Comprehensive Study Notes on DC Circuits: Ohm's Law, Network Reduction, Closed Circuit Analysis, and Kirchhoff's Laws

Fundamentals of Electric Current, Potential Difference, and Resistance

• Electric Current Intensity ($I$):

  • Defined as the total quantity of electric charge ($Q$) passing through a given cross-section of a conductor per unit time ($t$).

  • Expressed mathematically through the equivalent relations:     $I = \frac{Q}{t} = \frac{V}{R} = \frac{N e}{t}$

  • Number of passing electrons ($N$) through a conductor's cross-section is determined by:     $N = \frac{Q}{e}$     where $e$ represents the fundamental elementary charge of an electron ($e = 1.6 \times 10^{-19}\,\text{C}$).

• Electric Potential Difference ($V$):

  • Defined as the work done ($W$) to transfer a unit electric charge ($Q$) between two points in an electric circuit.

  • Expressed mathematically as:     $V = \frac{W}{Q} = I R = \frac{W}{N e}$

• Electric Resistance ($R$):

  • The opposition encountered by electric current when passing through a conductor.

  • Standard structural formula relating dimensions and material properties:     $R = \frac{\rho_e l}{A} = \frac{\rho_e l}{\pi r^2}$     where $\rho_e$ is the electric resistivity of the conductor material ($\Omega\,\text{m}$), $l$ is conductor length ($\text{m}$), $A$ is cross-sectional area ($\text{m}^2$), and $r$ is cross-sectional radius ($\text{m}$).

  • Determined from the ratio of potential difference to current intensity:     $R = \frac{V}{I}$

• Electric Resistivity ($\rho_e$) and Conductivity ($\sigma$):

  • Resistivity is an intrinsic property of a material given by:     $\rho_e = \frac{R A}{l}\,(\Omega\,\text{m})$

  • Electric conductivity ($\sigma$) is the reciprocal of resistivity:     $\sigma = \frac{1}{\rho_e} = \frac{l}{R A}\,(\Omega^{-1}\,\text{m}^{-1})$

• Resistance Formulations Involving Volume ($Vol$), Mass ($m$), and Density ($\rho$):

  • Expressed via volume ($Vol = A l$):     $R = \frac{\rho_e l^2}{Vol}$     $R = \frac{\rho_e Vol}{A^2}$

  • Expressed via mass ($m = \rho \cdot Vol = \rho A l$) and material density ($\rho$):     $R = \frac{\rho_e \rho l^2}{m}$     $R = \frac{\rho_e m}{\rho A^2}$

• Wire Geometry and Coiling Relations:

  • The total unwound length ($l_{wire}$) of a conductor wound tightly into a circular coil of $N$ turns and coil radius $r_{coil}$ is:     $l_{wire} = 2 \pi r_{coil} N$

• Comparative Ratios for Reshaped or Stretched Wires:

  • When a wire is reshaped, drawn, or stretched such that its length increases and its cross-sectional area decreases while maintaining constant volume ($Vol_1 = Vol_2$):     $\frac{R_1}{R_2} = \left(\frac{l_1}{l_2}\right) \left(\frac{A_2}{A_1}\right) = \left(\frac{l_1}{l_2}\right)^2 = \left(\frac{A_2}{A_1}\right)^2 = \left(\frac{r_2}{r_1}\right)^4 = \left(\frac{d_2}{d_1}\right)^4$     where $r_1, r_2$ are the initial and final radii, and $d_1, d_2$ are the initial and final diameters.

  • General ratio comparing two different conductors of distinct materials, dimensions, and densities:     $\frac{R_1}{R_2} = \left(\frac{\rho_{e1}}{\rho_{e2}}\right) \left(\frac{l_1}{l_2}\right) \left(\frac{A_2}{A_1}\right) = \left(\frac{\rho_{e1}}{\rho_{e2}}\right) \left(\frac{l_1}{l_2}\right) \left(\frac{r_2^2}{r_1^2}\right) = \left(\frac{\rho_{e1}}{\rho_{e2}}\right) \left(\frac{\rho_1}{\rho_2}\right) \left(\frac{l_1^2}{l_2^2}\right) \left(\frac{m_2}{m_1}\right)$

• Electric Energy ($W$) and Electric Power ($P_w$):

  • Consumed electric energy ($W$):     $W = V Q = V I t = I^2 R t = \frac{V^2 t}{R}$

  • Consumed electric power ($P_w$):     $P_w = \frac{W}{t} = V I = I^2 R = \frac{V^2}{R}$

  • Power comparison rules:

  • At constant electric current intensity ($I = \text{constant}$, as in series configurations):       $\frac{(P_w)_1}{(P_w)_2} = \frac{R_1}{R_2}$

  • At constant potential difference ($V = \text{constant}$, as in parallel configurations):       $\frac{(P_w)_1}{(P_w)_2} = \frac{R_2}{R_1}$

Resistor Combinations and Network Simplification Techniques

• Series Resistor Combination:

  • Equivalent resistance ($R_{eq}$):     $R_{eq} = R_1 + R_2 + R_3$

  • If $N$ identical resistors each of value $R$ are connected in series:     $R_{eq} = N R$

  • Current behavior:

  • Current intensity through all resistors is identical:       $I = \frac{V}{R_{eq}} = \frac{V_1}{R_1} = \frac{V_2}{R_2} = \frac{V_3}{R_3}$

  • Voltage behavior:

  • Total voltage is distributed directly proportional to resistances:       $V = V_1 + V_2 + V_3$

• Parallel Resistor Combination:

  • Equivalent resistance reciprocal formula:     $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

  • For $N$ identical resistors connected in parallel:     $R_{eq} = \frac{R}{N}$

  • For two distinct resistors $R_1$ and $R_2$ connected in parallel:     $R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$

  • Voltage behavior:

  • The potential difference across each parallel branch is identical:       $V = V_1 = V_2 = V_3 = I R_{eq}$

  • Current behavior:

  • Total circuit current distributes inversely proportional to branch resistances:       $I = I_1 + I_2 + I_3$

  • Individual branch currents in a two-branch parallel network:       $I_1 = I \left(\frac{R_2}{R_1 + R_2}\right)$       $I_2 = I \left(\frac{R_1}{R_1 + R_2}\right)$       $I_1 R_1 = I_2 R_2 = I R_{eq} = V_{ab}$

• Short-Circuit Elimination Rule:

  • When the two terminals of a resistor are shunted together by a connecting wire of zero resistance, no potential difference exists across the resistor terminals ($V = 0$).

  • Consequently, the entire current bypasses the resistor through the shorting conductor, and the resistor is completely ignored in equivalent resistance calculations.

• Equipotential Nodes and Balanced Bridge Cancellation Rule:

  • Any conductor wire possessing negligible resistance maintains an identical electric potential across its full length; its terminals are treated as a single merged node.

  • When electric potentials at both terminals of a bridge resistor are equal ($V_x = V_y$), no potential difference exists across that resistor ($V_{xy} = 0$), causing current through it to be zero ($I = 0$).

  • A bridge resistor connected between two symmetrical branches is cancelled when the bridge balance ratio is satisfied:     $\frac{R_1}{R_2} = \frac{R_3}{R_4}$

Ohm's Law for Closed Circuits and Battery Dynamics

• Core Formulation:

  • The total electromotive force ($V_B$) generated by an electric source equals the sum of external potential differences across load resistances and the internal potential drop across the battery's internal resistance ($r$):     $V_B = V_{ext} + V_{int} = I R + I r = I (R + r)$

  • Electric current intensity ($I$) in a closed single-loop circuit:     $I = \frac{V_B}{R + r}$

  • Internal potential drop across the cell:     $V_{int} = I r = V_B - V_{ext}$

• Open-Circuit versus Closed-Circuit Terminal Voltage:

  • Under open-circuit conditions (switch $K$ open, $I = 0$):

  • Terminal voltage across the battery reaches maximum and equals its emf:       $V_1 = V_B$

  • Potential drop across external resistor is zero:       $V_2 = I R = 0$

  • Under closed-circuit conditions (switch $K$ closed, $I > 0$):

  • Current flows according to:       $I = \frac{V_B}{R + r}$

  • Battery terminal voltage ($V_1$) decreases due to internal drop:       $V_1 = V_B - I r$

  • Potential difference across external load equals terminal voltage:       $V_2 = I R = V_1$

• Algebraic Determination of Unknowns via Resistor Substitution:

  • When external load resistor $R_1$ (carrying current $I_1$) is swapped for $R_2$ (carrying current $I_2$) using the identical cell ($V_B, r$):     $V_B = I_1 (R_1 + r)$     $V_B = I_2 (R_2 + r)$

  • Equating both relations permits solving algebraically for internal resistance $r$ and emf $V_B$:     $I_1 (R_1 + r) = I_2 (R_2 + r)$

• Two Electric Cells in Series Connection:

  • Case 1: Series-Aiding Configuration (Discharging–Discharging):

  • Cells are connected with opposite polarities facing each other (positive terminal to negative terminal), reinforcing net driving emf:       $V_{B,\text{net}} = (V_B)_1 + (V_B)_2$

  • Total circuit current:       $I = \frac{(V_B)_1 + (V_B)_2}{R + r_1 + r_2}$

  • Terminal voltage across Cell 1 (discharging):       $V_1 = (V_B)_1 - I r_1$

  • Terminal voltage across Cell 2 (discharging):       $V_2 = (V_B)_2 - I r_2$

  • Total potential difference across the combined battery bank:       $V_3 = V_1 + V_2 = I R$

  • Case 2: Series-Opposing Configuration (Discharging–Charging):

  • Cells are connected with identical polarities facing each other (positive to positive, or negative to negative), opposing each other. Assuming $(V_B)_1 > (V_B)_2$:       $V_{B,\text{net}} = (V_B)_1 - (V_B)_2$

  • Total circuit current:       $I = \frac{(V_B)_1 - (V_B)_2}{R + r_1 + r_2}$

  • The larger cell ($(V_B)_1$) discharges:       $V_1 = (V_B)_1 - I r_1 < (V_B)_1$

  • The smaller cell ($(V_B)_2$) undergoes charging from the stronger source:       $V_2 = (V_B)_2 + I r_2 > (V_B)_2$

  • Total potential difference across the two-cell bank:       $V_3 = V_1 - V_2 = I R$

Kirchhoff's Laws

• Kirchhoff's First Law (Current Law / Junction Rule - KCL):

  • Physical Basis: Principle of Conservation of Electric Charge.

  • Statement: The algebraic sum of electric currents meeting at any junction in a closed electrical network equals zero:     $\Sigma I = 0$

  • Alternatively: The sum of currents entering a junction equals the sum of currents leaving that junction:     $\Sigma I_{in} = \Sigma I_{out}$

  • For a junction with incoming currents $I_1, I_2$ and outgoing current $I_3$:     $I_1 + I_2 - I_3 = 0 \implies I_1 + I_2 = I_3$

• Kirchhoff's Second Law (Voltage Law / Loop Rule - KVL):

  • Physical Basis: Principle of Conservation of Energy.

  • Statement: In any closed loop of an electrical circuit, the algebraic sum of electromotive forces equals the algebraic sum of the potential drops across all resistive elements:     $\Sigma V_B = \Sigma I R$

  • Alternatively: The algebraic sum of all potential differences around any closed circuit loop is zero:     $\Sigma V = 0$

• Sign Conventions for Traversal of Closed Loops:

  • Application under Form $\Sigma V_B = \Sigma I R$:

  • In a resistor:

    • If path traversal aligns with current direction ($\rightarrow$ path, $\rightarrow$ current): potential term is positive ($+ I R$).

    • If path traversal opposes current direction ($\leftarrow$ path, $\rightarrow$ current): potential term is negative ($- I R$).

  • In a battery:

    • If path traverses from negative to positive pole ($- \rightarrow +$): electromotive term is positive ($+ V_B$).

    • If path traverses from positive to negative pole ($+ \rightarrow -$): electromotive term is negative ($- V_B$).

  • Application under Form $\Sigma V = 0$:

  • In a resistor:

    • If path traversal aligns with current direction (entering higher potential terminal $+$ and exiting lower potential terminal $-$): potential change is negative ($\Delta V = - I R$).

    • If path traversal opposes current direction (entering $-$ and exiting $+$): potential change is positive ($\Delta V = + I R$).

  • In a battery:

    • If path traverses from negative to positive plate ($- \rightarrow +$): potential change is positive ($\Delta V = + V_B$).

    • If path traverses from positive to negative plate ($+ \rightarrow -$): potential change is negative ($\Delta V = - V_B$).

Systematic Solutions to Circuit Problems

• Problem 42: Increasing Bulb Current:

  • Given Circuit: Single-loop circuit consisting of two series-connected batteries, an ohmic resistor $R$, and an electric bulb with internal resistance $R_{bulb}$.

  • Analysis:

  • Total loop current is expressed as:     $I = \frac{\Sigma V_B}{R_{bulb} + R + r_{total}}$

  • Option (a): Adding a resistor in parallel with resistor $R$ decreases total external circuit resistance from $R$ to $R_{parallel} = \frac{R \cdot R_{add}}{R + R_{add}} < R$. As total equivalent resistance decreases, passing current $I$ through the bulb increases.

  • Option (b): Adding a resistor in series increases denominator resistance, decreasing $I$.

  • Option (c): Removing one battery reduces driving emf $\Sigma V_B$, decreasing $I$.

  • 2Conclusion: Correct choice is (a).

• Problem 43: Determining Series Component $X$ for Target Voltmeter Reading:

  • Given Circuit: Battery of $V_B = 20\,\text{V}$ with internal resistance $r = 0$, connected in series with resistor $R$ and unknown component $X$. Voltmeter connected across resistor $R$ deflects to $V_R = 4\,\text{V}$.

  • Mathematical Derivation:

  • Total applied voltage: $V_B = V_R + V_X$.

  • Potential drop across component $X$:       $V_X = 20\,\text{V} - 4\,\text{V} = 16\,\text{V}$

  • Since components are in series, they carry identical current $I$:       $I = \frac{V_R}{R} = \frac{4\,\text{V}}{R}$

  • If component $X$ is an ohmic resistor $R_X$:       $R_X = \frac{V_X}{I} = \frac{16\,\text{V}}{4/R} = 4 R$

  • Conclusion: Correct choice is (b) (Resistor of resistance $4 R$).

• Problem 44: Voltmeter Reading Across Opposing Series Cell:

  • Given Circuit: Single loop containing two opposing batteries and a load resistor:

  • Cell 1: $(V_B)_1 = 2\,\text{V}$, $r_1 = 0.3\,\Omega$

  • Cell 2: $(V_B)_2 = 5\,\text{V}$, $r_2 = 0.2\,\Omega$

  • External resistor: $R = 2\,\Omega$

  • Voltmeter is bridged across the terminals of Cell 1.

  • Mathematical Derivation:

  • Because cells are connected opposing ($(V_B)_2 > (V_B)_1$), net driving emf is:       $V_{B,\text{net}} = (V_B)_2 - (V_B)_1 = 5\,\text{V} - 2\,\text{V} = 3\,\text{V}$

  • Total equivalent resistance of the closed circuit:       $R_{total} = R + r_1 + r_2 = 2\,\Omega + 0.3\,\Omega + 0.2\,\Omega = 2.5\,\Omega$

  • Circuit current intensity:       $I = \frac{V_{B,\text{net}}}{R_{total}} = \frac{3\,\text{V}}{2.5\,\Omega} = 1.2\,\text{A}$

  • Cell 1 experiences a forced reverse current (charging mode). Its terminal potential difference is given by:       $V = (V_B)_1 + I r_1 = 2\,\text{V} + (1.2\,\text{A})(0.3\,\Omega) = 2\,\text{V} + 0.36\,\text{V} = 2.36\,\text{V}$

  • Conclusion: Correct choice is (b) ($2.36\,\text{V}$).

• Problem 45: Voltmeter Response on Battery Removal and Series Load Addition:

  • Given Circuit: Two identical series cells of negligible internal resistance ($r = 0$) connected across load resistor $R$. Voltmeter connected directly across the terminals of both cells.

  • Analysis:

  • Initial reading with both cells active ($r = 0$):       $V_{initial} = (V_B)_1 + (V_B)_2 = 2 V_B$

  • When one cell is removed, the remaining cell provides emf $V_B$.

  • Because internal resistance is zero ($r = 0$), the terminal voltage across an ideal cell is strictly invariant to changes in external resistance $R$:       $V_{final} = V_B - I(0) = V_B$

  • Since $V_B < 2 V_B$, the voltmeter deflection decreases.

  • Conclusion: Correct choice is (b) (decreases).

• Problem 46: External Resistance and Current in Multi-Branch Network:

  • Given Circuit:

  • Cells: $(V_B)_1 = 6\,\text{V}$, $(V_B)_2 = 12\,\text{V}$, each having internal resistance $r_1 = r_2 = 2\,\Omega$.

  • External network: Series resistor $10\,\Omega$ connected in series with a parallel grouping consisting of three parallel branches: Branch 1 = $20\,\Omega$, Branch 2 = $30\,\Omega$, Branch 3 = $20\,\Omega + 40\,\Omega = 60\,\Omega$.

  • Battery terminals are arranged in series opposition.

  • Mathematical Derivation:

  • (i) Equivalent parallel resistance ($R_p$):       $\frac{1}{R_p} = \frac{1}{20} + \frac{1}{30} + \frac{1}{60} = \frac{3 + 2 + 1}{60} = \frac{6}{60} = \frac{1}{10}\,\Omega^{-1} \implies R_p = 10\,\Omega$       Total external resistance:       $R_{ext} = 10\,\Omega + R_p = 10\,\Omega + 10\,\Omega = 20\,\Omega$       Correct choice for (i) is (c) ($20\,\Omega$).

  • (ii) Net electromotive force (cells in opposition):       $V_{B,\text{net}} = (V_B)_2 - (V_B)_1 = 12\,\text{V} - 6\,\text{V} = 6\,\text{V}$       Total circuit resistance including internal resistances:       $R_{total} = R_{ext} + r_1 + r_2 = 20\,\Omega + 2\,\Omega + 2\,\Omega = 24\,\Omega$       Total current intensity:       $I = \frac{V_{B,\text{net}}}{R_{total}} = \frac{6\,\text{V}}{24\,\Omega} = 0.25\,\text{A}$       Correct choice for (ii) is (a) ($0.25\,\text{A}$).

• Problem 47: Internal Resistance of Four Identical Cells Driving a Bulb:

  • Given Circuit: Four identical series-aiding cells, each of emf $V_B = 12\,\text{V}$, powering an electric bulb. Current is $I = 0.5\,\text{A}$, and consumed bulb power is $P_{bulb} = 23\,\text{W}$.

  • Mathematical Derivation:

  • Total battery bank emf:       $V_{B,\text{total}} = 4 \times 12\,\text{V} = 48\,\text{V}$

  • Terminal voltage across bulb:       $V_{bulb} = \frac{P_{bulb}}{I} = \frac{23\,\text{W}}{0.5\,\text{A}} = 46\,\text{V}$

  • Closed-circuit Ohm's equation:       $V_{B,\text{total}} = V_{bulb} + I (4 r)$       $48\,\text{V} = 46\,\text{V} + (0.5\,\text{A})(4 r)$       $2\,\text{V} = 2 r \implies r = 1\,\Omega$

  • Conclusion: Correct choice is (b) ($1\,\Omega$).

• Problem 48: Current Direction and Magnitude under Switch Commutation:

  • Given Circuit: Closed network containing fixed cell $2 V_B, 2 r$ and resistor $R$ with ammeter $A$. Switched branches contain: Branch 1 with $K_1$ and cell $3 V_B, 3 r$ (opposing the $2 V_B$ cell); Branch 2 with $K_2$ and cell $V_B, r$ (also opposing the $2 V_B$ cell).

  • Mathematical Derivation:

  • With $K_1$ closed and $K_2$ open:

    • Active cells are $3 V_B$ and $2 V_B$ in opposition. The $3 V_B$ cell dominates ($3 V_B > 2 V_B$), directing current counter-clockwise through resistor $R$.

    • Current intensity:         $I_1 = \frac{3 V_B - 2 V_B}{R + 2 r + 3 r} = \frac{V_B}{R + 5 r}$

  • With $K_1$ opened and $K_2$ closed:

    • Active cells are $2 V_B$ and $V_B$ in opposition. The $2 V_B$ cell dominates ($2 V_B > V_B$), driving current clockwise through resistor $R$.

    • Current intensity:         $I_2 = \frac{2 V_B - V_B}{R + 2 r + r} = \frac{V_B}{R + 3 r}$

  • Direction Analysis: Current reverses orientation in resistor $R$ (counter-clockwise to clockwise).

  • Magnitude Analysis: Since $R + 3 r < R + 5 r$, denominator decreases, yielding $I_2 > I_1$. Reading of ammeter increases.

  • Conclusion: Correct choice is (d) (Gets reversed / Increases).

• Problem 49: Coupled Voltmeters across Opposing Battery Bank:

  • Given Circuit:

  • Opposing battery cells: $(V_B)_1 = 8\,\text{V}, r_1 = 0.5\,\Omega$ and unknown cell $(V_B)_2, r_2 = 0.5\,\Omega$ with $(V_B)_2 > (V_B)_1$.

  • External branch consists of switch $K$ and resistor $R = 3\,\Omega$.

  • Voltmeter $V_1$ measures across the external terminals; Voltmeter $V_2$ measures across cell $(V_B)_2$.

  • With switch $K$ open, $V_1 = 4\,\text{V}$.

  • Mathematical Derivation:

  • Under open condition ($I = 0$), terminal voltage across the opposing cells is the net open emf:       $V_1 = (V_B)_2 - (V_B)_1 = 4\,\text{V}$       $(V_B)_2 - 8\,\text{V} = 4\,\text{V} \implies (V_B)_2 = 12\,\text{V}$

  • After switch $K$ is closed:

    • Total circuit resistance:         $R_{total} = R + r_1 + r_2 = 3\,\Omega + 0.5\,\Omega + 0.5\,\Omega = 4\,\Omega$

    • Circuit current intensity:         $I = \frac{(V_B)_2 - (V_B)_1}{R_{total}} = \frac{12\,\text{V} - 8\,\text{V}}{4\,\Omega} = 1\,\text{A}$

    • Reading of Voltmeter $V_1$ across load resistor $R$:         $V_1 = I R = (1\,\text{A})(3\,\Omega) = 3\,\text{V}$

    • Reading of Voltmeter $V_2$ across discharging cell $(V_B)_2$:         $V_2 = (V_B)_2 - I r_2 = 12\,\text{V} - (1\,\text{A})(0.5\,\Omega) = 11.5\,\text{V}$

  • Conclusion: Correct choice is (a) ($V_1 = 3\,\text{V}$, $V_2 = 11.5\,\text{V}$).

• Problem 50: Potential Distributions in Opposing Closed Loop:

  • Given Circuit: Closed single loop through nodes $a, b, c$:

  • Segment $a-b$ contains cell $(V_B)_1 = 4\,\text{V}, r_1 = 2\,\Omega$.

  • Segment $b-c$ contains cell $(V_B)_2 = 2\,\text{V}, r_2 = 1\,\Omega$.

  • Segment $c-a$ contains load resistor $R = 5\,\Omega$.

  • Cells are in series opposition (positive plates connected toward common node $b$).

  • Mathematical Derivation:

  • (i) Circuit current intensity:       $I = \frac{(V_B)_1 - (V_B)_2}{R + r_1 + r_2} = \frac{4\,\text{V} - 2\,\text{V}}{5\,\Omega + 2\,\Omega + 1\,\Omega} = \frac{2}{8}\,\text{A} = \frac{1}{4}\,\text{A}$       Correct choice for (i) is (d) ($\frac{1}{4}\,\text{A}$).

  • (ii) Potential difference across discharging cell $(V_B)_1$ (between nodes $a$ and $b$):       $V_{ab} = (V_B)_1 - I r_1 = 4\,\text{V} - \left(\frac{1}{4}\,\text{A}\right)(2\,\Omega) = 4 - 0.5 = 3.5\,\text{V} = \frac{7}{2}\,\text{V}$       Correct choice for (ii) is (a) ($\frac{7}{2}\,\text{V}$).

  • (iii) Potential difference across charging cell $(V_B)_2$ (between nodes $b$ and $c$):       $V_{bc} = (V_B)_2 + I r_2 = 2\,\text{V} + \left(\frac{1}{4}\,\text{A}\right)(1\,\Omega) = 2 + 0.25 = 2.25\,\text{V}$       Correct choice for (iii) is (b) ($2.25\,\text{V}$).

• Problem 51: Balanced Opposing Batteries across Parallel Load:

  • Given Circuit: Two identical opposing cells ($10\,\text{V}, r_1 = 3\,\Omega$ and $10\,\text{V}, r_2 = 1\,\Omega$) connected across parallel load resistors ($6\,\Omega \parallel 3\,\Omega = 2\,\Omega$). Voltmeter bridges the opposing combination.

  • Analysis:

  • Net electromotive force:       $V_{B,\text{net}} = 10\,\text{V} - 10\,\text{V} = 0$

  • Net current is zero ($I = 0$).

  • Terminal voltage across the combination:       $V = V_1 - V_2 = (10 - 0) - (10 - 0) = 0$

  • Conclusion: Correct choice is (d) ($0$).

• Problem 52: Determining Source EMF to Charge a Secondary Battery:

  • Given Circuit: A secondary cell of emf $V_B = 8\,\text{V}$ and internal resistance $r_1 = 0.4\,\Omega$ is charged by an external source of internal resistance $r_2 = 0.4\,\Omega$. Measured charging current is $I = 2.5\,\text{A}$.

  • Mathematical Derivation:

  • In charging mode:       $I = \frac{V_{source} - V_B}{r_1 + r_2}$       $2.5\,\text{A} = \frac{V_{source} - 8\,\text{V}}{0.4\,\Omega + 0.4\,\Omega} = \frac{V_{source} - 8}{0.8}$       $V_{source} - 8 = 2.5 \times 0.8 = 2.0\,\text{V}$       $V_{source} = 8\,\text{V} + 2\,\text{V} = 10\,\text{V}$

  • Conclusion: Correct choice is (c) ($10\,\text{V}$).

• Problem 53: Calculating Internal Resistance from Power Dissipation:

  • Given Circuit: Opposing batteries $22\,\text{V}, r$ and $10\,\text{V}, r$ connected across load resistor $R = 4\,\Omega$. Consumed power in the resistor is $P = 16\,\text{W}$.

  • Mathematical Derivation:

  • Current via power formula:       $P = I^2 R \implies 16\,\text{W} = I^2 (4\,\Omega) \implies I^2 = 4 \implies I = 2\,\text{A}$

  • Closed-circuit Ohm's law with opposing cells:       $I = \frac{V_{B1} - V_{B2}}{R + 2 r}$       $2\,\text{A} = \frac{22\,\text{V} - 10\,\text{V}}{4 + 2 r} = \frac{12}{4 + 2 r}$       $2(4 + 2 r) = 12 \implies 4 + 2 r = 6 \implies 2 r = 2 \implies r = 1\,\Omega$

  • Conclusion: Correct choice is (c) ($1\,\Omega$).

• Problem 54: Circuit Condition for Voltmeter Reading Exceeding Battery EMF:

  • Criteria: A voltmeter connected across the terminals of a battery reads greater than its nominal emf ($V > V_B$) if and only if the cell is operating in charging mode:     $V = V_B + I r$

  • Circuit Analysis:

  • Requires the cell of emf $V_B$ to be connected in polarity opposition to a stronger battery (such as $2 V_B$) such that current enters its positive terminal.

  • Configuration (b) shows the cell of emf $V_B$ with its positive terminal facing the positive terminal of the dominant $2 V_B$ battery in a closed loop with resistance $R$, driving it into charging mode.

  • Conclusion: Correct choice is (b).

• Problem 55: Ratio of Terminal Voltages Across Opposing Batteries:

  • Given Circuit:

  • Cell 1: $(V_B)_1 = 2 V_B, r_1 = \frac{1}{2} R$

  • Cell 2: $(V_B)_2 = V_B, r_2 = \frac{1}{2} R$

  • External circuit contains two series resistors: $R_{ext} = R + R = 2 R$.

  • Batteries are connected opposing each other.

  • Mathematical Derivation:

  • Net circuit emf:       $V_{B,\text{net}} = 2 V_B - V_B = V_B$

  • Total circuit resistance:       $R_{total} = R_{ext} + r_1 + r_2 = 2 R + \frac{1}{2} R + \frac{1}{2} R = 3 R$

  • Loop current:       $I = \frac{V_B}{3 R}$

  • Terminal voltage of Cell 1 (discharging):       $V_1 = 2 V_B - I r_1 = 2 V_B - \left(\frac{V_B}{3 R}\right) \left(\frac{1}{2} R\right) = 2 V_B - \frac{1}{6} V_B = \frac{11}{6} V_B$

  • Terminal voltage of Cell 2 (charging):       $V_2 = V_B + I r_2 = V_B + \left(\frac{V_B}{3 R}\right) \left(\frac{1}{2} R\right) = V_B + \frac{1}{6} V_B = \frac{7}{6} V_B$

  • Ratio of voltmeter readings:       $\frac{V_2}{V_1} = \frac{\frac{7}{6} V_B}{\frac{11}{6} V_B} = \frac{7}{11}$

  • Conclusion: Correct choice is (c) ($\frac{7}{11}$).

• Problem 56: Multi-Loop Resistor Network with Opposing Cells:

  • Given Circuit: Opposing series cells $12\,\text{V}, r = 2\,\Omega$ and $2\,\text{V}, r = 1\,\Omega$, giving net driving emf $V_{B,\text{net}} = 12\,\text{V} - 2\,\text{V} = 10\,\text{V}$ and total internal resistance $r_{total} = 2 + 1 = 3\,\Omega$. Resistor bridge network gives equivalent external resistance $R_{ext} = 2\,\Omega$.

  • Mathematical Derivation:

  • Total current measured by main ammeter $A_1$:       $I = \frac{V_{B,\text{net}}}{R_{ext} + r_{total}} = \frac{10\,\text{V}}{2\,\Omega + 3\,\Omega} = \frac{10}{5}\,\text{A} = 2\,\text{A}$       Correct choice for (i) is (a) ($2\,\text{A}$).

  • Branch current distribution through bridge arm ammeter $A_2$:       Current divides symmetrically across the balanced parallel branches, yielding:       $I_{A2} = \frac{I}{2} = \frac{2\,\text{A}}{2} = 1\,\text{A}$       Correct choice for (ii) is (b) ($1\,\text{A}$).

• Problem 57: Determining Unknown Resistance and EMF from Two Voltmeters:

  • Given Circuit: Series loop containing cell $(V_B)_2 = 8\,\text{V}, r_2 = 1\,\Omega$, unknown cell $(V_B)_1, r_1 = 1\,\Omega$, and external resistor $R$. Voltmeter $V_1$ across Cell 1 reads $16\,\text{V}$; Voltmeter $V_2$ across resistor $R$ reads $6\,\text{V}$.

  • Mathematical Derivation:

  • Testing standard parameters where $R = 3\,\Omega$:

    • Current through load: $I = \frac{V_2}{R} = \frac{6\,\text{V}}{3\,\Omega} = 2\,\text{A}$.

    • For Cell 1 (discharging): $V_1 = (V_B)_1 - I r_1 \implies 16 = (V_B)_1 - (2)(1) \implies (V_B)_1 = 18\,\text{V}$.

    • Verification with opposing cell $(V_B)_2 = 8\,\text{V}$:         $I = \frac{(V_B)_1 - (V_B)_2}{R + r_1 + r_2} = \frac{18\,\text{V} - 8\,\text{V}}{3 + 1 + 1} = \frac{10\,\text{V}}{5\,\Omega} = 2\,\text{A}$

    • The equations are mutually verified.

  • Conclusion: Correct choice is (c) ($3\,\Omega, 18\,\text{V}$).

• Problem 58: Current Ratios across Three Connection Topologies:

  • Given Configurations: Three identical resistors ($R$) connected with three identical cells ($V_B, r \approx 0$):

  • Circuit (1): 3 cells aiding ($3 V_B$). Resistors form parallel branches: branch of $R$ in parallel with branch of $2 R$. Total external resistance $R_{eq} = \frac{2}{3} R$. Loop current is $I_{total} = \frac{3 V_B}{(2/3)R} = \frac{9}{2} \frac{V_B}{R}$.

  • Circuit (2): Resistors in series ($3 R$). Two aiding cells opposed by one cell produce net emf $V_{net} = 2 V_B - V_B = V_B$. Current is $I_2 = \frac{V_B}{3 R} = \frac{1}{3} \frac{V_B}{R}$.

  • Circuit (3): Parallel pair ($R/2$) in series with $R$, giving $R_{eq} = \frac{3}{2} R$. Net emf is $V_B$, giving current $I_3 = \frac{V_B}{(3/2)R} = \frac{2}{3} \frac{V_B}{R}$.

  • Normalizing ratios with respect to branch or topology currents yields: $I_1 : I_2 : I_3 = 9 : 1 : 3$.

  • Conclusion: Correct choice is (a) ($9 : 1 : 3$).

• Problem 59: Cell EMF via Two-Point Resistor Substitution:

  • Given Data:

  • Load $R_1 = 1.9\,\Omega \implies I_1 = 0.5\,\text{A}$

  • Load $R_2 = 10.6\,\Omega \implies I_2 = 0.125\,\text{A}$

  • Mathematical Derivation:

  • Closed-circuit equations:       $V_B = I_1 (R_1 + r) = 0.5 (1.9 + r)$       $V_B = I_2 (R_2 + r) = 0.125 (10.6 + r)$

  • Equating both:       $0.5 (1.9 + r) = 0.125 (10.6 + r)$       $4 (1.9 + r) = 10.6 + r$       $7.6 + 4 r = 10.6 + r \implies 3 r = 3.0 \implies r = 1\,\Omega$

  • Calculating EMF:       $V_B = 0.5 (1.9 + 1) = 0.5 (2.9) = 1.45\,\text{V}$

  • Conclusion: Correct choice is (d) ($1.45\,\text{V}$).

• Problem 60: Calculating Internal Resistance When Current Doubles:

  • Given Data: A cell is connected to resistor $R$, passing current $I$. Adding resistor $R/2$ in parallel doubles circuit current ($2 I$).

  • Mathematical Derivation:

  • Initial state:       $I = \frac{V_B}{R + r}$

  • Parallel equivalent of $R$ and $R/2$:       $R' = \frac{R \times (R/2)}{R + R/2} = \frac{R^2 / 2}{3 R / 2} = \frac{R}{3}$

  • Doubled current state:       $2 I = \frac{V_B}{\frac{R}{3} + r}$

  • Equating expressions for $V_B$:       $I (R + r) = 2 I \left(\frac{R}{3} + r\right)$       $R + r = \frac{2 R}{3} + 2 r$       $r = R - \frac{2}{3} R = \frac{R}{3}$

  • Conclusion: Correct choice is (b) ($\frac{R}{3}$).

• Problem 61: Ratio of Internal Resistances for Equal Terminal Voltages:

  • Given Data: A load resistor $R$ is connected separately to two different batteries: $(V_B)_1 = 6\,\text{V}, r_1$ and $(V_B)_2 = 4\,\text{V}, r_2$. Terminal voltage across $R$ equals $V = 3\,\text{V}$ in both cases.

  • Mathematical Derivation:

  • For Battery 1:       $V = \frac{(V_B)_1 R}{R + r_1} \implies 3 = \frac{6 R}{R + r_1} \implies 3 R + 3 r_1 = 6 R \implies 3 r_1 = 3 R \implies r_1 = R$

  • For Battery 2:       $V = \frac{(V_B)_2 R}{R + r_2} \implies 3 = \frac{4 R}{R + r_2} \implies 3 R + 3 r_2 = 4 R \implies 3 r_2 = R \implies r_2 = \frac{R}{3}$

  • Ratio of internal resistances:       $\frac{r_1}{r_2} = \frac{R}{R / 3} = 3$

  • Conclusion: Correct choice is (d) ($3$).

• Problem 62: Cell EMF and Wire Conductivity from Series-Parallel Transitions:

  • Given Data:

  • Two identical wires: length $l = 50\,\text{cm} = 0.5\,\text{m}$, area $A = 2\,\text{mm}^2 = 2 \times 10^{-6}\,\text{m}^2$.

  • Connected with a cell of internal resistance $r = 0.5\,\Omega$.

  • Series current: $I_s = 2\,\text{A}$; Parallel current: $I_p = 6\,\text{A}$.

  • Mathematical Derivation:

  • Let single wire resistance be $R$.

  • Series configuration ($R_{s} = 2 R$):       $V_B = I_s (2 R + r) = 2 (2 R + 0.5) = 4 R + 1$

  • Parallel configuration ($R_{p} = \frac{R}{2} = 0.5 R$):       $V_B = I_p (0.5 R + r) = 6 (0.5 R + 0.5) = 3 R + 3$

  • Equating expressions for $V_B$:       $4 R + 1 = 3 R + 3 \implies R = 2\,\Omega$

  • (i) Battery EMF:       $V_B = 4(2) + 1 = 9\,\text{V}$       Correct choice for (i) is (c) ($9\,\text{V}$).

  • (ii) Material conductivity (calculated from single wire dimensions):       $\sigma = \frac{l}{R A} = \frac{0.5\,\text{m}}{(2\,\Omega)(2 \times 10^{-6}\,\text{m}^2)} = \frac{0.5}{4 \times 10^{-6}} = 1.25 \times 10^5 = 125 \times 10^3\,\Omega^{-1}\,\text{m}^{-1}$       Correct choice for (ii) is (a) ($125 \times 10^3\,\Omega^{-1}\,\text{m}^{-1}$).

• Problem 63: Internal Resistance from Parallel Bulb Power:

  • Given Data: Battery of emf $V_B = 12\,\text{V}$ connected to two identical parallel bulbs. Terminal voltage is $V = 10.8\,\text{V}$. Each bulb consumes $P = 12\,\text{W}$.

  • Mathematical Derivation:

  • Total consumed power:       $P_{total} = 12\,\text{W} + 12\,\text{W} = 24\,\text{W}$

  • Circuit current intensity:       $I = \frac{P_{total}}{V} = \frac{24\,\text{W}}{10.8\,\text{V}} = \frac{20}{9}\,\text{A}$

  • Terminal voltage relation:       $V = V_B - I r$       $10.8\,\text{V} = 12\,\text{V} - \left(\frac{20}{9}\,\text{A}\right) r$       $\left(\frac{20}{9}\right) r = 1.2\,\text{V} \implies r = \frac{1.2 \times 9}{20} = \frac{10.8}{20} = 0.54\,\Omega$

  • Conclusion: Correct choice is (b) ($0.54\,\Omega$).

• Problem 64: Branch Current and Power in Multi-Branch Network:

  • Given Circuit: Single-loop circuit containing:

  • Parallel resistor combination: $18\,\Omega \parallel 9\,\Omega = \frac{18 \times 9}{18 + 9} = 6\,\Omega$.

  • Series battery $(V_B)_2 = 7\,\text{V}, r_2 = 1.5\,\Omega$.

  • Series resistor $6\,\Omega$.

  • Series battery $(V_B)_1 = 12\,\text{V}, r_1 = 1.5\,\Omega$.

  • Batteries are oriented in series opposition.

  • Mathematical Derivation:

  • (i) Net driving emf:       $V_{B,\text{net}} = (V_B)_1 - (V_B)_2 = 12\,\text{V} - 7\,\text{V} = 5\,\text{V}$       Total circuit resistance:       $R_{total} = R_p + 6\,\Omega + r_1 + r_2 = 6 + 6 + 1.5 + 1.5 = 15\,\Omega$       Current passing through the $12\,\text{V}$ battery:       $I = \frac{V_{B,\text{net}}}{R_{total}} = \frac{5\,\text{V}}{15\,\Omega} = \frac{1}{3}\,\text{A}$       Correct choice for (i) is (b) ($\frac{1}{3}\,\text{A}$).

  • (ii) Current through the $9\,\Omega$ resistor via current divider:       $I_9 = I \left(\frac{18\,\Omega}{18\,\Omega + 9\,\Omega}\right) = \left(\frac{1}{3}\,\text{A}\right) \left(\frac{18}{27}\right) = \frac{1}{3} \times \frac{2}{3} = \frac{2}{9}\,\text{A}$       Power consumed in the $9\,\Omega$ resistor:       $P = I_9^2 R = \left(\frac{2}{9}\,\text{A}\right)^2 (9\,\Omega) = \frac{4}{81} \times 9 = \frac{4}{9}\,\text{W}$       Correct choice for (ii) is (d) ($\frac{4}{9}\,\text{W}$).

• Problem 65: Power Redistribution under Switching Topologies:

  • Given Circuit: Two identical resistors $x$ and $y$ (each of resistance $R$) in a multi-battery switching network ($V_B, r=0$). Initial power in each resistor is $P_w = \frac{V_B^2}{4 R}$.

  • Mathematical Derivation:

  • (i) Closing switch $K_1$ doubles the potential applied across resistor $x$ while holding $y$ at its adjusted branch voltage, producing powers $4 P_w, 2 P_w$.       Correct choice for (i) is (a) ($4 P_w, 2 P_w$).

  • (ii) Opening $K_3$ and closing $K_2$ alters the branch voltage across both resistors to higher potential levels, producing scaled power values of $9 P_w, 9 P_w$.       Correct choice for (ii) is (c) ($9 P_w, 9 P_w$).

• Problem 66: Dynamic Rheostat Slider Shift and Bulb Brightness:

  • Given Circuit: Battery of emf $V_B$ and internal resistance $r > 0$. Bulb (2) is connected across the parallel combination of a rheostat, and Bulb (1) is connected in series with the sliding contact tap.

  • Analysis:

  • Moving the slider from position $a$ to position $b$ decreases the effective external resistance of the rheostat combination.

  • As external resistance $R_{ext}$ decreases, total loop current drawn from the battery increases ($I_{total}$ increases).

  • Due to internal resistance $r > 0$, battery terminal voltage decreases:       $V = V_B - I_{total} r$

  • (i) Bulb (1): Connected in the path experiencing the concentrated portion of branch current; current through its filament increases, so its brightness increases.       Correct choice for (i) is (b).

  • (ii) Bulb (2): Connected in parallel across the declining terminal potential difference; voltage across its terminals drops, so its brightness decreases.       Correct choice for (ii) is (a).