Kirchhoff's Rules

Why the rules are needed, and how to set a problem up

• Real circuits contain several cells and resistors interconnected in ways that no amount of series-and-parallel reduction will simplify. Kirchhoff's two rules handle any such network.
• Label a current with a symbol and an arrow in every branch. If the answer comes out positive, the current really flows the way the arrow points; if negative, it flows the other way — do not redraw, just report the sign.
• Label each source's positive and negative electrodes and the current through it. Between the terminals of a cell, $V = V(\text{P}) - V(\text{N}) = \varepsilon - Ir$ when $I$ flows from N to P inside the cell, and $V = \varepsilon + Ir$ if the labelled current runs from P to N.
• Use the junction rule first, at every junction, so that as few unknowns as possible are left; then write loop equations, one for each independent loop, until you have as many equations as unknowns.

The two rules, with their justifications

• Junction rule. At any junction, the sum of the currents entering the junction equals the sum of the currents leaving it; equivalently the algebraic sum of the currents meeting at a junction is zero, $\Sigma I = 0$.
• Justification: conservation of charge. When currents are steady there is no accumulation of charge at any junction or at any point of a line, so the rate at which charge flows in must equal the rate at which it flows out. It applies equally to a point in an unbranched wire, and bending or re-routing the wire changes nothing.
• Loop rule. The algebraic sum of the changes in potential around any closed loop involving resistors and cells is zero; equivalently, around any closed loop, $\Sigma\varepsilon = \Sigma IR$.
• Justification: conservation of energy. Electric potential is a single-valued function of position, so on going round a closed loop and returning to the starting point the total change of potential must be zero. Equivalently, the net work done by the electric force on a charge taken round a closed loop is zero.

Sign conventions to state before you write an equation

• Go round each loop in one fixed sense and do not change it halfway.
• Only independent loops give new information: applying the loop rule to the remaining loops of a network reproduces equations you already have.

Series and parallel resistors as a special case

• Series: the junction rule forces the same current $I$ through each; the loop rule gives $V = IR_1 + IR_2 + \dots$, so

$$R_s = R_1 + R_2 + R_3 + \dots$$

• Parallel: the loop rule forces the same $V$ across each; the junction rule gives $I = V/R_1 + V/R_2 + \dots$, so

$$\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots, \qquad R_p = \dfrac{R_1R_2}{R_1+R_2}\ \text{for two}$$

• $R_s$ exceeds the largest resistance in the group; $R_p$ is less than the smallest. In series $V \propto R$; in parallel $I \propto 1/R$.
• Two points joined by a wire of zero resistance are at the same potential; anything connected between two such points carries no current and can be ignored when finding the equivalent resistance.

Worked examples of the kind CBSE sets

• Symmetry, then the loop rule (NCERT's cube). Twelve equal resistors $R$ form the edges of a cube, and a battery of emf $\varepsilon$ is across a body diagonal. By symmetry the current $3I$ from the battery splits equally into $I$ along each of the three edges at the entry corner, then into $I/2$ along each of the six middle edges, then recombines. Round the loop from one corner to the opposite one, $-IR - \dfrac{1}{2}IR - IR + \varepsilon = 0$, so $\varepsilon = \dfrac{5}{2}IR$ and

$$R_{eq} = \dfrac{\varepsilon}{3I} = \dfrac{5R}{6}$$

• With $R = 1\ \Omega$ and $\varepsilon = 10\ \text{V}$, the total current is $3I = 10/(5/6) = 12\ \text{A}$, so $I = 4\ \text{A}$.
• Two meshes with two cells. With $I_1$ through the 12 V branch (internal resistance $2\ \Omega$), $I_2$ through the 6 V branch, and $(I_1+I_2)$ through $R = 4\ \Omega$: mesh 1 gives $2I_1 + 4(I_1+I_2) = 12$, i.e. $3I_1 + 2I_2 = 6$; mesh 2 gives $4(I_1+I_2) = 6$, i.e. $2I_1 + 2I_2 = 3$. Solving, $I_1 = 3\ \text{A}$ and $I_2 = -1.5\ \text{A}$, so the current through $R$ is $1.5\ \text{A}$ and $P = (1.5)^2\times4 = 9\ \text{W}$.
• Two meshes with three resistors. With $I_1$ in the $20\ \Omega$ branch, $I_2$ in the $10\ \Omega$ branch and $(I_1 - I_2)$ in the $40\ \Omega$ branch: loop ABCDA gives $-80 + 20I_1 + 40(I_1-I_2) = 0$, i.e. $3I_1 - 2I_2 = 4$; loop DCFED gives $-40(I_1-I_2) + 10I_2 - 40 = 0$, i.e. $-4I_1 + 5I_2 = 4$. Solving, $I_1 = I_2 = 4\ \text{A}$, so the $40\ \Omega$ resistor carries zero current and the $20\ \Omega$ resistor carries 4 A.
• A branch carrying no current. If no current flows in one arm, the same current runs round the outer loop, and the loop rule applied to the smaller loop gives that current directly; the potential difference between two points is then found by walking from one to the other and adding the emfs and $IR$ drops with their signs.
• A network of equal resistors. If a network between A and B reduces to four resistors $r$, $r$, $2r$, $2r$ in parallel, $1/R_{AB} = 1/r + 1/r + 1/2r + 1/2r = 3/r$, so $R_{AB} = r/3$. With the cell's own internal resistance $r$ in series, $R = r + r/3 = 4r/3$, so $I = 3E/4r$ and the power consumed in the network is $P = I^2R_{AB} = (3E/4r)^2(r/3) = 3E^2/16r$.

Standard results

• $\Sigma I = 0$ at a junction (conservation of charge); $\Sigma\varepsilon = \Sigma IR$ round a loop (conservation of energy).
• Number of independent loop equations = number of unknown currents left after the junction rule.
• $R_s = \Sigma R_i$; $1/R_p = \Sigma(1/R_i)$; $R_p = R_1R_2/(R_1+R_2)$ for two.
• Cube of twelve equal resistors across a body diagonal: $R_{eq} = 5R/6$.
• A negative current means the arrow was drawn the wrong way; the magnitude is still correct.