Comprehensive Study Guide to DC Electric Circuits: Ohm's Law, Network Topology, and Kirchhoff's Laws

Fundamental Principles of Electric Current and Resistance

• Definition of Electric Current Intensity ($I$):

  • Electric current intensity represents the rate of flow of electric charge ($Q$) passing through a given cross-sectional area of a conductor per unit time ($t$):         $I = \frac{Q}{t}$

  • The standard SI unit of current intensity is the Ampere ($\text{A}$), equivalent to Coulombs per second ($\text{C}\cdot\text{s}^{-1}$).

• Ohm's Law and Invariance of Physical Conductor Parameters:

  • Ohm's Law states that the current intensity passing through an ideal ohmic conductor is directly proportional to the potential difference ($V$) applied across its terminals, provided temperature and other physical conditions remain constant:         $V = I \times R$

  • The electrical resistance ($R$) of an ohmic conductor depends exclusively on its intrinsic material property (resistivity, $\rho_e$) and geometrical parameters (length $L$ and cross-sectional area $A$):         $R = \rho_e \frac{L}{A}$

  • Independence from Operating Variables: Electrical resistance and cross-sectional area ($A$) do not depend on the applied electromotive force ($V_B$) or the resulting current intensity ($I$).

  • Application: If a current of intensity $I$ flows through a conductor of length $L$ and cross-sectional area $A$, and the battery is subsequently replaced such that the current increases to $3I$, the cross-sectional area remains strictly $A$.

Resistivity, Conductivity, and Graphical Relationships

• Constitutive Formulas for Resistance and Conductivity:

  • The resistance of a uniform conductor is defined by:         $R = \rho_e \frac{L}{A} = \frac{L}{\sigma \times A}$         where $\rho_e$ is the electrical resistivity ($\Omega\cdot\text{m}$) and $\sigma$ is the electrical conductivity ($\Omega^{-1}\cdot\text{m}^{-1}$ or $\text{S}\cdot\text{m}^{-1}$), with $\sigma = \frac{1}{\rho_e}$.

  • In terms of volume ($\text{Vol} = A \times L$), the resistance can be rewritten:         $R = \rho_e \frac{L^2}{\text{Vol}} = \frac{L^2}{\sigma \times \text{Vol}}$

• Analysis of Resistance versus Reciprocal Volume ($R$ vs. $1/\text{Vol}$):

  • When plotting electric resistance ($R$) on the vertical axis against the reciprocal of volume ($\frac{1}{\text{Vol}}$) on the horizontal axis for wires of identical length ($L$) but made of different materials:         $\text{Slope} = \frac{\Delta Y}{\Delta X} = \frac{R}{1/\text{Vol}} = R \times \text{Vol} = R \times A \times L$

  • Since length $L$ is constant across all wires:         $\text{Slope} \propto R \times A$

  • From the definition of conductivity:         $\sigma = \frac{L}{R \times A} \implies \sigma \propto \frac{1}{R \times A}$

  • Combining these proportionalities reveals that conductivity is inversely proportional to the slope of the line:         $\sigma \propto \frac{1}{\text{Slope}}$

  • Since $\text{Slope} = \tan(\theta)$, a higher inclination angle $\theta$ corresponds to a steeper slope and consequently a lower conductivity:         $\theta_4 > \theta_3 > \theta_2 > \theta_1 \implies \text{Slope}_4 > \text{Slope}_3 > \text{Slope}_2 > \text{Slope}_1$         $\therefore \sigma_1 > \sigma_2 > \sigma_3 > \sigma_4$

• Analysis of Resistance versus Reciprocal Area ($R$ vs. $1/A$):

  • For two groups of wires, $X$ and $Y$, each having uniform and identical length ($L$) but composed of different materials, the relationship between $R$ and $\frac{1}{A}$ is linear:         $\text{Slope} = \frac{R}{1/A} = R \times A = \rho_e \times L$

  • For a constant unit length ($L = 1\,\text{m}$), the slope directly equals the resistivity:         $\text{Slope} = \rho_e = \tan(\theta)$

  • Comparing the graphical trajectories:         $\theta_X > \theta_Y \implies \tan(\theta_X) > \tan(\theta_Y) \implies \text{Slope}_X > \text{Slope}_Y$         $\therefore \rho_{e(X)} > \rho_{e(Y)}$

  • Evaluation at Constant Resistance: Drawing a horizontal line at a constant value of $R$ intersects the curves such that:         $\left(\frac{1}{A}\right)_X < \left(\frac{1}{A}\right)_Y \implies A_X > A_Y$

Conductor Geometry and Proportional Resistance Scaling

• General Scaling Ratio for Wires of Circular Cross-Section:

  • For a cylindrical wire with radius $r$ and diameter $D$, the cross-sectional area is $A = \pi r^2 = \frac{\pi D^2}{4}$.

  • The ratio of resistances between two conductors (labeled 1 and 2) is formulated as:         $\frac{R_1}{R_2} = \frac{\rho_{e1}}{\rho_{e2}} \times \frac{L_1}{L_2} \times \frac{A_2}{A_1} = \frac{\rho_{e1}}{\rho_{e2}} \times \frac{L_1}{L_2} \times \left(\frac{r_2}{r_1}\right)^2 = \frac{\rho_{e1}}{\rho_{e2}} \times \frac{L_1}{L_2} \times \left(\frac{D_2}{D_1}\right)^2$

• Case Study: Scaling of Current Across Unequal Geometries and Supplies:

  • Consider Wire (1) with length $L_1 = 3\ell$, radius $r_1 = r$, powered by ideal battery $(V_B)_1$ ($r = 0$), and Wire (2) with length $L_2 = 2\ell$, radius $r_2 = 2r$, powered by ideal battery $(V_B)_2 = 2(V_B)_1$ ($r = 0$), both constructed from the same material ($\rho_{e1} = \rho_{e2}$).

*   Step 1: Calculate the resistance ratio:

        $\frac{R_1}{R_2} = \frac{1}{1} \times \frac{3\ell}{2\ell} \times \left(\frac{2r}{r}\right)^2 = \frac{3}{2} \times 4 = \frac{6}{1}$ * Step 2: Relate current intensity to potential difference and resistance ($I = \frac{V_B}{R}$):         $\frac{I_1}{I_2} = \frac{(V_B)_1}{(V_B)_2} \times \frac{R_2}{R_1} = \frac{(V_B)_1}{2(V_B)_1} \times \frac{1}{6} = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$

• Case Study: Determining Unknown Conductor Length from Diameter and Resistance Ratios:

  • Two wires composed of the identical material have diameters such that $D_1 = 3D_2$ and resistances such that $R_2 = 4R_1$.

  • Applying the resistance scaling equation:         $\frac{R_1}{R_2} = 1 \times \frac{L_1}{L_2} \times \left(\frac{D_2}{D_1}\right)^2$         $\frac{R_1}{4R_1} = \frac{L_1}{L_2} \times \left(\frac{D_2}{3D_2}\right)^2$         $\frac{1}{4} = \frac{L_1}{L_2} \times \frac{1}{9}$         $\frac{L_2}{L_1} = \frac{4}{9} \implies L_2 = \frac{4}{9} L_1$

• Case Study: Ranking Conductors by Resistance Based on Dimensions:

  • Consider four uniform conductors fabricated from identical material ($\rho_e = \text{constant}$) with specified dimensions:

    • Conductor A: Length $2L$, Area $2A$

    • Conductor B: Length $L$, Area $3A$

    • Conductor C: Length $2L$, Area $A$

    • Conductor D: Length $L$, Area $4A$

*   Assuming normalized base units ($L = 1\,\text{m}, A = 1\,\text{m}^2, \rho_e = 1\,\Omega\cdot\text{m}$):

        $R_A = \frac{2}{2} = 1\,\Omega$         $R_B = \frac{1}{3}\,\Omega$         $R_C = \frac{2}{1} = 2\,\Omega$         $R_D = \frac{1}{4}\,\Omega$ * Arranging in ascending order (least to greatest electrical resistance):         $R_D < R_B < R_A < R_C \implies \text{D} \rightarrow \text{B} \rightarrow \text{A} \rightarrow \text{C}$

• Case Study: Resistance Ratio of Constant-Length Wires with Specified Area Multiples:

  • For two copper wires of identical length ($L_1 = L_2$), where the cross-sectional area of the second wire is three times that of the first wire ($A_2 = 3A_1$):         $\frac{R_1}{R_2} = \frac{\rho_{e1}}{\rho_{e2}} \times \frac{L_1}{L_2} \times \frac{A_2}{A_1} = 1 \times 1 \times \frac{3A_1}{A_1} = \frac{3}{1}$

Electrical Power and Rating of Appliances

• Mathematical Formulations of Consumed Electrical Power ($P_W$):

  • Electric power dissipated or consumed in a pure resistive conductor is defined by three interrelated forms:         $P_W = V \times I = I^2 \times R = \frac{V^2}{R}$

• Power versus Square of Current Intensity ($P_W$ vs. $I^2$):

  • When the consumed power $P_W$ is graphed against $I^2$:         $\text{Slope} = \frac{\Delta P_W}{\Delta (I^2)} = \frac{P_W}{I^2}$

  • Substituting Joule's law of heating ($P_W = I^2 R$):         $\frac{P_W}{I^2} = R$

  • The slope of the linear plot corresponds directly to the electrical resistance ($R$) of the conductor.

• Physical Significance of Appliance Nameplate Ratings:

  • An electrical device marked with the nominal rating $(100\,\text{V} - 80\,\text{W})$ indicates the operating parameters under standard designed potential difference:

    • Rated current intensity:             $I = \frac{P}{V} = \frac{80}{100} = 0.8\,\text{A}$

    • Internal operating resistance:             $R = \frac{V^2}{P} = \frac{(100)^2}{80} = \frac{10000}{80} = 125\,\Omega$

  • Interpretation: When a potential difference of $100\,\text{V}$ is maintained across the lamp terminals, a current of $0.8\,\text{A}$ flows through it.

Resistor Network Topology and Equivalent Resistance

• Fundamental Laws of Combination:

  • Series Connection: Current remains constant across each component; potential differences add linearly:         $R_{\text{eq}} = R_1 + R_2 + \dots + R_n$

  • Parallel Connection: Potential difference remains identical across all branches; currents divide inversely proportional to branch resistance:         $\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}$

  • For two parallel resistors:         $R_{\text{eq}} = \frac{R_1 \times R_2}{R_1 + R_2}$

• Comparative Permutations of Four Equal Resistors ($R = 1\,\Omega$):

*   *Configuration (1):* Three resistors in parallel, connected in series with the fourth resistor:

        $(R_{\text{eq}})_1 = \frac{1}{3} + 1 = \frac{4}{3}\,\Omega \approx 1.33\,\Omega$ * Configuration (2): Four resistors connected entirely in series:         $(R_{\text{eq}})_2 = 1 + 1 + 1 + 1 = 4\,\Omega$ * Configuration (3): Two resistors in series placed in parallel with one resistor, followed by one resistor in series:         $(R_{\text{eq}})_3 = \frac{2 \times 1}{2 + 1} + 1 = \frac{2}{3} + 1 = \frac{5}{3}\,\Omega \approx 1.67\,\Omega$ * Configuration (4): Three resistors in series connected in parallel with the fourth resistor:         $(R_{\text{eq}})_4 = \frac{3 \times 1}{3 + 1} = \frac{3}{4}\,\Omega = 0.75\,\Omega$ * Ranking from least to highest equivalent resistance:         $(R_{\text{eq}})_4 < (R_{\text{eq}})_1 < (R_{\text{eq}})_3 < (R_{\text{eq}})_2 \implies 4 < 1 < 3 < 2$

• Alternative Permutations of Identical Resistors ($R = 1\,\Omega$):

*   *Group (1):* Two resistors in parallel, in series with one resistor:

        $(R_{\text{eq}})_1 = \frac{1}{2} + 1 = \frac{3}{2}\,\Omega = 1.5\,\Omega$ * Group (2): Three resistors in series:         $(R_{\text{eq}})_2 = 3 \times 1 = 3\,\Omega$ * Group (3): Three resistors in parallel:         $(R_{\text{eq}})_3 = \frac{1}{3}\,\Omega \approx 0.33\,\Omega$ * Group (4): Two resistors in series, in parallel with one resistor:         $(R_{\text{eq}})_4 = \frac{2 \times 1}{2 + 1} = \frac{2}{3}\,\Omega \approx 0.67\,\Omega$ * Ascending sequence: $(R_{\text{eq}})_3 < (R_{\text{eq}})_4 < (R_{\text{eq}})_1 < (R_{\text{eq}})_2 \implies 3 < 4 < 1 < 2$.

• Closed Continuous Wire Geometries (Rectangular Copper Loop):

  • Consider a uniform copper wire shaped into a rectangle $(kyxm)$ where length is twice the width ($L = 2W$).

  • Assign resistances to each segment: width segments $mk = xy = W = 1\,\Omega$, and length segments $ky = mx = L = 2\,\Omega$. Total perimeter resistance is $1 + 2 + 1 + 2 = 6\,\Omega$.

  • Connecting across $(m, k)$: The segment $mk$ ($1\,\Omega$) is in parallel with the remaining perimeter ($2 + 1 + 2 = 5\,\Omega$):         $R_{\text{eq}} = \frac{5 \times 1}{5 + 1} = \frac{5}{6}\,\Omega \approx 0.83\,\Omega$

  • Connecting across $(k, y)$: The segment $ky$ ($2\,\Omega$) is in parallel with the remaining perimeter ($1 + 2 + 1 = 4\,\Omega$):         $R_{\text{eq}} = \frac{4 \times 2}{4 + 2} = \frac{8}{6}\,\Omega \approx 1.33\,\Omega$

  • Connecting across $(x, y)$: Identical to terminals $(m, k)$, yielding $R_{\text{eq}} \approx 0.83\,\Omega$.

  • Connecting across diagonally opposite terminals $(k, x)$: Path $k-y-x$ has resistance $2 + 1 = 3\,\Omega$; path $k-m-x$ has resistance $1 + 2 = 3\,\Omega$. These two equal branches are in parallel:         $R_{\text{eq}} = \frac{3}{2} = 1.5\,\Omega$

  • Conclusion: The greatest equivalent resistance is obtained when the electric source is connected between terminals $(k, x)$.

• Transition Between Series and Parallel Combinations:

  • Problem Statement: Two resistors satisfy $R_1 = 3R_2$. When placed in parallel, their equivalent resistance is $3\,\Omega$. Determine their equivalent resistance when placed in series.

  • Solution Procedure:         $R_{\text{eq(parallel)}} = \frac{R_1 \times R_2}{R_1 + R_2} \implies 3 = \frac{3R_2 \times R_2}{3R_2 + R_2} = \frac{3R_2^2}{4R_2} = \frac{3R_2}{4}$         $3 = \frac{3R_2}{4} \implies R_2 = 4\,\Omega$         $R_1 = 3 \times 4 = 12\,\Omega$         $R_{\text{eq(series)}} = R_1 + R_2 = 12 + 4 = 16\,\Omega$

  • Three-Resistor Scaling: Three resistors are proportional according to $R_1 = 3R$, $R_2 = 4R$, and $R_3 = 6R$. When placed in parallel, they yield $R_{\text{eq}} = 4\,\Omega$:         $\frac{1}{R_{\text{eq}}} = \frac{1}{3R} + \frac{1}{4R} + \frac{1}{6R} = \frac{4 + 3 + 2}{12R} = \frac{9}{12R} = \frac{3}{4R}$         $\frac{1}{4} = \frac{3}{4R} \implies 4R = 12 \implies R = 3\,\Omega$         $R_1 = 3(3) = 9\,\Omega, \quad R_2 = 4(3) = 12\,\Omega, \quad R_3 = 6(3) = 18\,\Omega$         $R_{\text{eq(series)}} = 9 + 12 + 18 = 39\,\Omega$

Short-Circuit Phenomena and Switch Actions in Resistor Networks

• Physics of Short Circuits:

  • A zero-resistance conducting wire connected in parallel across any resistive element forces the potential difference across that element to zero ($\Delta V = 0$).

  • As a consequence of Ohm's Law, zero current enters the resistive branch ($I = \frac{\Delta V}{R} = 0$), effectively cancelling and removing the component from the active equivalent circuit.

• Analysis of Parallel Resistors with Central Bypass Path:

*   In a circuit bounded by terminals $a$ and $b$, two sets of parallel $10\,\Omega$ resistors are connected in series.
*   A direct zero-resistance wire bridges terminal $a$ to the intermediate junction between the two sets.
*   *Result:* The two left-hand $10\,\Omega$ resistors have both of their respective terminals connected to the same electrical node ($a$), rendering them completely short-circuited.
*   The remaining active network consists solely of the right-hand pair of $10\,\Omega$ resistors in parallel:

        $R_{\text{eq}} = \frac{10}{2} = 5\,\Omega$

• Switch Closure Short-Circuiting Specific Circuit Elements:

*   Consider a circuit comprising a battery ($V_B, r = 0$), an ammeter, a fixed resistor of $4\,\Omega$, and a switch $K$ connected in parallel across a $2\,\Omega$ resistor.
*   *State 1: Key $K$ is Open:*
    *   The $4\,\Omega$ and $2\,\Omega$ resistors are in series:

            $R_{\text{eq}} = 4 + 2 = 6\,\Omega$ * Given the ammeter reads $1\,\text{A}$:             $I = \frac{V_B}{R_{\text{eq}} + r} \implies 1 = \frac{V_B}{6 + 0} \implies V_B = 6\,\text{V}$ * State 2: Key $K$ is Closed: * The switch creates a short circuit across the $2\,\Omega$ resistor, bypassing it completely ($R_{2\Omega} = 0$). * The new equivalent resistance is strictly $R_{\text{eq}} = 4\,\Omega$. * The new ammeter reading is:             $I = \frac{V_B}{R_{\text{eq}} + r} = \frac{6}{4 + 0} = 1.5\,\text{A}$

• Bridge Symmetry and Short-Circuited Bridge Resistors:

*   In a multi-resistor bridge network where a central conducting link connects identical potential nodes across symmetric branches of equal resistors ($R$):
*   The central transverse resistors experience zero potential drop and carry no current.
*   The circuit simplifies to an upper branch containing two series resistors ($R_{\text{upper}} = 2R$) in parallel with a lower branch containing two series resistors ($R_{\text{lower}} = 2R$):

        $R_{\text{eq}} = \frac{R_{\text{branch}}}{n} = \frac{2R}{2} = R$

• Comparison of Topology-Dependent Current Intensities:

• Evaluate three circuits powered by identical ideal batteries ($V_B = 1\,\text{V}, r = 0$) using identical resistors ($R = 1\,\Omega$):

  • Circuit (1): A zero-resistance bypass wire shorts out the left resistor, leaving only one active resistor $R$:         $R_{\text{eq1}} = R = 1\,\Omega \implies I_1 = \frac{1}{1} = 1\,\text{A}$

  • Circuit (2): Two resistors connected in series:         $R_{\text{eq2}} = 2R = 2\,\Omega \implies I_2 = \frac{1}{2} = 0.5\,\text{A}$

  • Circuit (3): Two parallel resistors in series with a third resistor:         $R_{\text{eq3}} = \frac{R}{2} + R = 1.5\,\Omega \implies I_3 = \frac{1}{1.5} = \frac{2}{3}\,\text{A} \approx 0.667\,\text{A}$

• Relative Current Comparison: $I_1 > I_3 > I_2$, validating that $I_1 > I_3$.

Terminal Voltage and Ohm's Law for Closed Circuits

• Fundamental Closed-Circuit Formulation:

  • For a real source with electromotive force $V_B$ and internal resistance $r$ supplying current $I$ to an external load $R_{\text{eq}}$, the total current is governed by Ohm's Law for a closed circuit:         $I = \frac{V_B}{R_{\text{eq}} + r}$

  • The terminal potential difference ($V$) across a discharging battery is reduced by the internal voltage drop ($Ir$):         $V = V_B - I \times r$

  • When the circuit is opened ($I = 0$), the terminal voltage reaches its maximum, equal to the electromotive force ($V = V_B$).

  • The potential difference across the external circuit load ($R_{\text{eq}}$) obeys:         $V = I \times R_{\text{eq}}$

• Voltmeter Ratios Across Circuit Boundaries:

  • For a closed circuit where voltmeter $V_1$ monitors the battery terminals and voltmeter $V_2$ monitors an external resistor $R$:         $V_1 = V_B - I \times r$         $V_2 = I \times R$         $\frac{V_1}{V_2} = \frac{V_B - I \times r}{I \times R}$

• Determination of Battery Electromotive Force and Internal Resistance:

• Case Study 1: A voltmeter across a cell reads $18\,\text{V}$ when switch $K$ is open, and $15\,\text{V}$ when switch $K$ is closed across an external resistance $R = 5\,\Omega$.

  • Step 1: Open circuit establishes $V_B$:         $I = 0 \implies V_B = 18\,\text{V}$

  • Step 2: Closed circuit current calculation:         $I = \frac{V}{R} = \frac{15}{5} = 3\,\text{A}$

  • Step 3: Internal resistance derivation:         $V = V_B - I \times r \implies 15 = 18 - 3r \implies 3r = 3 \implies r = 1\,\Omega$

*   *Case Study 2:* A voltmeter across a battery reads $14\,\text{V}$ with switch $K$ open, and $8\,\text{V}$ when switch $K$ is closed across two parallel resistors of $6\,\Omega$ and $3\,\Omega$.
    *   Open switch: $V_B = 14\,\text{V}$.
    *   Closed switch external resistance:

            $R_{\text{eq}} = \frac{6 \times 3}{6 + 3} = 2\,\Omega$ * Circuit current:             $I = \frac{V}{R_{\text{eq}}} = \frac{8}{2} = 4\,\text{A}$ * Internal resistance:             $V = V_B - I \times r \implies 8 = 14 - 4r \implies 4r = 6 \implies r = 1.5\,\Omega$

*   *Case Study 3 (Dual Circuit Comparison):* An identical cell ($V_B, r$) is connected to two independent circuits:
    *   Circuit 1 ($R_1 = 9\,\Omega$): Ammeter reads $I_1 = 1.2\,\text{A}$
    *   Circuit 2 ($R_2 = 11\,\Omega$): Ammeter reads $I_2 = 1.0\,\text{A}$
    *   Equating expressions for electromotive force:

            $V_B = I_1 (R_1 + r) = 1.2(9 + r) = 10.8 + 1.2r$             $V_B = I_2 (R_2 + r) = 1.0(11 + r) = 11.0 + r$             $10.8 + 1.2r = 11.0 + r \implies 0.2r = 0.2 \implies r = 1\,\Omega$             $V_B = 11.0 + 1 = 12\,\text{V}$

• Graphical Analysis of Terminal Voltage versus Current ($V$ vs. $I$):

• The relationship $V = V_B - I \times r$ describes a straight line with a negative slope:     $\text{Slope} = \frac{\Delta V}{\Delta I} = -r$

• The vertical axis intercept ($I = 0$) represents the electromotive force $V_B$.

• Example: Given internal resistance $r = 0.5\,\Omega$ and an operating coordinate on the line at $I = 4\,\text{A}$, $V = 8\,\text{V}$:     $V = V_B - I \times r$     $8 = V_B - (4 \times 0.5)$     $8 = V_B - 2 \implies V_B = 10\,\text{V}$

Variable Resistors, Rheostats, and Internal Resistance Analysis

• Effect of Changing Rheostat Resistance on Terminal Voltmeters:

  • Consider a circuit consisting of a battery ($V_B, r$) in series with a fixed resistor $R$ and a variable rheostat $S$.

  • Scenario A: Variable Resistance Increases ($S \uparrow$):

    • Total circuit resistance increases: $R_{\text{eq}} = R + S \uparrow$

    • Current intensity decreases: $I = \frac{V_B}{R_{\text{eq}} + r} \downarrow$

    • Voltmeter across the battery terminals ($V_2 = V_B - I r$): As $I$ decreases, the internal drop $I r$ decreases; therefore, $V_2$ increases.

    • Voltmeter across the variable component ($V_1$): Although current decreases, the resistance $S$ increases at a greater rate. Alternatively, expressed via Kirchhoff's loop:             $V_1 = V_{\text{battery}} - V_R = (V_B - Ir) - IR = V_B - I(R + r)$             Since current $I$ decreases, the total deduction $I(R + r)$ decreases; therefore, $V_1$ increases.

  • Scenario B: Variable Resistance Decreases ($S \downarrow$):

    • Total resistance decreases, driving an increase in total current ($I \uparrow$).

    • The internal voltage drop increases ($I r \uparrow$).

    • Terminal voltage drops: $V = V_B - I r \downarrow$.

• Ideal Battery Clamping ($r = 0$):

*   If a variable resistor $R$ is connected directly across an ideal battery having zero internal resistance ($r = 0$):

        $V_1 = V_B$         $V_2 = V_B$ * When the resistance of $R$ is varied, the branch current changes ($I = \frac{V_B}{R}$), but the terminal voltage across the source ($V_1$) and across the resistor ($V_2$) remains clamped and unchanged ($V_1 = V_2 = V_B$).

• Switch Dynamics and Meter Fluctuations in Multi-Branch Systems:

*   Consider a circuit containing a battery ($V_B, r = 0$), a series branch with voltmeter $V_3$ across a series component $R$, a series resistor $2R$ monitored by $V_2$, and a parallel pair of $2R$ resistors bridged by switch $K$ and monitored by $V_1$.
*   *Action: Switch $K$ is closed:*
    1.  The switch short-circuits both parallel $2R$ resistors, causing the potential difference across them to vanish: $V_1 \rightarrow 0\,\text{V}$ (**becomes zero**).
    2.  Total external circuit resistance drops:

            $R_{\text{total}} = 2R + 0 + R = 3R$ 3. Current intensity increases ($I \uparrow$). 4. Voltmeter $V_2$ monitors fixed resistor $2R$: $V_2 = I \times 2R$. As $I$ increases, $V_2$ increases. 5. Voltmeter $V_3$ across the battery/load branch ($V_3 = V_B - I R$): As $I$ increases, $V_3$ decreases.

• Ammeter and Voltmeter Dual Responses upon Opening a Branch:

  • In a multi-branch closed circuit with internal resistance $r > 0$, opening a switch $K$ removes a parallel branch:

    1. Total equivalent resistance ($R_{\text{eq}}$) increases.

    2. Total main current ($I_{\text{total}}$) decreases.

    3. The battery terminal voltmeter ($V = V_B - I_{\text{total}} r$) increases due to smaller internal drop.

    4. Branch current through an individual parallel branch ($I_{\text{branch}}$) monitored by an ammeter decreases because the branch now receives a divided share of a significantly reduced total current.

Multi-Battery Systems: Charging vs. Discharging Cells

• Differentiation Between Source (Discharging) and Sink (Charging) Batteries:

  • When two batteries with electromotive forces $V_{B1}$ and $V_{B2}$ are connected with opposing polarities (positive connected to positive, or negative to negative):

    • Discharging Battery (Source): The battery with the larger electromotive force ($V_{B2} > V_{B1}$) drives current out of its positive terminal:             $V_2 = V_{B2} - I \times r_2$             Here, terminal voltage is strictly less than electromotive force ($V_2 < V_{B2}$).

    • Charging Battery (Sink/Load): The battery with the smaller electromotive force ($V_{B1}$) has current forced into its positive terminal:             $V_1 = V_{B1} + I \times r_1$             Here, terminal voltage is strictly greater than electromotive force ($V_1 > V_{B1}$).

  • Net Circuit Current in Opposing Systems:         $I_{\text{net}} = \frac{V_{B2} - V_{B1}}{R_{\text{eq}} + r_1 + r_2}$

• Quantitative Worked Example:

  • Given $V_{B2} = 10\,\text{V}$, $r_2 = 2\,\Omega$; $V_{B1} = 8\,\text{V}$, $r_1 = 1\,\Omega$; and a series resistor $R = 2\,\Omega$ across which a voltmeter reads $V_3 = 0.8\,\text{V}$.

    • Step 1: Determine circuit current:             $I = \frac{V_3}{R} = \frac{0.8}{2} = 0.4\,\text{A}$             (Verification via net equation:)             $I = \frac{10 - 8}{2 + 1 + 2} = \frac{2}{5} = 0.4\,\text{A}$

    • Step 2: Terminal voltage of charging battery ($V_1$):             $V_1 = V_{B1} + I \times r_1 = 8 + (0.4 \times 1) = 8.4\,\text{V}$

    • Step 3: Terminal voltage of discharging battery ($V_2$):             $V_2 = V_{B2} - I \times r_2 = 10 - (0.4 \times 2) = 9.2\,\text{V}$

• Parallel Multi-Battery Branch Analysis:

• A network contains a primary battery of $12\,\text{V}$ connected in opposition to a secondary battery $V_B$, feeding parallel branches with resistances $R$ and $2R$.

• The primary $12\,\text{V}$ battery functions as the driving source; therefore, $V_B$ must be a charging battery, dictating that $V_B < 12\,\text{V}$ (identifying $V_B = 6\,\text{V}$).

• For the parallel branch resistors, current division requires:     $R_2 = R, \quad R_3 = 2R \implies I_2 = 2 \times I_3$

• By junction conservation: $I_1 = I_2 + I_3$.

• Substituting the physical condition $I_2 = 2I_3$:     $I_1 = 2I_3 + I_3 = 3I_3$

• Checking values: $I_1 = 3\,\text{A}$, $I_2 = 2\,\text{A}$, and $I_3 = 1\,\text{A}$ consistently satisfies $I_2 = 2I_3$ and $I_1 = I_2 + I_3$.

Kirchhoff's First Law (Current Law / Junction Rule)

• Statement and Conservation Principle:

  • Kirchhoff's First Law (KCL) is based on the Law of Conservation of Electric Charge.

  • At any junction (node) in an electric circuit, the sum of currents entering the junction must equal the sum of currents leaving the junction:         $\sum I_{\text{in}} = \sum I_{\text{out}} \iff \sum I = 0$

• Conventional Current versus Electron Flow Conventions:

  • Conventional current flow is defined in the direction of positive charge motion (from high potential to low potential, or out of the battery's positive terminal).

  • Electron flow is opposite to conventional current. A stream of electrons moving away from a node is physically equivalent to a conventional current entering that node.

• Mathematical Formulations at Complex Nodes:

• Case 1 (Pure Electron Flow Vectors): Five conductors meet at node $X$. If arrows represent the physical motion of electrons:

  • Electrons entering node $X$: $I_1, I_3, I_4$

  • Electrons leaving node $X$: $I_2, I_5$

  • By charge conservation:         $\sum I_{\text{in}} = \sum I_{\text{out}} \implies I_1 + I_3 + I_4 = I_2 + I_5$

  • Rearranging into standard algebraic summation format:         $-I_1 - I_3 - I_4 + I_2 + I_5 = 0$

• Case 2 (Mixed Conventions): At a common node, currents $I_1, I_2, I_3$ represent conventional current directions, while $I_4$ represents the direction of electron motion:

  • Conventional $I_1$ enters the node.

  • Conventional $I_2$ enters the node.

  • Conventional $I_3$ leaves the node.

  • Electron current $I_4$ leaves the node, which corresponds to an incoming conventional current ($+I_4$).

  • Summing incoming and outgoing conventional currents:         $\sum I_{\text{in}} = \sum I_{\text{out}} \implies I_1 + I_2 + I_4 = I_3$

• Case 3 (Direct Node Evaluation): At node $B$, three currents enter ($2\,\text{A}, 3\,\text{A}, I$) and one current exits ($6\,\text{A}$):     $\sum I_{\text{in}} = \sum I_{\text{out}} \implies 2 + 3 + I = 6 \implies 5 + I = 6 \implies I = 1\,\text{A}$

Kirchhoff's Second Law (Voltage Law / Loop Rule) and Multi-Loop Circuit Analysis

• Statement and Conservation Principle:

  • Kirchhoff's Second Law (KVL) is based on the Law of Conservation of Energy.

  • The algebraic sum of electromotive forces ($V_B$) around any closed loop is equal to the algebraic sum of the potential drops ($I \times R$) across the resistances in that loop:         $\sum V_B = \sum (I \times R) \iff \sum \Delta V = 0$

• Application in Three-Branch DC Networks:

*   A network consists of three parallel horizontal branches:
    *   Top: Resistor $R$, battery $2V_B$, current $I_1$
    *   Middle: Resistor $R$, battery $V_B$, current $I_2$
    *   Bottom: Resistor $R$, battery $V_B$, current $I_3$
*   *Loop Analysis (Lower Loop):*

        $V_B + (I_2 \times R) - (I_3 \times R) - V_B = 0$         $I_2 \times R = I_3 \times R \implies I_2 = I_3$ * Node Analysis (Right Junction):         $I_1 = I_2 + I_3 = I_2 + I_2 = 2I_2$ * Current Ratio Calculation:         $\frac{I_2}{I_1} = \frac{I_2}{2I_2} = \frac{1}{2}$

• Multi-Branch Current Determination:

  • Consider a dual-mesh circuit with three vertical branches containing negligible battery internal resistances:

    • Left branch: Resistor $8\,\Omega$, battery $6\,\text{V}$, carrying current $I_2$

    • Center branch: Battery $4\,\text{V}$, carrying upward current $I_3$

    • Right branch: Resistor $10\,\Omega$, battery $8\,\text{V}$, carrying current $I_1$

  • Right Mesh KVL:         $10 I_1 - 8 - 4 = 0 \implies 10 I_1 = 12 \implies I_1 = 1.2\,\text{A}$

  • Left Mesh KVL:         $8 I_2 - 6 - 4 = 0 \implies 8 I_2 = 10 \implies I_2 = 1.25\,\text{A}$

  • Junction KCL (Upper Node):         $I_3 = I_1 + I_2 = 1.2 + 1.25 = 2.45\,\text{A}$

• Derivation of Loop Equations along Specified Paths:

*   In a multi-loop circuit, path $adcba$ defines a closed loop spanning an upper branch (battery $12\,\text{V}$, resistor $2\,\Omega$, current $I_1$), a middle branch (battery $8\,\text{V}$, resistor $1\,\Omega$, current $I_2$), and a lower branch (resistor $10\,\Omega$, current $I_3$).
*   Applying KVL around the loop formed by branches $ad$ and $ab$:

        $\sum V_B = \sum (IR) \implies 8 - 12 = I_2(1) - I_1(2)$         $-4 = I_2 - 2I_1 \implies 2I_1 - I_2 - 4 = 0$ * Applying KCL at node $a$:         $\sum I_{\text{in}} = \sum I_{\text{out}} \implies I_1 + I_2 = I_3 \implies I_2 = I_3 - I_1$ * Substituting for $I_2$:         $2I_1 - (I_3 - I_1) - 4 = 0 \implies 3I_1 - I_3 - 4 = 0$

• Branch Current Solution Given Current Constraint:

*   In a circuit with batteries $V_{B1} = 10\,\text{V}$ and $V_{B2} = 20\,\text{V}$, resistors $R_1 = 10\,\Omega$, $R_3 = 40\,\Omega$, and $R_2$, given the condition that $I_3 = -2I_1$:
*   Express $I_1$ in terms of $I_3$:

        $I_1 = -0.5 I_3$ * Apply KVL around the left loop:         $\sum V_B = \sum (IR) \implies 10 = I_1 R_1 + I_3 R_3$         $10 = 10 I_1 + 40 I_3$ * Substitute $I_1 = -0.5 I_3$:         $10 = 10(-0.5 I_3) + 40 I_3$         $10 = -5 I_3 + 40 I_3 = 35 I_3$         $I_3 = \frac{10}{35} = \frac{2}{7}\,\text{A}$

• Null-Current Condition in Bridge Branches (Vanishing Ammeter):

*   Consider a circuit where an ammeter is placed in the left branch containing battery $(V_B)_2$ and resistor $2R$. The central branch has resistor $3R$, and the right branch contains battery $(V_B)_1 = 6\,\text{V}$ ($r = 0$) and resistor $R$.
*   *Condition:* The reading of the ammeter vanishes ($I_3 = 0$).
*   *KCL Analysis:* Since the left branch current is zero, the central branch and right branch share identical current:

        $I_1 = I_2 = I$ * Right Loop KVL: Trace the loop containing the right and central branches:         $(V_B)_1 - I R - I(3R) = 0$         $6 - 4 I R = 0 \implies I R = \frac{6}{4} = 1.5$ * Left Loop KVL: Trace the left loop containing $(V_B)_2$, $2R$, and $3R$:         $(V_B)_2 - I_2(3R) + I_3(2R) = 0$         $(V_B)_2 - 3 I R + 0 = 0$         $(V_B)_2 = 3 I R = 3 \times 1.5 = 4.5\,\text{V}$

• Solving Simultaneous Multi-Loop Equations:

*   To determine the current through the $8\,\Omega$ resistor in a circuit with three active branches:
    *   Branch 1 (Top): $10\,\text{V}$, $6\,\Omega$, carrying current $I_2$
    *   Branch 2 (Middle): $12\,\text{V}$, $4\,\Omega$, carrying current $I_1$
    *   Branch 3 (Bottom): Resistor $8\,\Omega$, carrying current $I_3$
*   *Equation 1 (KCL at junction):*