Comprehensive Study Notes on Electricity, Resistance, and Circuits

Electric Current and Circuit

• Definition of Electric Current:

  • Electric current is the flow of electric charges through a conductor, such as a metallic wire.
  • Quantitatively, electric current is defined as the rate of flow of electric charges, represented as the amount of charge flowing across a particular cross-sectional area per unit time.
  • If a net electric charge $Q$ flows across a cross-section of a conductor in time $t$, the current $I$ is expressed as:     $I = \frac{Q}{t}$
  • Alternatively, the net charge is given by:     $Q = I t$

• Electric Circuit:

  • An electric circuit is a continuous and closed conducting path through which an electric current flows.
  • If the path is broken at any point, or if a switch is turned off, the circuit becomes open and the current stops flowing.
  • In a handheld torch, cells or a battery placed in proper sequence provide the flow of charges through the miniature bulb to make it glow.

• Direction of Electric Current:

  • In metallic conductors, negatively charged subatomic particles known as electrons constitute the actual flow of physical charge.
  • When the phenomenon of electricity was first discovered, electrons had not yet been identified; electric current was originally considered to be the flow of positive charges.
  • Historically, the direction of motion of positive charge was defined as the direction of current.
  • Conventional Current Direction: By international convention, the direction of electric current in an electric circuit is taken as opposite to the direction of the flow of electrons. Electric current flows externally from the positive terminal of a cell/battery to the negative terminal, whereas electrons drift from the negative terminal to the positive terminal.

• Units of Charge and Current:

  • SI Unit of Electric Charge: The coulomb ($\text{C}$).
  • One electron possesses an intrinsic negative charge of $1.6 \times 10^{-19}\,\text{C}$.
  • A net charge of $1\,\text{C}$ is equivalent to the collective charge contained in approximately $6 \times 10^{18}$ electrons, calculated via $n = \frac{1}{1.6 \times 10^{-19}} \approx 6.25 \times 10^{18}$.
  • SI Unit of Electric Current: The ampere ($\text{A}$), named in honor of the French scientist André-Marie Ampère ($1775-1836$).
  • One ampere corresponds to the transfer of one coulomb of charge per second across a cross-section:       $1\,\text{A} = \frac{1\,\text{C}}{1\,\text{s}} = 1\,\text{C}\,\text{s}^{-1}$
  • Submultiples of Current:
  • Milliampere: $1\,\text{mA} = 10^{-3}\,\text{A}$
  • Microampere: $1\,\mu\text{A} = 10^{-6}\,\text{A}$

• Ammeter:

  • An ammeter is an instrument designed to measure electric current flowing through a branch or circuit.
  • It has a very low internal resistance and must always be connected in series with the component whose current is being evaluated.

• Microscopic Flow and Drift Speed:

  • In a solid metallic conductor, lattice atoms are closely packed together, yet conduction electrons move freely among the lattice ions almost as if moving in a vacuum.
  • When a potential gradient exists, electrons move through the wire with an average net drift speed, which is around $1\,\text{mm}\,\text{s}^{-1}$ in a typical copper wire.
  • Despite this slow physical migration, an electric bulb illuminates almost instantaneously when the switch is engaged because the electromagnetic field travels along the circuit at near the speed of light, initiating electron movement across all sections simultaneously.

• Worked Example: Charge Calculation:

  • Problem: A current of $0.5\,\text{A}$ is drawn by a filament of an electric bulb for $10\,\text{minutes}$. Find the amount of electric charge that flows through the circuit.
  • Solution:
  • Current, $I = 0.5\,\text{A}$
  • Time, $t = 10\,\text{min} = 10 \times 60\,\text{s} = 600\,\text{s}$
  • Net charge, $Q = I t$
  • $Q = 0.5\,\text{A} \times 600\,\text{s} = 300\,\text{C}$

Electric Potential and Potential Difference

• Concept of Electric Potential Difference:

  • Free electric charges do not move through a metallic conductor spontaneously; gravitational force has no practical role in driving their motion.
  • Electrons migrate only when there is an unequal distribution of electric pressure between two terminals, termed the electric potential difference.
  • A chemical cell or battery creates this potential difference between its terminals via internal chemical reactions, even under open-circuit conditions when no current is drawn.
  • Connecting the cell to an external conducting loop enables the potential difference to exert an electrostatic force on charges, producing current. To maintain this current, the cell expends its stored chemical energy.

• Mathematical Formulation:

  • The electric potential difference $V$ between two points in an energized circuit is defined as the work done $W$ to transport a unit positive charge $Q$ from one point to the other:     $\text{Potential Difference } (V) = \frac{\text{Work Done } (W)}{\text{Charge } (Q)}$$V = \frac{W}{Q}$
  • Rearranging for total work done yields:     $W = V Q$

• Units and Measurement:

  • SI Unit of Potential Difference: The volt ($\text{V}$), named after the Italian physicist Alessandro Volta ($1745-1827$).
  • Definition of One Volt: One volt represents the potential difference between two points in a current-carrying conductor when $1\,\text{joule}$ of work is performed in transferring a charge of $1\,\text{coulomb}$ from one point to the other:     $1\,\text{volt} = \frac{1\,\text{joule}}{1\,\text{coulomb}}$$1\,\text{V} = 1\,\text{J}\,\text{C}^{-1}$
  • Voltmeter: An instrument utilized to measure potential difference between two points in an active circuit. It possesses extremely high internal resistance and is always connected in parallel across the pair of points being tested.

• Worked Example: Work Done Calculation:

  • Problem: How much work is done in moving a charge of $2\,\text{C}$ across two points having a potential difference of $12\,\text{V}$?
  • Solution:
  • Amount of charge, $Q = 2\,\text{C}$
  • Potential difference, $V = 12\,\text{V}$
  • Work done, $W = V Q$
  • $W = 12\,\text{V} \times 2\,\text{C} = 24\,\text{J}$

Circuit Diagram and Standard Symbols

• Circuit Diagrams: A circuit diagram is a schematic representation of an electrical network using standardized pictorial symbols to represent each functional electrical component.

• Summary of Conventional Circuit Symbols:
  • Electric Cell: Depicted with a pair of parallel lines—a longer thin line representing the positive terminal and a shorter, thicker line representing the negative terminal.
  • Battery: Formed by multiple cells linked in series, alternating long and short parallel lines.
  • Plug Key / Switch (Open): Indicated by empty parentheses ( ) along the conduction line.
  • Plug Key / Switch (Closed): Indicated by parentheses with a center dot (•) indicating continuity.
  • Wire Joint: A junction where perpendicular or branch conducting lines intersect with a prominent connection dot.
  • Wires Crossing Without Joining: A continuous line with an arched loop over an intersecting line to show lack of physical electrical contact.
  • Electric Bulb: A loop representing a filament within an enclosure, or a bulb schematic icon.
  • Resistor of Resistance $R$: A recurring zig-zag line denoting fixed electrical resistance.
  • Variable Resistance or Rheostat: A zig-zag line overlaid with a diagonal crossing arrow or a top-mounted slider terminal arrow.
  • Ammeter: A circle bearing the letter A with polarity terminals marked ($+\text{ and } -$).
  • Voltmeter: A circle bearing the letter V with polarity terminals marked ($+\text{ and } -$).

Ohm's Law and Electrical Resistance

• Discovery and Statement of Ohm's Law:
  • In $1827$, the German physicist Georg Simon Ohm ($1787-1854$) established the mathematical relationship between the current flowing through a metallic wire and the potential difference applied across it.
  • Ohm's Law: The potential difference ($V$) across the ends of a given metallic conductor in an electric circuit is directly proportional to the electric current ($I$) flowing through it, provided its temperature and other physical conditions remain constant:     $V \propto I$$\frac{V}{I} = \text{constant} = R$$V = I R$$I = \frac{V}{R}$

• Experimental Verification:

  • In an experiment utilizing a nichrome wire $XY$ of length $0.5\,\text{m}$ (an alloy composed of Nickel, Chromium, Manganese, and Iron), an ammeter, a voltmeter, and four identical cells rated at $1.5\,\text{V}$ each are connected:
  • With 1 cell ($V = 1.5\,\text{V}$): $I = 0.5\,\text{A} \implies \frac{V}{I} = 3\,\text{V}\,\text{A}^{-1}$
  • With 2 cells ($V = 3.0\,\text{V}$): $I = 1.0\,\text{A} \implies \frac{V}{I} = 3\,\text{V}\,\text{A}^{-1}$
  • With 3 cells ($V = 4.5\,\text{V}$): $I = 1.5\,\text{A} \implies \frac{V}{I} = 3\,\text{V}\,\text{A}^{-1}$
  • With 4 cells ($V = 6.0\,\text{V}$): $I = 2.0\,\text{A} \implies \frac{V}{I} = 3\,\text{V}\,\text{A}^{-1}$
  • A plot of potential difference $V$ versus current $I$ yields a straight line passing through the origin, demonstrating that the ratio $\frac{V}{I}$ is strictly constant.

• Electrical Resistance ($R$):

  • Resistance is the intrinsic property of a conductor by virtue of which it opposes or impedes the flow of electric charge through it.
  • SI Unit of Resistance: The ohm, denoted by the Greek capital letter $\Omega$.
  • Definition of One Ohm: If the potential difference across the two ends of a conductor is $1\,\text{V}$ and the resulting current passing through it is $1\,\text{A}$, the resistance of the conductor is exactly $1\,\Omega$:     $1\,\text{ohm} = \frac{1\,\text{volt}}{1\,\text{ampere}}$$1\,\Omega = 1\,\text{V}\,\text{A}^{-1}$
  • Current Dependence: The current through a conductor is inversely proportional to its resistance ($I \propto \frac{1}{R}$). If resistance is doubled, current is reduced to half its original value, assuming constant voltage.

• Component Classification by Conductance:

  • Resistor: A component constructed to introduce an appreciable, controlled value of electrical resistance into a circuit.
  • Rheostat (Variable Resistor): An adjustable electromechanical device utilized to continuously regulate current within a network without changing the voltage source.
  • Good Conductor: A substance offering very low resistance to the passage of electrons (e.g., silver, copper, aluminum).
  • Poor Conductor: A substance offering high resistance to electric current (e.g., specific alloys, dirty metals).
  • Insulator: A material that possesses exceedingly high resistance, permitting virtually no charge transfer (e.g., rubber, glass, ebonite).
  • Physical Mechanism: Inside a material, migrating electrons collide with surrounding stationary atoms or ions. These electrostatic attractions and collisions retard electron drift, giving rise to resistance.

Factors Affecting Electrical Resistance and Resistivity

• Experimental Observations of Resistance Factors:

  • For a nichrome wire of length $l$, let the circuit current be $1\,\text{A}$.
  • When the length is doubled to $2l$, the ammeter reading falls to $0.5\,\text{A}$, demonstrating that resistance doubles when length doubles.
  • When replacing the wire with a thicker nichrome wire of identical length $l$ having twice the cross-sectional area, the current reading increases to $2\,\text{A}$, indicating that resistance is halved.
  • When replacing the wire with a copper wire of identical length $l$ and cross-sectional area $A$, the ammeter reading changes substantially, demonstrating dependence on material composition.

• Governing Proportionalities:

  • Resistance $R$ of a uniform conductor is directly proportional to its length ($l$):     $R \propto l$
  • Resistance $R$ is inversely proportional to its cross-sectional area ($A$):     $R \propto \frac{1}{A}$
  • Combining these geometric dependencies:     $R \propto \frac{l}{A}$$R = \rho \frac{l}{A}$

• Electrical Resistivity ($\rho$):

  • The proportionality constant $\rho$ (rho) is defined as the electrical resistivity of the conductor material.
  • Resistivity is a characteristic intrinsic physical property of a material that is independent of its geometric dimensions (length and area).
  • SI Unit of Resistivity: The ohm-meter ($\Omega\,\text{m}$).
  • Temperature Dependence: Both electrical resistance and electrical resistivity vary systematically with temperature.

• Material Classes and Resistivity Values (at $20^\circ\text{C}$):

  • Conductors (Metals): Possess very low resistivity ranging from $10^{-8}\,\Omega\,\text{m}$ to $10^{-6}\,\Omega\,\text{m}$.
  • Silver: $1.60 \times 10^{-8}\,\Omega\,\text{m}$
  • Copper: $1.62 \times 10^{-8}\,\Omega\,\text{m}$
  • Aluminium: $2.63 \times 10^{-8}\,\Omega\,\text{m}$
  • Tungsten: $5.20 \times 10^{-8}\,\Omega\,\text{m}$
  • Nickel: $6.84 \times 10^{-8}\,\Omega\,\text{m}$
  • Iron: $10.0 \times 10^{-8}\,\Omega\,\text{m}$
  • Chromium: $12.9 \times 10^{-8}\,\Omega\,\text{m}$
  • Mercury: $94.0 \times 10^{-8}\,\Omega\,\text{m}$
  • Manganese: $1.84 \times 10^{-6}\,\Omega\,\text{m}$
  • Alloys: Exhibit higher resistivities than their constituent metals and do not oxidize (burn) readily at elevated temperatures.
  • Constantan ($\text{alloy of Cu and Ni}$): $49 \times 10^{-6}\,\Omega\,\text{m}$
  • Manganin ($\text{alloy of Cu, Mn, and Ni}$): $44 \times 10^{-6}\,\Omega\,\text{m}$
  • Nichrome ($\text{alloy of Ni, Cr, Mn, and Fe}$): $100 \times 10^{-6}\,\Omega\,\text{m}$
  • Insulators: Characterized by extremely high resistivities ranging between $10^{10}\,\Omega\,\text{m}$ and $10^{17}\,\Omega\,\text{m}$.
  • Glass: $10^{10} - 10^{14}\,\Omega\,\text{m}$
  • Hard rubber: $10^{13} - 10^{16}\,\Omega\,\text{m}$
  • Ebonite: $10^{15} - 10^{17}\,\Omega\,\text{m}$
  • Diamond: $10^{12} - 10^{13}\,\Omega\,\text{m}$
  • Paper (dry): $10^{12}\,\Omega\,\text{m}$

• Engineering Applications Based on Resistivity:

  • Electrical Heating Appliances: Alloys like nichrome are used as heating elements in electric irons and toasters because of their high resistivity and high oxidation resistance at elevated temperatures.
  • Incandescent Filament Lamps: Tungsten is selected exclusively for bulb filaments because of its high melting point ($3380^\circ\text{C}$) and stability under prolonged incandescence.
  • Transmission Cables: Highly conductive metals such as copper and aluminium are employed for electrical power transmission lines to minimize resistive energy dissipation.

• Worked Examples: Resistance and Resistivity:

  • Problem 1 (Current Demand Variation):
  • (a) How much current will an electric bulb draw from a $220\,\text{V}$ source, if the resistance of bulb filament is $1200\,\Omega$?       $I = \frac{V}{R} = \frac{220\,\text{V}}{1200\,\Omega} \approx 0.18\,\text{A}$
  • (b) How much current will an electric heater draw from a $220\,\text{V}$ source, if the resistance of heater coil is $100\,\Omega$?       $I = \frac{V}{R} = \frac{220\,\text{V}}{100\,\Omega} = 2.2\,\text{A}$
  • Conclusion: Devices draw vastly different currents from the same voltage source depending on their internal resistance.
  • Problem 2 (Linearity with Voltage Change):
  • The potential difference between the terminals of an electric heater is $60\,\text{V}$ when it draws a current of $4\,\text{A}$. What current will the heater draw if the potential difference is increased to $120\,\text{V}$?
  • Resistance of heater coil:       $R = \frac{V}{I} = \frac{60\,\text{V}}{4\,\text{A}} = 15\,\Omega$
  • Current at $120\,\text{V}$:       $I = \frac{V}{R} = \frac{120\,\text{V}}{15\,\Omega} = 8\,\text{A}$
  • Problem 3 (Calculating Resistivity and Identifying Material):
  • Resistance of a metal wire of length $1\,\text{m}$ is $26\,\Omega$ at $20^\circ\text{C}$. If diameter $d = 0.3\,\text{mm}$, determine the resistivity and predict the material.
  • Cross-sectional diameter: $d = 0.3\,\text{mm} = 3 \times 10^{-4}\,\text{m}$
  • Cross-sectional area:       $A = \frac{\pi d^2}{4}$
  • Resistivity formula:       $\rho = \frac{R A}{l} = \frac{R \pi d^2}{4 l}$
  • Substitution:       $\rho = \frac{26\,\Omega \times \pi \times (3 \times 10^{-4}\,\text{m})^2}{4 \times 1\,\text{m}} = 1.84 \times 10^{-6}\,\Omega\,\text{m}$
  • Reference comparison reveals this resistivity matches manganese at $20^\circ\text{C}$.
  • Problem 4 (Geometric Reconfiguration):
  • A wire of given material having length $l$ and area of cross-section $A$ has a resistance of $4\,\Omega$. What would be the resistance of another wire of the same material having length $\frac{l}{2}$ and area of cross-section $2 A$?
  • Initial resistance: $R_1 = \rho \frac{l}{A} = 4\,\Omega$
  • New resistance:       $R_2 = \rho \frac{\frac{l}{2}}{2 A} = \frac{1}{4} \left(\rho \frac{l}{A}\right)$$R_2 = \frac{1}{4} R_1 = \frac{1}{4} \times 4\,\Omega = 1\,\Omega$

System of Resistors in Series

• Series Circuit Characteristics:

  • Two or more resistors are joined in series when they are connected end-to-end sequentially along a single conduction path.
  • The current $I$ remains identical through every component in the series string; the reading of an ammeter placed anywhere between individual resistors does not change.
  • The total potential difference $V$ supplied across the ends of the combination is equal to the algebraic sum of the potential drops across each individual resistor:     $V = V_1 + V_2 + V_3$

• Derivation of Equivalent Resistance ($R_s$):

  • Let the potential difference across resistors $R_1$, $R_2$, and $R_3$ carrying current $I$ be $V_1$, $V_2$, and $V_3$ respectively.
  • Applying Ohm's law to each individual resistor:     $V_1 = I R_1$$V_2 = I R_2$$V_3 = I R_3$
  • Applying Ohm's law to the complete equivalent circuit:     $V = I R_s$
  • Substituting into the sum of voltages:     $I R_s = I R_1 + I R_2 + I R_3$
  • Dividing both sides by common current $I$ yields:     $R_s = R_1 + R_2 + R_3$
  • General Principle: When multiple resistors are linked in series, the total equivalent resistance $R_s$ is strictly equal to the arithmetic sum of the individual component resistances, and is therefore always greater than any single resistor in the series branch.

• Worked Example: Series Combination:

  • Problem: An electric lamp of resistance $20\,\Omega$ and a conductor of $4\,\Omega$ resistance are connected in series to a $6\,\text{V}$ battery. Calculate:
  • (a) Total circuit resistance
  • (b) Current flowing through the circuit
  • (c) Potential difference across the electric lamp and the separate conductor
  • Solution:
  • (a) Total resistance:       $R_s = R_1 + R_2 = 20\,\Omega + 4\,\Omega = 24\,\Omega$
  • (b) Current through circuit:       $I = \frac{V}{R_s} = \frac{6\,\text{V}}{24\,\Omega} = 0.25\,\text{A}$
  • (c) Potential difference across lamp ($R_1$):       $V_1 = I R_1 = 0.25\,\text{A} \times 20\,\Omega = 5\,\text{V}$
  • Potential difference across conductor ($R_2$):       $V_2 = I R_2 = 0.25\,\text{A} \times 4\,\Omega = 1\,\text{V}$
  • Verification: Total applied voltage is preserved as $V_1 + V_2 = 5\,\text{V} + 1\,\text{V} = 6\,\text{V}$.

System of Resistors in Parallel

• Parallel Circuit Characteristics:

  • Resistors are connected in parallel when they are linked simultaneously across two common electrical nodes.
  • The potential difference $V$ across each branch in a parallel arrangement is identical and equals the total voltage applied across the network.
  • The total current $I$ entering the junction is divided among the branches, such that the main line current is equal to the sum of the separate branch currents:     $I = I_1 + I_2 + I_3$

• Derivation of Equivalent Resistance ($R_p$):

  • Let $R_p$ represent the equivalent resistance of the parallel combination across which potential difference $V$ exists.
  • By Ohm's law, total current is:     $I = \frac{V}{R_p}$
  • Branch currents through resistors $R_1$, $R_2$, and $R_3$ are given by:     $I_1 = \frac{V}{R_1}$$I_2 = \frac{V}{R_2}$$I_3 = \frac{V}{R_3}$
  • Substituting these into the current conservation expression:     $\frac{V}{R_p} = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3}$
  • Factoring out $V$ gives:     $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$
  • General Principle: The reciprocal of the equivalent resistance of a group of resistors joined in parallel equals the sum of the reciprocals of the individual resistances. The equivalent resistance $R_p$ is always smaller than the smallest individual resistance in the combination.

• Comparison: Series vs. Parallel Circuits:

  • Disadvantages of Series Circuits:
  • The electric current is constrained to be constant across all devices. Consequently, it is unfeasible to place an electric bulb and a high-power electric heater in series, as they require substantially different operating currents.
  • Component interdependency is total: if a single component fails or a fault develops, the entire path is broken, causing all downstream devices to shut off (e.g., locating a single failed bulb in a long string of festive decorative lights is extremely tedious).
  • Advantages of Parallel Circuits:
  • A parallel configuration divides the total current among different branches according to each appliance's individual resistance, allowing components with diverse current demands to operate independently at full source voltage.
  • The effective combined resistance of the system decreases, ensuring optimal power delivery from the source.
  • If one load or appliance disconnects or fails, current to the parallel branches is uninterrupted.

• Worked Examples: Parallel and Complex Resistor Networks:

  • Problem 1 (Parallel Branch Currents and Equivalent Resistance):
  • Resistors $R_1 = 5\,\Omega$, $R_2 = 10\,\Omega$, and $R_3 = 30\,\Omega$ are connected in parallel across a $12\,\text{V}$ battery.
  • (a) Current through each resistor:       $I_1 = \frac{V}{R_1} = \frac{12\,\text{V}}{5\,\Omega} = 2.4\,\text{A}$$I_2 = \frac{V}{R_2} = \frac{12\,\text{V}}{10\,\Omega} = 1.2\,\text{A}$$I_3 = \frac{V}{R_3} = \frac{12\,\text{V}}{30\,\Omega} = 0.4\,\text{A}$
  • (b) Total current entering network:       $I = I_1 + I_2 + I_3 = 2.4\,\text{A} + 1.2\,\text{A} + 0.4\,\text{A} = 4\,\text{A}$
  • (c) Total equivalent resistance $R_p$:       $\frac{1}{R_p} = \frac{1}{5} + \frac{1}{10} + \frac{1}{30} = \frac{6 + 3 + 1}{30} = \frac{10}{30} = \frac{1}{3}\,\Omega^{-1}$$R_p = 3\,\Omega$
  • Problem 2 (Compound Series-Parallel Circuit):
  • A network has two parallel clusters connected in series across a $12\,\text{V}$ battery: cluster 1 has $R_1 = 10\,\Omega$ and $R_2 = 40\,\Omega$; cluster 2 has $R_3 = 30\,\Omega$, $R_4 = 20\,\Omega$, and $R_5 = 60\,\Omega$.
  • Calculate (a) total equivalent resistance, and (b) total circuit current.
  • Equivalent resistance of cluster 1 ($R'$):       $\frac{1}{R'} = \frac{1}{10} + \frac{1}{40} = \frac{4 + 1}{40} = \frac{5}{40} = \frac{1}{8}\,\Omega^{-1} \implies R' = 8\,\Omega$
  • Equivalent resistance of cluster 2 ($R''$):       $\frac{1}{R''} = \frac{1}{30} + \frac{1}{20} + \frac{1}{60} = \frac{2 + 3 + 1}{60} = \frac{6}{60} = \frac{1}{10}\,\Omega^{-1} \implies R'' = 10\,\Omega$
  • Total circuit resistance ($R$):       $R = R' + R'' = 8\,\Omega + 10\,\Omega = 18\,\Omega$
  • Total current ($I$):       $I = \frac{V}{R} = \frac{12\,\text{V}}{18\,\Omega} \approx 0.67\,\text{A}$

Heating Effect of Electric Current (Joule's Heating)

• Source of Energy and Dissipation Mechanism:

  • A battery supplies electrical energy by establishing an electrostatic potential difference that accelerates free conduction electrons across a circuit.
  • While a fraction of this electrical energy is converted into useful mechanical work (e.g., driving the motor of an electric fan), the remaining portion is dissipated as thermal energy due to continuous collisions between moving charge carriers and the atomic lattice.
  • If a circuit is purely resistive—containing only passive resistors connected to an energy source—the source energy is converted entirely into heat.

• Derivation of Joule's Heating Equation:

  • Consider current $I$ flowing across a resistor with resistance $R$ subjected to potential difference $V$ for time $t$.
  • The quantity of charge moved through the potential difference is:     $Q = I t$
  • The total electrostatic work performed by the source in transferring charge $Q$ through potential $V$ is:     $W = V Q$
  • Therefore, power input supplied to the circuit is:     $P = \frac{W}{t} = \frac{V Q}{t} = V I$
  • Total energy supplied by the source in duration $t$ must be dissipated as heat $H$:     $H = P t = V I t$
  • Applying Ohm's law ($V = I R$) gives:     $H = (I R) I t$$H = I^2 R t$

• Joule's Law of Heating:

  • The heat generated in a resistor is directly proportional to:
  1. The square of the current for a given resistance ($H \propto I^2$).
  2. The resistance for a given current ($H \propto R$).
  3. The time for which the current passes through the resistor ($H \propto t$).
  • In situations where a load is driven by a constant voltage source $V$, the relation $I = \frac{V}{R}$ is evaluated first before using $H = I^2 R t$, or alternatively expressed directly as $H = \frac{V^2}{R} t$.

• Worked Examples: Heating Effect:

  • Problem 1 (Electric Iron at Variable Power Ratings):
  • An electric iron connected to a $220\,\text{V}$ supply consumes energy at a rate of $840\,\text{W}$ at maximum heating, and $360\,\text{W}$ at minimum heating. Find the current and resistance for both operational states.
  • Since $P = V I$, current is $I = \frac{P}{V}$, and resistance is $R = \frac{V}{I}$.
  • (a) At maximum heating rate ($P = 840\,\text{W}$):       $I = \frac{840\,\text{W}}{220\,\text{V}} \approx 3.82\,\text{A}$$R = \frac{V}{I} = \frac{220\,\text{V}}{3.82\,\text{A}} \approx 57.60\,\Omega$
  • (b) At minimum heating rate ($P = 360\,\text{W}$):       $I = \frac{360\,\text{W}}{220\,\text{V}} \approx 1.64\,\text{A}$$R = \frac{V}{I} = \frac{220\,\text{V}}{1.64\,\text{A}} \approx 134.15\,\Omega$
  • Problem 2 (Determining Potential Difference from Heat Dissipation):
  • A heat rate of $100\,\text{J}$ per second is produced in a $4\,\Omega$ resistor. Determine the potential difference across the resistor.
  • Given parameters: $H = 100\,\text{J}$, $R = 4\,\Omega$, $t = 1\,\text{s}$
  • Using $H = I^2 R t$ to determine current:       $I = \sqrt{\frac{H}{R t}} = \sqrt{\frac{100\,\text{J}}{4\,\Omega \times 1\,\text{s}}} = \sqrt{25} = 5\,\text{A}$
  • Applying Ohm's law to find potential difference:       $V = I R = 5\,\text{A} \times 4\,\Omega = 20\,\text{V}$

Practical Applications of the Heating Effect

• Thermal Domestic Appliances:

  • Joule heating is exploited in commercial and household appliances including electric laundry irons, electric toasters, electric ovens, electric kettles, and room heaters.
  • These devices incorporate heating elements formed of high-resistivity alloys (such as nichrome) capable of operating at red-hot temperatures without oxidising or melting.

• Incandescent Electric Bulb:

  • Light is emitted when an electric current heats a thin metallic filament to extreme incandescence.
  • Filament Requirements: High melting point and structural resilience at elevated temperatures. Tungsten is utilized due to its very high melting point of $3380^\circ\text{C}$.
  • The filament is mounted on an insulating support to minimize thermal conduction loss.
  • Bulbs are evacuated and filled with chemically non-reactive gases, such as nitrogen and argon, to suppress tungsten vaporization and prolong filament lifespan.
  • Most electrical energy consumed by the filament is emitted as thermal radiation (heat), with only a minor fraction converted into visible light.

• Electric Fuse:
  • A fuse is an electrical safety component designed to protect building circuits and delicate equipment from overcurrent damage caused by short-circuiting or overloading.
  • Placement: Always placed in series with the live line leading to the device.
  • Composition: Comprises a calibrated segment of wire made of a metal or alloy with a specific, relatively low melting point (e.g., aluminium, copper, iron, or lead).
  • Working Principle: If a current exceeds the rated threshold, excessive Joule heat ($I^2 R t$) raises the wire's temperature above its melting point, causing it to liquefy and sever the circuit, interrupting current flow.
  • Construction: The active fuse wire is enclosed in a protective cartridge composed of porcelain or equivalent ceramic material featuring metallic end-caps.
  • Standard Domestic Ratings: Fuses are produced with discrete current ratings such as $1\,\text{A}$, $2\,\text{A}$, $3\,\text{A}$, $5\,\text{A}$, and $10\,\text{A}$.
  • Worked Sizing Calculation:
  • An electric iron consuming $1\,\text{kW}$ ($1000\,\text{W}$) operated on a standard $220\,\text{V}$ supply line draws:       $I = \frac{P}{V} = \frac{1000\,\text{W}}{220\,\text{V}} \approx 4.54\,\text{A}$
  • To support this steady operational current without spurious tripping while still shielding the network from major surges, a $5\,\text{A}$ fuse must be deployed.

Electric Power and Commercial Electrical Energy

• Concept and Mathematical Formulas for Electric Power:

  • Electric power $P$ is the rate at which electrical work is done, or the rate at which electrical energy is consumed or dissipated within an electric circuit:     $P = \frac{W}{t}$
  • From the Joule heating relation ($W = V I t$):     $P = V I$
  • Using Ohm's law substitutions ($V = I R$ or $I = \frac{V}{R}$):     $P = I^2 R = \frac{V^2}{R}$

• Units of Electric Power:

  • SI Unit of Electric Power: The watt ($\text{W}$).
  • Definition of One Watt: One watt is the power consumed by an electrical load that draws a current of $1\,\text{A}$ when operated across a potential difference of $1\,\text{V}$:     $1\,\text{W} = 1\,\text{volt} \times 1\,\text{ampere} = 1\,\text{V}\,\text{A} = 1\,\text{J}\,\text{s}^{-1}$
  • Multiple Unit: Because the watt is relatively small for industrial and household indexing, the kilowatt ($\text{kW}$) is frequently used:     $1\,\text{kW} = 1000\,\text{W}$

• Electrical Energy and Commercial Billing Units:

  • Electrical energy $E$ represents the product of power input and the duration of its consumption:     $E = P t$
  • Unit of Electrical Energy: The watt-hour ($\text{W}\,\text{h}$), defined as the energy consumed when $1\,\text{watt}$ of power is used continuously for $1\,\text{hour}$.
  • Commercial Unit of Electrical Energy: The kilowatt-hour ($\text{kW}\,\text{h}$), conventionally referred to on utility electricity bills simply as a 'unit'.
  • Conversion of $1\,\text{kW}\,\text{h}$ into Joules:$1\,\text{kW}\,\text{h} = 1000\,\text{W} \times 3600\,\text{s}$$1\,\text{kW}\,\text{h} = 3.6 \times 10^6\,\text{W}\,\text{s} = 3.6 \times 10^6\,\text{J}$
  • Electron Conservation Principle: Electrons themselves are never destroyed or consumed in an operational circuit; consumers pay utilities solely for the energy required to impart drift velocity and push those electrons through connected gadgets against resistance.

• Worked Examples: Power and Energy Costs:

  • Problem 1 (Bulb Power from Generator):
  • An electric bulb is connected to a $220\,\text{V}$ generator. The resulting current drawn is $0.50\,\text{A}$. Determine the power of the bulb.
  • Calculation:       $P = V I = 220\,\text{V} \times 0.50\,\text{A} = 110\,\text{J}\,\text{s}^{-1} = 110\,\text{W}$
  • Problem 2 (Monthly Refrigerator Energy Cost):
  • An electric refrigerator rated at $400\,\text{W}$ operates $8.0\,\text{hours/day}$. Calculate the total energy cost to operate it for an entire month ($30\,\text{days}$) at an electricity tariff rate of $\text{Rs } 3.00$ per $\text{kW}\,\text{h}$.
  • Total energy consumption over 30 days:       $E = 400\,\text{W} \times 8.0\,\frac{\text{hours}}{\text{day}} \times 30\,\text{days} = 96000\,\text{W}\,\text{h}$
  • Converting to commercial units ($\text{kW}\,\text{h}$):       $E = \frac{96000\,\text{W}\,\text{h}}{1000} = 96\,\text{kW}\,\text{h}$
  • Cost of operating refrigerator:       $\text{Total Cost} = 96\,\text{kW}\,\text{h} \times \text{Rs } 3.00\,\text{per } \text{kW}\,\text{h} = \text{Rs } 288.00$