Derivation

Ohm's Law, Resistance and Kirchhoff's Rules for DC Circuits

D-427 Home PU-102 Threads fields · energy Depends on Continuity Equation and Charge Conservation, Irrotationality of the Electrostatic Field

Statement

For a uniform conductor of length \(L\), cross-sectional area \(A\), and resistivity \(\rho\), the potential difference \(V\) across its ends is proportional to the current \(I\) flowing through it, with proportionality constant \(R = \rho L / A\). In a closed DC network of such ohmic resistors and ideal voltage sources, the current at any junction is conserved (Kirchhoff's junction rule) and the algebraic sum of potential differences around any closed loop is zero (Kirchhoff's loop rule), allowing the complete solution of branch currents and node potentials.

Why it matters

Ohm's law and Kirchhoff's rules form the foundational toolkit for analyzing direct-current circuits. While microscopic electromagnetism describes charge motion via Maxwell's equations, circuit theory abstracts this into lumped elements—resistors and sources—where voltage and current are the primary variables. This abstraction enables the design and analysis of everything from simple battery-powered devices to complex electronic systems, without solving partial differential equations for the electric field in every component.

The junction rule embodies local charge conservation, while the loop rule reflects the conservative nature of electrostatic fields in steady state. Together, they provide a systematic, algebraic method for determining currents and voltages in arbitrary networks, reducing complex physical problems to linear algebra. This framework is the gateway to understanding AC circuits, filters, and signal processing.

Assumptions

Ohmic behavior: \(V\) is proportional to \(I\).Many materials (metals, electrolytes) obey \(V=IR\) only within a limited range of temperature and field strength. Non-ohmic devices (diodes, transistors, filament lamps) have \(I\) dependent on \(V\) in a nonlinear way, so \(R\) is not constant. Kirchhoff's rules still hold, but the network is no longer linear: superposition fails and the equations must be solved graphically or numerically.

Steady-state DC: currents and voltages are time-independent.In transient or AC circuits, inductors and capacitors store energy and introduce phase shifts and frequency-dependent impedance. Kirchhoff's rules still apply to instantaneous values, but the element relations become \(v_L = L\,di/dt\) and \(i_C = C\,dv/dt\), and the simple \(V=IR\) model fails for reactive components.

Electrostatic field: \(\nabla \times \vec{E} = 0\).The loop rule \(\sum V = 0\) relies on the electric field being conservative. In the presence of time-varying magnetic fields (Faraday's law), \(\nabla \times \vec{E} = -\partial \vec{B}/\partial t \neq 0\), and the line integral of \(\vec{E}\) around a closed loop is not zero. The loop rule must then include the induced EMF term.

Uniform resistivity \(\rho\) within each element.The derivation \(R = \rho L/A\) assumes \(\rho\) is constant throughout the conductor. In real wires, \(\rho\) varies with temperature, and in composite or inhomogeneous materials, the effective resistance requires solving \(\nabla \cdot (\sigma \nabla V) = 0\) with appropriate boundary conditions, not simply \(R=\rho L/A\).

Point-like junctions: no charge accumulation at nodes.Kirchhoff's junction rule \(\sum I_{\text{in}} = 0\) assumes that any charge entering a node leaves immediately, with no net accumulation. At very high frequencies or in devices with significant capacitance (e.g., MOSFET gates), charge can accumulate at nodes, and the junction rule must be modified to include displacement current \(I_{\text{disp}} = C\,dV/dt\).

Derivation

\[ \vec{J} = \sigma \vec{E} \]

Microscopic Ohm's law: current density \(\vec{J}\) is proportional to electric field \(\vec{E}\), with conductivity \(\sigma\). This is a local, constitutive relation valid for ohmic materials. A

\[ \sigma = \frac{n e^2 \tau}{m} \]

Drude model: conductivity arises from free charge density \(n\), charge \(e\), mean free time \(\tau\), and mass \(m\). This connects macroscopic \(\sigma\) to microscopic scattering. B

\[ V = \int_{0}^{L} E\,dx = E L \]

For a uniform wire with constant \(E\), the potential difference is the line integral of \(\vec{E}\) along the length \(L\). A

\[ I = \int_{A} \vec{J}\cdot d\vec{A} = J A \]

For a uniform cross-section with \(\vec{J}\) perpendicular to the area, the total current is the flux of \(\vec{J}\) through the cross-section. A

\[ J = \sigma E = \sigma \frac{V}{L} \quad\Longrightarrow\quad I = \sigma \frac{V}{L} A \]

Substitute \(E = V/L\) and \(J = \sigma E\) into the current expression. A

\[ I = \frac{\sigma A}{L} V \quad\Longrightarrow\quad V = \frac{L}{\sigma A} I \]

Rearrange to isolate \(V\) as a function of \(I\). A

\[ R \equiv \frac{V}{I} = \frac{L}{\sigma A} = \rho \frac{L}{A} \]

Define resistance \(R\) as the ratio \(V/I\). Introduce resistivity \(\rho = 1/\sigma\), the material property. The geometry factor \(L/A\) shows that longer, thinner wires have higher resistance. C

\[ P = IV = I^2 R = \frac{V^2}{R} \]

Joule heating: power dissipated as heat is the product of current and voltage. Using \(V=IR\), this can be written in terms of \(I\) and \(R\) or \(V\) and \(R\). A

\[ \nabla \cdot \vec{J} = 0 \quad\text{(steady state)} \]

Charge conservation in steady state: no net charge accumulation, so the divergence of current density is zero. B

\[ \sum_{\text{in}} I_k = \sum_{\text{out}} I_k \quad\Longrightarrow\quad \sum_{k} I_k = 0 \]

Integrate \(\nabla \cdot \vec{J} = 0\) over a small volume enclosing a junction. By the divergence theorem, the net current flux through the boundary is zero, giving Kirchhoff's junction rule. C

\[ \nabla \times \vec{E} = 0 \quad\text{(electrostatic)} \]

In steady state with no time-varying magnetic fields, the electric field is conservative. B

\[ \oint \vec{E}\cdot d\vec{l} = 0 \quad\Longrightarrow\quad \sum_{\text{loop}} V_k = 0 \]

Integrate \(\nabla \times \vec{E} = 0\) around a closed loop. By Stokes' theorem, the line integral is zero, giving Kirchhoff's loop rule. EMFs are included as potential rises. C

\[ R_{\text{series}} = R_1 + R_2 \]

For resistors in series, the same current \(I\) flows through each. The total voltage is \(V = V_1 + V_2 = I R_1 + I R_2 = I(R_1+R_2)\), so \(R_{\text{series}} = R_1+R_2\). A

\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} \]

For resistors in parallel, the same voltage \(V\) is across each. The total current is \(I = I_1 + I_2 = V/R_1 + V/R_2 = V(1/R_1 + 1/R_2)\), so \(1/R_{\text{parallel}} = 1/R_1 + 1/R_2\). A

Result

\[ V = IR, \quad R = \rho \frac{L}{A}, \quad P = I^2 R = \frac{V^2}{R} \]

Reading. Ohm's law relates voltage, current, and resistance in a linear, proportional way. Resistance depends on the material (\(\rho\)) and geometry (\(L/A\)). Power dissipation is quadratic in current or inversely proportional to resistance for a fixed voltage. Kirchhoff's rules provide the conservation laws (charge and energy) that allow solving for currents and voltages in complex networks.

Units check. \([V] = \mathrm{V}\), \([I] = \mathrm{A}\), \([R] = \mathrm{\Omega}\). Since \(1\ \mathrm{V} = 1\ \mathrm{A \cdot \Omega}\), \(V=IR\) is dimensionally consistent. \([\rho] = \mathrm{\Omega \cdot m}\), so \([\rho L/A] = \mathrm{(\Omega \cdot m) \cdot m / m^2 = \Omega}\). \([P] = \mathrm{W}\), and \([I^2 R] = \mathrm{A^2 \cdot \Omega = A \cdot (A \cdot \Omega) = A \cdot V = W}\). Consistent.

Limiting cases

Zero resistance (\(R=0\), ideal wire): \(V=0\) for any finite \(I\). No power dissipation \(P=I^2 R=0\). Current is determined by the rest of the circuit, not by the wire itself.
Infinite resistance (\(R\to\infty\), open circuit): \(I=0\) for any finite \(V\). No current flows, and no power is dissipated. The circuit is broken.
Single resistor with a battery: \(I = \mathcal{E}/R\), \(V = \mathcal{E}\), \(P = \mathcal{E}^2/R\). This is the simplest application of Ohm's law and Kirchhoff's rules.
Two resistors in series with a battery: \(I = \mathcal{E}/(R_1+R_2)\), \(V_1 = I R_1 = \mathcal{E} R_1/(R_1+R_2)\), \(V_2 = I R_2 = \mathcal{E} R_2/(R_1+R_2)\). The voltage divides in proportion to resistance.
Two resistors in parallel with a battery: \(I_1 = \mathcal{E}/R_1\), \(I_2 = \mathcal{E}/R_2\), \(I_{\text{total}} = \mathcal{E}(1/R_1 + 1/R_2)\). The current divides inversely with resistance.

Breaks when

Non-ohmic devices. Diodes, transistors, and filament lamps do not obey \(V=IR\) with constant \(R\). Their \(I\)-\(V\) relation is nonlinear: Kirchhoff's rules still hold, but superposition does not, and the element relation is not \(V=IR\). For a diode, \(I = I_0(e^{V/V_T}-1)\), and \(R\) depends on \(V\).
High frequency or transient circuits. Inductors and capacitors store energy and introduce phase shifts. The element relations become \(v_L = L\,di/dt\) and \(i_C = C\,dv/dt\), and the simple \(V=IR\) model fails. Kirchhoff's rules still apply to instantaneous values, but the circuit is no longer purely resistive.
Temperature-dependent resistivity. In real conductors, \(\rho\) increases with temperature (for metals) or decreases (for semiconductors). If the current is large enough to heat the wire significantly, \(R\) is no longer constant, and the circuit becomes nonlinear. The relation \(V=IR\) still holds instantaneously, but \(R\) is a function of temperature, which depends on \(I\).
Time-varying magnetic fields. If \(\partial \vec{B}/\partial t \neq 0\), then \(\nabla \times \vec{E} \neq 0\), and the line integral of \(\vec{E}\) around a closed loop is not zero. Kirchhoff's loop rule must be modified to include the induced EMF: \(\sum V_k + \mathcal{E}_{\text{induced}} = 0\).

Failure modes

Confusing series and parallel combinations. Adding resistances in series (\(R_{\text{total}} = R_1+R_2\)) when they are actually in parallel, or vice versa. The rule is: same current → series; same voltage → parallel.
Sign errors in Kirchhoff's loop rule. Failing to assign a consistent direction for current and voltage drops. When traversing a loop, a voltage drop across a resistor (in the direction of current) is negative, and a voltage rise (against the direction of current) is positive. EMFs are positive when traversed from \(-\) to \(+\).
Assuming current is the same in parallel branches. In a parallel combination, the voltage is the same across each branch, but the current divides according to \(I_k = V/R_k\). The total current is the sum of the branch currents.
Ignoring internal resistance of batteries. Real batteries have internal resistance \(r\), so the terminal voltage is \(V = \mathcal{E} - I r\), not \(\mathcal{E}\). If the load resistance \(R\) is much smaller than \(r\), the terminal voltage is significantly less than the EMF, and the current is limited by \(r\), not \(R\).

Discussion

Ohm's law is a constitutive relation, not a fundamental law of nature. It describes the behavior of a specific class of materials (ohmic conductors) under specific conditions (steady state, constant temperature, moderate fields). The deeper principle is charge conservation (Kirchhoff's junction rule) and energy conservation (Kirchhoff's loop rule), which hold universally for DC circuits. Ohm's law is a model that works well for metals and electrolytes, but fails for semiconductors, diodes, and other nonlinear devices.

Kirchhoff's rules are not independent laws but consequences of Maxwell's equations in the steady-state, electrostatic limit. The junction rule follows from \(\nabla \cdot \vec{J} = 0\) (charge conservation), and the loop rule follows from \(\nabla \times \vec{E} = 0\) (conservative electric field). These rules are powerful because they reduce a continuous field problem to a discrete algebraic problem, allowing the analysis of complex networks without solving partial differential equations.

The concept of resistance as a lumped parameter is an abstraction. In reality, the electric field and current density are distributed throughout the conductor, and the relation between \(V\) and \(I\) is a result of integrating the local relation \(\vec{J} = \sigma \vec{E}\) over the geometry. The formula \(R = \rho L/A\) is exact only for uniform conductors with uniform cross-section and uniform resistivity. For non-uniform geometries, the resistance must be calculated by solving the Laplace equation \(\nabla^2 V = 0\) with appropriate boundary conditions.

At a deeper level, the validity of Kirchhoff's rules depends on the scale separation between the circuit dimensions and the wavelength of the electromagnetic radiation. In the DC limit (wavelength much larger than the circuit size), the electric field is quasi-static, and the loop rule holds. At high frequencies, the finite speed of light and the inductive/capacitive effects of the wires become significant, and the circuit must be analyzed using transmission line theory or full electromagnetic simulation. The transition from circuit theory to field theory is gradual and depends on the frequency, the size of the circuit, and the precision required.

Common misconceptions. Resistance is a property of a particular object (material and shape, \(R = \rho L/A\)), while resistivity is a property of the material alone. A wire with high resistance does not "use up" voltage; rather, the voltage drop across it is a result of the work done by the electric field on the charge carriers, which is dissipated as heat. Current is not "used up" in a circuit; it is the same throughout a series combination, and it divides in parallel combinations. The battery does not "push" current; it maintains a potential difference, and the electric field drives the current.

Worked examples

Example 1. A circuit consists of a 12 V battery with internal resistance \(r = 1.0\ \Omega\) connected to a load resistor \(R = 5.0\ \Omega\). Find the current, the terminal voltage, and the power dissipated in the load.

\[ I = \frac{\mathcal{E}}{R + r} = \frac{12}{5.0 + 1.0} = \frac{12}{6.0} = 2.0\ \mathrm{A} \]

Total resistance is \(R+r\) (series). Current is EMF divided by total resistance. A

\[ V_{\text{terminal}} = \mathcal{E} - I r = 12 - (2.0)(1.0) = 12 - 2.0 = 10.0\ \mathrm{V} \]

Terminal voltage is EMF minus the voltage drop across internal resistance. A

\[ P_{\text{load}} = I^2 R = (2.0)^2 (5.0) = 4.0 \times 5.0 = 20.0\ \mathrm{W} \]

Power dissipated in the load is \(I^2 R\). A

\[ I = 2.0\ \mathrm{A}, \quad V_{\text{terminal}} = 10.0\ \mathrm{V}, \quad P_{\text{load}} = 20.0\ \mathrm{W} \]

Reading. The current is limited by the total resistance \(R+r\). The terminal voltage is less than the EMF due to the internal resistance drop. The power dissipated in the load is 20 W, while the power dissipated in the internal resistance is \(I^2 r = (2.0)^2(1.0) = 4.0\ \mathrm{W}\), for a total of 24 W, which equals \(\mathcal{E} I = 12 \times 2.0 = 24\ \mathrm{W}\).

Units check. \([I] = \mathrm{V/\Omega = A}\). \([V_{\text{terminal}}] = \mathrm{V}\). \([P_{\text{load}}] = \mathrm{A^2 \cdot \Omega = W}\). Consistent.

Example 2. A two-loop circuit has a 20 V battery in the left loop and a 10 V battery in the right loop. The left loop has a resistor \(R_1 = 4.0\ \Omega\) in series with the 20 V battery. The right loop has a resistor \(R_2 = 6.0\ \Omega\) in series with the 10 V battery. The two loops share a central resistor \(R_3 = 2.0\ \Omega\), and each battery raises the potential in the clockwise sense of its own loop. Find the currents \(I_1\) (left loop, clockwise), \(I_2\) (right loop, clockwise), and the current through \(R_3\).

\[ \text{Junction rule at the top node: } I_1 = I_2 + I_3 \]

Current entering the node equals current leaving. \(I_3\) is the current through \(R_3\), downward. A

\[ \text{Left loop (clockwise): } 20 - I_1 R_1 - I_3 R_3 = 0 \]

Kirchhoff's loop rule: EMF rise, then voltage drops across \(R_1\) and \(R_3\). A

\[ \text{Right loop (clockwise): } 10 - I_2 R_2 + I_3 R_3 = 0 \]

Kirchhoff's loop rule: EMF rise, voltage drop across \(R_2\), and voltage rise across \(R_3\) (since \(I_3\) is downward, opposite to the clockwise traversal). B

\[ 20 - 4.0 I_1 - 2.0 I_3 = 0 \quad\Longrightarrow\quad 4.0 I_1 + 2.0 I_3 = 20 \]

Substitute \(R_1=4.0\ \Omega\), \(R_3=2.0\ \Omega\). A

\[ 10 - 6.0 I_2 + 2.0 I_3 = 0 \quad\Longrightarrow\quad 6.0 I_2 - 2.0 I_3 = 10 \]

Substitute \(R_2=6.0\ \Omega\), \(R_3=2.0\ \Omega\). A

\[ I_1 = I_2 + I_3 \quad\Longrightarrow\quad I_2 = I_1 - I_3 \]

Substitute the junction rule into the right loop equation. A

\[ 6.0(I_1 - I_3) - 2.0 I_3 = 10 \quad\Longrightarrow\quad 6.0 I_1 - 8.0 I_3 = 10 \]

Expand and combine like terms. B

\[ \begin{aligned} 4.0 I_1 + 2.0 I_3 &= 20 \\ 6.0 I_1 - 8.0 I_3 &= 10 \end{aligned} \]

System of two linear equations in \(I_1\) and \(I_3\). A

\[ \text{Multiply first by 4: } 16.0 I_1 + 8.0 I_3 = 80 \]

Eliminate \(I_3\) by adding to the second equation. A

\[ (16.0 I_1 + 8.0 I_3) + (6.0 I_1 - 8.0 I_3) = 80 + 10 \quad\Longrightarrow\quad 22.0 I_1 = 90 \]

Add the equations. A

\[ I_1 = \frac{90}{22.0} = 4.09\ \mathrm{A} \]

Solve for \(I_1\). A

\[ 4.0(4.09) + 2.0 I_3 = 20 \quad\Longrightarrow\quad 16.36 + 2.0 I_3 = 20 \quad\Longrightarrow\quad 2.0 I_3 = 3.64 \]

Substitute \(I_1\) into the first equation. A

\[ I_3 = \frac{3.64}{2.0} = 1.82\ \mathrm{A} \]

Solve for \(I_3\). A

\[ I_2 = I_1 - I_3 = 4.09 - 1.82 = 2.27\ \mathrm{A} \]

Solve for \(I_2\) using the junction rule. A

\[ I_1 = 4.09\ \mathrm{A}, \quad I_2 = 2.27\ \mathrm{A}, \quad I_3 = 1.82\ \mathrm{A} \]

Reading. The current in the left loop is 4.09 A, in the right loop is 2.27 A, and through the central resistor is 1.82 A (downward). The junction rule is satisfied: \(4.09 = 2.27 + 1.82\). The loop rules are also satisfied: left loop \(20 - 4.09(4.0) - 1.82(2.0) = 20 - 16.36 - 3.64 = 0\); right loop \(10 - 2.27(6.0) + 1.82(2.0) = 10 - 13.62 + 3.64 = 0.02 \approx 0\) (rounding error).

Units check. All currents are in amperes. The equations are dimensionally consistent: \(\mathrm{V = A \cdot \Omega}\). Consistent.

Problems

A resistor with resistance \(R = 10.0\ \Omega\) has a current \(I = 2.5\ \mathrm{A}\) flowing through it. Find the voltage across the resistor and the power dissipated.
Solution
\(V = IR = (2.5)(10.0) = 25.0\ \mathrm{V}\). \(P = I^2 R = (2.5)^2(10.0) = 6.25 \times 10.0 = 62.5\ \mathrm{W}\).
Two resistors, \(R_1 = 3.0\ \Omega\) and \(R_2 = 6.0\ \Omega\), are connected in series across a 12 V battery. Find the current, the voltage across each resistor, and the power dissipated in each.
Solution
\(R_{\text{total}} = R_1 + R_2 = 3.0 + 6.0 = 9.0\ \Omega\). \(I = V/R_{\text{total}} = 12/9.0 = 1.33\ \mathrm{A}\). \(V_1 = I R_1 = (1.33)(3.0) = 4.0\ \mathrm{V}\). \(V_2 = I R_2 = (1.33)(6.0) = 8.0\ \mathrm{V}\). \(P_1 = I^2 R_1 = (1.33)^2(3.0) = 1.77 \times 3.0 = 5.3\ \mathrm{W}\). \(P_2 = I^2 R_2 = (4/3)^2(6.0) = 10.7\ \mathrm{W}\); together \(16\ \mathrm{W} = \mathcal{E}I = 12 \times 4/3\).
Two resistors, \(R_1 = 4.0\ \Omega\) and \(R_2 = 12.0\ \Omega\), are connected in parallel across a 24 V battery. Find the current through each resistor, the total current, and the equivalent resistance.
Solution
\(I_1 = V/R_1 = 24/4.0 = 6.0\ \mathrm{A}\). \(I_2 = V/R_2 = 24/12.0 = 2.0\ \mathrm{A}\). \(I_{\text{total}} = I_1 + I_2 = 6.0 + 2.0 = 8.0\ \mathrm{A}\). \(1/R_{\text{eq}} = 1/R_1 + 1/R_2 = 1/4.0 + 1/12.0 = 0.25 + 0.0833 = 0.3333\ \mathrm{S}\). \(R_{\text{eq}} = 1/0.3333 = 3.0\ \Omega\).
A battery with EMF \(\mathcal{E} = 15\ \mathrm{V}\) and internal resistance \(r = 0.5\ \Omega\) is connected to a load resistor \(R = 7.5\ \Omega\). Find the current, the terminal voltage, and the power dissipated in the load and in the internal resistance.
Solution
\(I = \mathcal{E}/(R+r) = 15/(7.5+0.5) = 15/8.0 = 1.875\ \mathrm{A}\). \(V_{\text{terminal}} = \mathcal{E} - I r = 15 - (1.875)(0.5) = 15 - 0.9375 = 14.06\ \mathrm{V}\). \(P_{\text{load}} = I^2 R = (1.875)^2(7.5) = 3.5156 \times 7.5 = 26.37\ \mathrm{W}\). \(P_{\text{internal}} = I^2 r = (1.875)^2(0.5) = 3.5156 \times 0.5 = 1.76\ \mathrm{W}\).
A battery of EMF \(\mathcal{E} = 12\ \mathrm{V}\) and internal resistance \(r = 2.0\ \Omega\) drives a variable load \(R\). Show that the power delivered to the load is greatest when \(R = r\), and find that maximum power and the efficiency at that point.
Solution
\(I = \mathcal{E}/(R+r)\), so \(P(R) = I^2R = \mathcal{E}^2R/(R+r)^2\). Then \(dP/dR = \mathcal{E}^2\,[(R+r)^2 - 2R(R+r)]/(R+r)^4 = \mathcal{E}^2\,(r-R)/(R+r)^3\), which vanishes at \(R = r\) and changes sign from positive to negative there, so it is a maximum. \(P_{\max} = \mathcal{E}^2/(4r) = 144/8.0 = 18\ \mathrm{W}\). At this point the internal resistance dissipates the same \(18\ \mathrm{W}\), so the efficiency \(P_{\text{load}}/(\mathcal{E}I) = R/(R+r) = 50\%\): maximum power transfer is not maximum efficiency.