts ($\text{V}$), current in amperes ($\text{A}$), and resulting power in watts ($\text{W}$).

  • Step 2: Convert Power to Kilowatts ($\text{kW}$):

  • Divide power in watts ($\text{W}$) by the metric conversion factor of $1000\,\text{W/kW}$:       $\text{Power in kW} = \frac{\text{Power in W}}{1000\,\text{W/kW}}$

  • Step 3: Calculate Consumed Electrical Energy ($\text{kWh}$):

  • Multiply power in kilowatts ($\text{kW}$) by the total elapsed operational time in hours ($\text{h}$):       $\text{Energy (kWh)} = \text{Power (kW)} \times \text{Time (h)}$

  • Step 4: Compute Monetary Cost:

  • Multiply total energy consumed in kilowatt-hours by the specific energy tariff rate per kilowatt-hour:       $\text{Cost} = \text{Energy (kWh)} \times \text{Unit Price per kWh}$

• Worked Example — Slow Cooker Consumption:

  • Problem: A slow cooker operates on an electric current of $4\,\text{A}$ drawn from a $110\,\text{V}$ circuit for a duration of $2\,\text{h}$. Calculate the total electrical energy consumed in kilowatt-hours ($\text{kWh}$).

  • Step 1 — Determine Power:
        $P = 110\,\text{V} \times 4\,\text{A} = 440\,\text{W}$

  • Step 2 — Convert Watts to Kilowatts:
        $\frac{440\,\text{W}}{1000\,\text{W/kW}} = 0.44\,\text{kW}$

  • Step 3 — Multiply by Operational Hours:
        $0.44\,\text{kW} \times 2\,\text{h} = 0.88\,\text{kWh}$

• Practical Energy and Operational Cost Problems:

  • Problem 1 (Electric Stove Energy): An electric stove draws $20\,\text{A}$ of current on a $220\,\text{V}$ circuit for $1\,\text{h}$.

  • Power: $P = 220\,\text{V} \times 20\,\text{A} = 4400\,\text{W} = 4.4\,\text{kW}$

  • Energy consumed: $4.4\,\text{kW} \times 1\,\text{h} = 4.4\,\text{kWh}$

  • Problem 2 (Electric Stove Operating Cost): Total cost to run the electric stove in Problem 1 at an energy tariff of $\text{R}0.08$ per $\text{kWh}$:

  • $\text{Cost} = 4.4\,\text{kWh} \times \text{R}0.08/\text{kWh} = \text{R}0.352$ (or $35.2\,\text{cents}$)

  • Problem 3 (Microwave Oven Energy): A microwave oven draws $6\,\text{A}$ of current on a $110\,\text{V}$ circuit for $2\,\text{h}$.

  • Power: $P = 110\,\text{V} \times 6\,\text{A} = 660\,\text{W} = 0.66\,\text{kW}$

  • Energy consumed: $0.66\,\text{kW} \times 2\,\text{h} = 1.32\,\text{kWh}$

  • Problem 4 (Microwave Oven Operating Cost): Total cost to operate the microwave oven in Problem 3 at an energy rate of $\text{R}0.08$ per $\text{kWh}$:

  • $\text{Cost} = 1.32\,\text{kWh} \times \text{R}0.08/\text{kWh} = \text{R}0.1056$ (or $10.56\,\text{cents}$)

  • Problem 5 (Refrigerator Daily Consumption): A refrigerator operates at $10\,\text{A}$ on a $220\,\text{V}$ circuit for an average run time of $18\,\text{h}$ per day.

  • Power: $P = 220\,\text{V} \times 10\,\text{A} = 2200\,\text{W} = 2.2\,\text{kW}$

  • Daily energy consumed: $2.2\,\text{kW} \times 18\,\text{h} = 39.6\,\text{kWh}$

  • Problem 6 (Refrigerator Daily Running Cost): Daily operational cost of running the refrigerator in Problem 5 at an electricity rate of $14\,\text{cents}$ per $\text{kWh}$:

  • $\text{Cost} = 39.6\,\text{kWh} \times 14\,\text{cents/kWh} = 554.4\,\text{cents} = \text{R}5.544$

  • Problem 7 (Household Monthly Electricity Bill): A residential electricity meter displays a reading of $73602\,\text{kWh}$ on June 1 and $76167\,\text{kWh}$ on July 1. The unit cost of electricity is $15\,\text{cents}$ per $\text{kWh}$.

  • Total monthly energy consumption:
          $76167\,\text{kWh} - 73602\,\text{kWh} = 2565\,\text{kWh}$

  • Total electricity bill for June:
          $2565\,\text{kWh} \times 15\,\text{cents/kWh} = 38475\,\text{cents} = \$384.75$ (or $\text{R}384.75$)

  • Problem 8 (Lighting Efficiency and Cost Savings): An office space is illuminated by three $100\,\text{W}$ incandescent bulbs for $8\,\text{h}$ per day. The fixtures are retrofitted with $60\,\text{W}$ bulbs. Electricity costs $12\,\text{cents}$ per $\text{kWh}$.

  • Initial total power: $3 \times 100\,\text{W} = 300\,\text{W} = 0.3\,\text{kW}$

  • Retrofitted total power: $3 \times 60\,\text{W} = 180\,\text{W} = 0.18\,\text{kW}$

  • Power saved:
          $0.3\,\text{kW} - 0.18\,\text{kW} = 0.12\,\text{kW}$

  • Daily electrical energy saved:
          $0.12\,\text{kW} \times 8\,\text{h} = 0.96\,\text{kWh}$

  • Monetary savings achieved per day:
          $0.96\,\text{kWh} \times 12\,\text{cents/kWh} = 11.52\,\text{cents}$

Electric Current and Charge

• Definition of Electric Current:

  • An electric current is defined fundamentally as a flow of charge.

• Conventional Current vs. Electron Flow:

  • In a conducting circuit, negatively charged electrons physically move away from the negative terminal and toward the positive terminal of a power source.

  • Conventional current ($I$) is defined as the flow of positive charge from the positive terminal to the negative terminal (represented conceptually as empty positive "spaces" migrating toward the negative terminal).

• Mathematical Formulation of Current:

  • Conventional current equals the rate of flow of charge past a given cross-sectional boundary:     $I = \frac{Q}{t}$

  • Variable definitions:

  • $I$ = conventional current, measured in amperes ($\text{A}$)

  • $Q$ = quantity of charge flowing, measured in coulombs ($\text{C}$)

  • $t$ = elapsed time, measured in seconds ($\text{s}$)

• Unit Definition of the Ampere:

  • One ampere ($1\,\text{A}$) is the constant current flowing when one coulomb ($1\,\text{C}$) of charge passes a specific point in an electrical conductor in one second ($1\,\text{s}$):     $1\,\text{A} = 1\,\text{C/s}$

• Unit Definition of the Coulomb:

  • One coulomb ($1\,\text{C}$) is the quantity of electric charge transported past a point when a steady current of one ampere ($1\,\text{A}$) flows for a duration of one second ($1\,\text{s}$):     $Q = I \times t$     $1\,\text{C} = 1\,\text{A} \times 1\,\text{s}$

Potential Difference (Voltage) and Resistance

• Potential Difference (Voltage — $V$):

  • Potential difference is the work done or energy expended by electrical cells to move charge between two points in an electrical circuit:     $V = \frac{W}{Q}$

  • Variable definitions:

  • $V$ = potential difference, measured in volts ($\text{V}$)

  • $W$ = electrical work or energy transferred, measured in joules ($\text{J}$)

  • $Q$ = quantity of electrical charge moved, measured in coulombs ($\text{C}$)

  • One volt ($1\,\text{V}$) represents the transfer of one joule ($1\,\text{J}$) of energy per one coulomb ($1\,\text{C}$) of charge displaced:     $1\,\text{V} = 1\,\text{J/C}$

• Electrical Resistance ($R$):

  • Resistance is a quantitative measure of the opposition experienced by moving charge carriers passing through a circuit component.

  • Microscopic cause: Electrical resistance arises directly due to collisions between mobile conduction electrons and the vibrating, stationary positive metal ions forming the crystalline lattice of the conductor.

• Ohm's Law:

  • The resistance of a conductor, maintained at a constant temperature, is measured by the ratio of the potential difference across it to the current flowing through it (the "push-to-flow" ratio):     $R = \frac{V}{I}$

  • Variable definitions:

  • $R$ = resistance, measured in ohms ($\Omega$)

  • $V$ = potential difference, measured in volts ($\text{V}$)

  • $I$ = current, measured in amperes ($\text{A}$)

  • Definition of the Ohm: One ohm ($1\,\Omega$) is the resistance of an electrical conductor when a potential difference of one volt ($1\,\text{V}$) applied across its terminals produces a current of one ampere ($1\,\text{A}$).

• Physical Determinants of Electrical Resistance:

  • Conductor resistance depends fundamentally upon four physical parameters:

  • Temperature: For metallic conductors, resistance increases as temperature rises due to increased thermal lattice vibrations, leading to more frequent electron-ion collisions.

  • Length: Resistance is directly proportional to conductor length ($R \propto L$).

  • Thickness (Cross-Sectional Area): Resistance is inversely proportional to conductor thickness or cross-sectional area ($R \propto \frac{1}{A}$).

  • Type of Metal: Inherent resistivity varies across different materials based on atomic structure and free electron density.

Circuit Symbols and Measuring Instruments

• Standard Circuit Symbols:

  • Standard graphical representations are utilized for electrical schematic drafting:

  • Cell: Pair of parallel unequal lines (long thin positive line, short thick negative line).

  • Alternating Source: Circle enclosing a sinusoidal wave symbol.

  • Filament Bulb: Circle enclosing an "X".

  • Switch: Two terminal circles connected by a hinged lever (open or closed).

  • Fuse: Rectangular box with an axial connecting wire traversing through its center.

  • Resistor: Hollow elongated rectangle.

  • Variable Resistor (Rheostat): Hollow rectangle crossed diagonally by an arrow indicating adjustability.

  • Voltmeter: Circle enclosing the uppercase letter "V".

  • Ammeter: Circle enclosing the uppercase letter "A".

• Measurement Instruments (Meters):

  • Ammeters:

  • Function: Measure electric current strength flowing through a circuit.

  • Connection: Always connected strictly in series within the path of interest.

  • Internal Resistance: Designed with almost zero internal resistance ($R \approx 0\,\Omega$) so that insertion of the meter does not appreciably alter or restrict the current flowing through the circuit.

  • Voltmeters:

  • Function: Measure electrical potential difference between two separate points.

  • Connection: Always connected strictly in parallel across the component being measured.

  • Internal Resistance: Designed with extremely high internal resistance ($R \to \infty$) so that the meter draws virtually no current from the main circuit.

Resistors in Series and Parallel Networks

• Resistors Connected in Series:

  • Configuration: Components connected end-to-end along a single continuous conducting loop.

  • Path: There is only one path available for current flow.

  • Current Behavior: Current remains constant throughout all series components:     $I_{\text{TOT}} = I_1 = I_2 = I_3 = \dots$

  • Potential Difference Distribution: Total source potential difference splits across the individual resistors in direct proportion to their resistances:     $V_{\text{TOT}} = V_1 + V_2 + V_3 + \dots$

  • Total Equivalent Resistance: Total resistance equals the direct algebraic sum of the individual resistances:     $R_{\text{TOT}} = R_1 + R_2 + R_3 + \dots$

  • Circuit Reliability: If any single series component fails or burns out, the circuit is broken completely and all current flow ceases instantly.

• Worked Example — Resistors in Series:

  • Scenario: A $12\,\text{V}$ battery is connected across three series resistors with values of $1\,\Omega$, $2\,\Omega$, and $3\,\Omega$.

  • (a) Total Resistance:
        $R_{\text{TOT}} = R_1 + R_2 + R_3 = 1\,\Omega + 2\,\Omega + 3\,\Omega = 6\,\Omega$

  • (b) Circuit Current:
        $I = \frac{V}{R_{\text{TOT}}} = \frac{12\,\text{V}}{6\,\Omega} = 2\,\text{A}$

  • (c) Potential Difference Across Each Resistor:
        $V_{1\Omega} = I \times R_1 = 2\,\text{A} \times 1\,\Omega = 2\,\text{V}$     $V_{2\Omega} = I \times R_2 = 2\,\text{A} \times 2\,\Omega = 4\,\text{V}$     $V_{3\Omega} = I \times R_3 = 2\,\text{A} \times 3\,\Omega = 6\,\text{V}$

  • Verification: $V_{\text{TOT}} = 2\,\text{V} + 4\,\text{V} + 6\,\text{V} = 12\,\text{V}$

• Resistors Connected in Parallel:

  • Configuration: Multiple components connected across common electrical junctions.

  • Path: There is more than one path available for current flow.

  • Potential Difference Behavior: Potential difference across all parallel branches is identical and constant:     $V_{\text{parallel}} = V_1 = V_2 = V_3 = \dots$

  • Total Equivalent Resistance ($R_{\text{eq}}$): The reciprocal of equivalent resistance equals the sum of reciprocals of individual resistances:     $\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots$

  • Resistance Property: The equivalent resistance of a parallel combination is always strictly less than the smallest individual resistor in the combination because parallel branches create additional pathways for charge flow.

  • Current Division: Total main circuit current splits into each parallel branch inversely proportional to branch resistance:     $I_{\text{branch}} = \frac{V_{\text{parallel}}}{R_{\text{branch}}}$ or $I_{\text{branch}} = \frac{I_{\text{main}} \times R_{\text{eq}}}{R_{\text{branch}}}$     $I_{\text{main}} = I_1 + I_2 + I_3 + \dots$

  • Circuit Reliability: If one branch component fails or blows, the circuit loop remains intact through all alternate parallel branches, permitting current to continue flowing uninterrupted (e.g., standard domestic household circuits).

• Worked Example — Resistors in Parallel:

  • Scenario: A $12\,\text{V}$ battery is connected across three parallel resistors with ratings of $10\,\Omega$, $15\,\Omega$, and $30\,\Omega$.

  • (a) Equivalent Resistance ($R_{\text{eq}}$):
        $\frac{1}{R_{\text{eq}}} = \frac{1}{10} + \frac{1}{15} + \frac{1}{30} = \frac{3 + 2 + 1}{30} = \frac{6}{30}$     $R_{\text{eq}} = \frac{30}{6} = 5\,\Omega$

  • (b) Total Ammeter Reading ($I_{\text{main}}$):
        $I = \frac{V}{R_{\text{eq}}} = \frac{12\,\text{V}}{5\,\Omega} = 2.4\,\text{A}$

  • (c) Current Through the $10\,\Omega$ Resistor:
        $I_{10\Omega} = \frac{V_{\text{parallel}}}{R_{10\Omega}} = \frac{12\,\text{V}}{10\,\Omega} = 1.2\,\text{A}$

  • Branch current check:
          $I_{15\Omega} = \frac{12\,\text{V}}{15\,\Omega} = 0.8\,\text{A}$       $I_{30\Omega} = \frac{12\,\text{V}}{30\,\Omega} = 0.4\,\text{A}$       $I_{\text{TOT}} = 1.2\,\text{A} + 0.8\,\text{A} + 0.4\,\text{A} = 2.4\,\text{A}$

• Combinations of Electrochemical Cells:

  • Cells in Series: Total potential difference equals the sum of potential differences of the individual cells:     $V_{\text{TOT}} = V_1 + V_2 + \dots$

  • Example: A $2\,\text{V}$ cell connected in series with a $1\,\text{V}$ cell provides:       $V_{\text{TOT}} = 2\,\text{V} + 1\,\text{V} = 3\,\text{V}$

  • Cells in Parallel: Total potential difference equals the potential difference of the largest cell in the parallel bank:     $V_{\text{TOT}} = V_{\text{largest}}$

  • Example: A $2\,\text{V}$ cell connected in parallel with a $1\,\text{V}$ cell provides:       $V_{\text{TOT}} = 2\,\text{V}$

Direct Current (D.C.) and Alternating Current (A.C.)

• Direct Current (D.C.):

  • Definition: Electric current in which electrons drift unidirectionally through a conductor, moving from negative to positive.

  • Typical sources: Electrochemical cells, chemical batteries, and direct current power supplies.

  • Waveform: A continuous horizontal line above zero on both voltage-time ($V$-$t$) and current-time ($I$-$t$) oscilloscope traces.

• Alternating Current (A.C.):

  • Definition: Electric current in which moving charge carriers regularly reverse their direction of flow at fixed temporal intervals.

  • Utility Grid Standards: In South Africa, the national grid frequency supplied by ESKOM operates at $50\,\text{Hz}$ ($50\,\text{cycles/second}$).

  • Biological Perception Limit: The human eye can discern a maximum of $44\,\text{oscillations/second}$; consequently, visual lighting running at $50\,\text{Hz}$ appears perfectly continuous without perceptible flicker.

  • Waveform: A sinusoidal trace displaying periodic oscillation above and below the zero-axis on voltage-time ($V$-$t$) and current-time ($I$-$t$) traces.

Domestic Electrical Installations and Household Circuits

• Utility Supply Characteristics:

  • The electrical utility (ESKOM) supplies residential properties with a nominal $220\,\text{V}$ alternating current.

  • Supply enters the home through an external service cable comprising two primary conductors passing through the utility meter to the main distribution box:

  • Live Wire (L): High-potential conductor operating at an alternating potential of $220\,\text{V}$.

  • Neutral Wire (N): Return conductor maintained at ground potential ($0\,\text{V}$).

  • Earth Wire (E): Low-resistance safety conductor connected directly to a buried earth ground electrode.

• Internal Distribution Box Architecture:

  • Incoming power passes through a paired main switch capable of isolating both Live and Neutral simultaneously.

  • Sub-circuits originate from shared busbars in the distribution box, each governed by dedicated circuit breakers or trip switches tailored to expected load capacities:

  • Stove Circuit: Heavy-duty dedicated branch protected by a $60\,\text{A}$ breaker.

  • Wall Socket Plugs (Main Circuit): Power circuits protected by a $30\,\text{A}$ breaker feeding three-pin wall sockets with connected earth conductors.

  • Lighting Circuit: Protected by a $15\,\text{A}$ breaker supplying ceiling illumination fittings.

  • Main Circuit Breaker: Overall system safety protected by a master $60\,\text{A}$ trip switch to prevent wiring from overheating and causing structural fires.

• Safety and Structural Wiring Rules:

  • Parallel Arrangement: All wall plugs, sockets, and lighting fixtures are wired in parallel across the Live and Neutral rails. This ensures that every individual appliance receives the full operational potential difference of $220\,\text{V}$ and can operate independently.

  • Switch Placement: All control switches must be situated on the Live wire. This guarantees that opening a switch fully de-energizes the appliance from the $220\,\text{V}$ supply, mitigating electrocution risks during servicing.

Summary of Physical Quantities, Units, and Triangle Relations

• Physical Quantities and SI Units:

  • Charge: Symbol $Q$, Unit name coulomb, Unit symbol $\text{C}$

  • Current strength: Symbol $I$, Unit name ampere, Unit symbol $\text{A}$

  • Time: Symbol $t$, Unit name second, Unit symbol $\text{s}$

  • Potential difference: Symbol $V$, Unit name volt, Unit symbol $\text{V}$

  • Resistance: Symbol $R$, Unit name ohm, Unit symbol $\Omega$

  • Work: Symbol $W$, Unit name joule, Unit symbol $\text{J}$

  • Energy: Symbol $W$, Unit name joule, Unit symbol $\text{J}$

• Fundamental Electrical Formula Triangles:

  • Charge-Current-Time Triangle ($Q = I \cdot t$):     $Q = I \times t$     $I = \frac{Q}{t}$     $t = \frac{Q}{I}$

  • Work-Charge-Voltage Triangle ($W = Q \cdot V$):     $W = Q \times V$     $V = \frac{W}{Q}$     $Q = \frac{W}{V}$

  • Ohm's Law Triangle ($V = I \cdot R$):     $V = I \times R$     $I = \frac{V}{R}$     $R = \frac{V}{I}$

Comprehensive Practice Problems and Worked Solutions

• Theoretical Review Questions:

  • Question 1: What is the function of a cell in an electric circuit?

  • Answer: A cell acts as an electrical energy source; it performs work on charges to create and maintain an electrical potential difference, driving charge through the circuit.

  • Question 2: What is an electric current?

  • Answer: An electric current is a continuous flow of electrical charge.

  • Question 3: In what unit is an electric current measured?

  • Answer: The ampere ($\text{A}$).

  • Question 4: How would you connect an ammeter in a circuit?

  • Answer: An ammeter is connected strictly in series with the component whose current is to be measured.

  • Question 5: Does an ammeter have a high or low resistance? Explain why this must be so.

  • Answer: An ammeter has an extremely low resistance (approaching zero). This ensures it introduces negligible total resistance into the branch, preventing it from reducing the current it is intended to measure.

  • Question 6: Which current direction do we use?

  • Answer: Conventional current direction (from the positive terminal to the negative terminal).

  • Question 7: Define a coulomb charge.

  • Answer: One coulomb is the quantity of electric charge that flows past a given cross-section of a conductor when a steady current of one ampere is sustained for one second ($1\,\text{C} = 1\,\text{A} \times 1\,\text{s}$).

• Charge, Current, and Time Calculations (Question 8):

  • Row a: Given $I = 2.5\,\text{A}$, $t = 6\,\text{s}$.     $Q = I \times t = 2.5\,\text{A} \times 6\,\text{s} = 15\,\text{C}$

  • Row b: Given $I = 0.6\,\text{A}$, $t = 30\,\text{s}$.     $Q = I \times t = 0.6\,\text{A} \times 30\,\text{s} = 18\,\text{C}$

  • Row c: Given $Q = 250\,\text{C}$, $t = 200\,\text{s}$.     $I = \frac{Q}{t} = \frac{250\,\text{C}}{200\,\text{s}} = 1.25\,\text{A}$

  • Row d: Given $Q = 300\,\text{C}$, $t = 1\,\text{minute} = 60\,\text{s}$.     $I = \frac{Q}{t} = \frac{300\,\text{C}}{60\,\text{s}} = 5\,\text{A}$

  • Row e: Given $Q = 18\,\text{C}$, $I = 0.2\,\text{A}$.     $t = \frac{Q}{I} = \frac{18\,\text{C}}{0.2\,\text{A}} = 90\,\text{s}$

  • Row f: Given $Q = 0.8\,\text{C}$, $I = 16\,\text{mA} = 0.016\,\text{A}$.     $t = \frac{Q}{I} = \frac{0.8\,\text{C}}{0.016\,\text{A}} = 50\,\text{s}$

• Single-Variable Circuit Current Calculations:

  • Question 9: A current of $15\,\text{A}$ flows in an electric heater for $12\,\text{s}$. Calculate the charge passing a point in the circuit:     $Q = I \times t = 15\,\text{A} \times 12\,\text{s} = 180\,\text{C}$

  • Question 10: A tape recorder draws $0.01\,\text{A}$ for $5\,\text{minutes}$ ($5 \times 60 = 300\,\text{s}$). Calculate the charge drawn from the battery:     $Q = I \times t = 0.01\,\text{A} \times 300\,\text{s} = 3\,\text{C}$

  • Question 11: If $150\,\text{C}$ passes through a circuit cross-section in $50\,\text{s}$, calculate the current:     $I = \frac{Q}{t} = \frac{150\,\text{C}}{50\,\text{s}} = 3\,\text{A}$

  • Question 12: A radio operates for $120\,\text{s}$ with $60\,\text{C}$ passing a point. Calculate current:     $I = \frac{Q}{t} = \frac{60\,\text{C}}{120\,\text{s}} = 0.5\,\text{A}$

  • Question 13: A resistor carries a current of $0.8\,\text{A}$. Calculate the time required for $200\,\text{C}$ to pass through:     $t = \frac{Q}{I} = \frac{200\,\text{C}}{0.8\,\text{A}} = 250\,\text{s}$

• Potential Difference and Work Concepts:

  • Question 14: What is meant by the electrical potential difference between two points on a conductor?

  • Answer: It is the amount of work done or energy transformed per unit electric charge as it moves between those two specified points ($V = \frac{W}{Q}$).

  • Question 15: What unit is used to measure electrical potential difference?

  • Answer: The volt ($\text{V}$).

  • Question 16: Which apparatus is used to measure potential difference, and how must it be connected?

  • Answer: A voltmeter, connected in parallel across the two points or components of interest.

  • Question 17: Calculate the potential difference between points $X$ and $Y$ if $120\,\text{J}$ of work is required to move $40\,\text{C}$ between them:     $V = \frac{W}{Q} = \frac{120\,\text{J}}{40\,\text{C}} = 3\,\text{V}$

• Potential Difference, Work, and Charge Calculations (Question 18):

  • Row a: Given $W = 88\,\text{J}$, $Q = 4\,\text{C}$.     $V = \frac{W}{Q} = \frac{88\,\text{J}}{4\,\text{C}} = 22\,\text{V}$

  • Row b: Given $V = 6\,\text{V}$, $Q = 25\,\text{C}$.     $W = V \times Q = 6\,\text{V} \times 25\,\text{C} = 150\,\text{J}$

  • Row c: Given $V = 9\,\text{V}$, $W = 54\,\text{J}$.     $Q = \frac{W}{V} = \frac{54\,\text{J}}{9\,\text{V}} = 6\,\text{C}$

  • Row d: Given $V = 4.5\,\text{V}$, $W = 2.5\,\text{J}$.     $Q = \frac{W}{V} = \frac{2.5\,\text{J}}{4.5\,\text{V}} = \frac{5}{9} \approx 0.56\,\text{C}$

• Work and Energy Applications:

  • Question 19: Calculate the potential difference across a resistor requiring $20\,\text{J}$ of work to move $8\,\text{C}$ through it:     $V = \frac{W}{Q} = \frac{20\,\text{J}}{8\,\text{C}} = 2.5\,\text{V}$

  • Question 20: Calculate work done when moving $1.5\,\text{C}$ through a potential difference of $220\,\text{V}$:     $W = Q \times V = 1.5\,\text{C} \times 220\,\text{V} = 330\,\text{J}$

  • Question 21: If $60\,\text{J}$ of work is performed to move charge through a potential difference of $10\,\text{V}$, calculate the charge moved:     $Q = \frac{W}{V} = \frac{60\,\text{J}}{10\,\text{V}} = 6\,\text{C}$

• Resistance, Current, and Potential Difference Calculations (Question 22):

  • Row a: Given $I = 6\,\text{A}$, $V = 24\,\text{V}$.     $R = \frac{V}{I} = \frac{24\,\text{V}}{6\,\text{A}} = 4\,\Omega$

  • Row b: Given $R = 25\,\Omega$, $I = 3.2\,\text{A}$.     $V = I \times R = 3.2\,\text{A} \times 25\,\Omega = 80\,\text{V}$

  • Row c: Given $I = 0.2\,\text{A}$, $V = 16\,\text{V}$.     $R = \frac{V}{I} = \frac{16\,\text{V}}{0.2\,\text{A}} = 80\,\Omega$

  • Row d: Given $R = 4.5\,\Omega$, $V = 0.9\,\text{V}$.     $I = \frac{V}{R} = \frac{0.9\,\text{V}}{4.5\,\Omega} = 0.2\,\text{A}$

• Ohm's Law Applications:

  • Question 23: When the potential difference across a resistor is $6\,\text{V}$, current is $2\,\text{A}$. Determine resistance:     $R = \frac{V}{I} = \frac{6\,\text{V}}{2\,\text{A}} = 3\,\Omega$

  • Question 24: Calculate potential difference required to drive $6\,\text{A}$ through a $12\,\Omega$ resistor:     $V = I \times R = 6\,\text{A} \times 12\,\Omega = 72\,\text{V}$

  • Question 25: A potential difference of $200\,\text{V}$ is applied to an $80\,\Omega$ conductor. Calculate current:     $I = \frac{V}{R} = \frac{200\,\text{V}}{80\,\Omega} = 2.5\,\text{A}$

  • Question 26: Across a resistor with potential difference $30\,\text{V}$, a charge of $120\,\text{C}$ flows in $15\,\text{s}$.

  • (a) Current:
          $I = \frac{Q}{t} = \frac{120\,\text{C}}{15\,\text{s}} = 8\,\text{A}$

  • (b) Resistance:
          $R = \frac{V}{I} = \frac{30\,\text{V}}{8\,\text{A}} = 3.75\,\Omega$

  • Question 27: A potential difference of $20\,\text{V}$ yields a current of $0.4\,\text{A}$.

  • (a) Charge flowing per minute ($t = 60\,\text{s}$):
          $Q = I \times t = 0.4\,\text{A} \times 60\,\text{s} = 24\,\text{C}$

  • (b) Value of the resistor:
          $R = \frac{V}{I} = \frac{20\,\text{V}}{0.4\,\text{A}} = 50\,\Omega$

  • Question 28: A potential difference of $24\,\text{V}$ is applied across a $60\,\Omega$ resistor.

  • (a) Charge passing through in $10\,\text{s}$:
          $I = \frac{V}{R} = \frac{24\,\text{V}}{60\,\Omega} = 0.4\,\text{A}$       $Q = I \times t = 0.4\,\text{A} \times 10\,\text{s} = 4\,\text{C}$

  • (b) Work done in this time:
          $W = Q \times V = 4\,\text{C} \times 24\,\text{V} = 96\,\text{J}$

  • Question 29: A potential difference of $6\,\text{V}$ is maintained across a $20\,\Omega$ resistor.

  • (a) Charge passing per minute ($t = 60\,\text{s}$):
          $I = \frac{V}{R} = \frac{6\,\text{V}}{20\,\Omega} = 0.3\,\text{A}$       $Q = I \times t = 0.3\,\text{A} \times 60\,\text{s} = 18\,\text{C}$

  • (b) Work done in this time:
          $W = Q \times V = 18\,\text{C} \times 6\,\text{V} = 108\,\text{J}$

Complex Circuit Network Analysis

• Series Circuit Analysis (Question 30):

  • Three resistors ($R_1 = 6\,\Omega$, $R_2 = 1\,\Omega$, and $R_3 = 3\,\Omega$) are wired in series across a $20\,\text{V}$ DC source.

• (a) Total Effective Resistance:
      $R_{\text{TOT}} = R_1 + R_2 + R_3 = 6\,\Omega + 1\,\Omega + 3\,\Omega = 10\,\Omega$

• (b) Circuit Current:
      $I = \frac{V}{R_{\text{TOT}}} = \frac{20\,\text{V}}{10\,\Omega} = 2\,\text{A}$

• (c) Potential Difference Across the $3\,\Omega$ Resistor:
      $V_{3\Omega} = I \times R_3 = 2\,\text{A} \times 3\,\Omega = 6\,\text{V}$

  • Parallel Resistor Network Analysis (Question 32):

• Network (a): Resistors of $6\,\Omega$ and $24\,\Omega$ in parallel:     $\frac{1}{R_{\text{eq}}} = \frac{1}{6} + \frac{1}{24} = \frac{4 + 1}{24} = \frac{5}{24}$     $R_{\text{eq}} = \frac{24}{5} = 4.8\,\Omega$

• Network (b): Resistors of $12\,\Omega$, $6\,\Omega$, and $4\,\Omega$ in parallel:     $\frac{1}{R_{\text{eq}}} = \frac{1}{12} + \frac{1}{6} + \frac{1}{4} = \frac{1 + 2 + 3}{12} = \frac{6}{12} = \frac{1}{2}$     $R_{\text{eq}} = 2\,\Omega$

• Network (c): Resistors of $20\,\Omega$, $5\,\Omega$, and $4\,\Omega$ in parallel:     $\frac{1}{R_{\text{eq}}} = \frac{1}{20} + \frac{1}{5} + \frac{1}{4} = \frac{1 + 4 + 5}{20} = \frac{10}{20} = \frac{1}{2}$     $R_{\text{eq}} = 2\,\Omega$

  • Compound and Geometric Resistor Combinations (Question 33):

• Combination (d): Top parallel branch comprises a series combination of $6\,\Omega$ and $4\,\Omega$; bottom branch comprises a $30\,\Omega$ resistor.

  • Resistance of top branch:
          $R_{\text{top}} = 6\,\Omega + 4\,\Omega = 10\,\Omega$

  • Total equivalent resistance of the parallel combination:
          $\frac{1}{R_{\text{eq}}} = \frac{1}{10} + \frac{1}{30} = \frac{3 + 1}{30} = \frac{4}{30}$       $R_{\text{eq}} = \frac{30}{4} = 7.5\,\Omega$

• Combination (e): Square closed-loop network with terminal leads connected across opposite diagonal vertices. The perimeter consists of resistors of $4\,\Omega$, $2\,\Omega$, $2\,\Omega$, and $4\,\Omega$.

  • Current entering the node splits into two parallel paths:

    • Path 1: $4\,\Omega + 2\,\Omega = 6\,\Omega$

    • Path 2: $4\,\Omega + 2\,\Omega = 6\,\Omega$

  • Total equivalent resistance:
          $R_{\text{eq}} = \frac{6\,\Omega \times 6\,\Omega}{6\,\Omega + 6\,\Omega} = \frac{36}{12} = 3\,\Omega$

• Combination (f): Triangular (delta) network connected across the bottom base terminals. The base is a $3\,\Omega$ resistor; the upper apex path consists of a $2\,\Omega$ and a $4\,\Omega$ resistor in series.

  • Upper path series resistance:
          $R_{\text{upper}} = 2\,\Omega + 4\,\Omega = 6\,\Omega$

  • Base path resistance:
          $R_{\text{base}} = 3\,\Omega$

  • Total equivalent resistance:
          $R_{\text{eq}} = \frac{6\,\Omega \times 3\,\Omega}{6\,\Omega + 3\,\Omega} = \frac{18}{9} = 2\,\Omega$

  • Series-Parallel Network Analysis (Question 34):

• A circuit consists of a $10\,\Omega$ resistor connected in series with a parallel combination of a $6\,\Omega$ and a $3\,\Omega$ resistor.

• (a) Total Resistance of the Circuit:

  • Calculate parallel bank equivalent resistance:
          $R_{\text{parallel}} = \frac{6\,\Omega \times 3\,\Omega}{6\,\Omega + 3\,\Omega} = \frac{18}{9} = 2\,\Omega$

  • Add the series resistor:
          $R_{\text{TOT}} = 10\,\Omega + R_{\text{parallel}} = 10\,\Omega + 2\,\Omega = 12\,\Omega$

• (b) Readings on Main Ammeters ($A_1$ and $A_2$):

  • Located on the main branch, both ammeters record total circuit current:
          $I_{\text{main}} = \frac{V_{\text{supply}}}{R_{\text{TOT}}} = \frac{V_{\text{supply}}}{12\,\Omega}$

• (c) Reading on Voltmeter $V_2$:

  • Measures potential difference across the $10\,\Omega$ series resistor:
          $V_2 = I_{\text{main}} \times 10\,\Omega$

• (d) Reading on Voltmeters $V_3$ and $V_4$:

  • Connected across the parallel network; both register the identical potential drop across the parallel combination:
          $V_3 = V_4 = I_{\text{main}} \times 2\,\Omega = V_{\text{parallel}}$

• (e) Readings on Branch Ammeters ($A_3$ and $A_4$):

  • Current divides through the parallel branches inversely according to branch resistance:
          $I_{A3} = \frac{V_{\text{parallel}}}{6\,\Omega}$       $I_{A4} = \frac{V_{\text{parallel}}}{3\,\Omega}$

  • Ratio check: Current through $A_4$ is twice that through $A_3$ ($I_{A4} = 2 \times I_{A3}$), and $I_{A3} + I_{A4} = I_{\text{main}}$.

  • Dual-Control Doorbell Circuit Design (Question 35):

• Design goal: Connect a single audible doorbell unit, a common DC electrical power source, and two push-button switches (front door and back door) so that each switch can independently activate the doorbell.

• Circuit configuration:

  • The front door push-button switch and back door push-button switch are wired in parallel with each other.

  • This parallel switch sub-assembly is wired in series with the single doorbell unit and the DC power source.

  • Functional operation: Closing the front door switch completes the conducting circuit loop independently of the back door switch; similarly, closing the back door switch completes the loop independently. If both switches are closed simultaneously, the circuit remains closed and the bell rings normally.