Direct Current Circuits

Lessons

Senior 4 · Physics

Direct Current Circuits

5 lessons

Lesson 4 of 5

Kirchhoff's Laws

25 min

When series and parallel are not enough

A solar panel charges a battery while the battery also runs a lamp; two batteries of different e.m.f. are connected to the same resistor network. Such circuits cannot be simplified into series and parallel groups, because more than one source drives current through the same components. In 1845 the German physicist Gustav Kirchhoff stated two rules that solve any d.c. circuit. They are nothing more than the conservation of charge and the conservation of energy, written for circuits.

Kirchhoff's first law: the junction rule

Charge cannot be created, destroyed or stored at a junction of wires. So the charge flowing in each second equals the charge flowing out each second:

\sum I_{\text{in}} = \sum I_{\text{out}}

Figure 4.1: Currents at a junction.

In Figure 4.1, $5.0 + 2.0 = 3.0 + 1.5 + I_{5}$ , so $I_5 = 2.5\,\mathrm{A}$ out of the junction.

Kirchhoff's second law: the loop rule

A coulomb that travels once round any closed loop of a circuit returns to its starting point with the same energy it started with. So the energy it gains from e.m.f.s must equal the energy it gives up in resistances:

\sum E = \sum IR \quad\text{around any closed loop}

Equivalently, the algebraic sum of the potential changes around any closed loop is zero.

Definition

Kirchhoff's first (junction) law: the sum of the currents entering any junction equals the sum of the currents leaving it (conservation of charge). Kirchhoff's second (loop) law: in any closed loop of a circuit, the sum of the e.m.f.s equals the sum of the products $I R$ (conservation of energy). Example: in a single loop with a 9.0 V battery and resistors of $1.0\,\Omega$ and $2.0\,\Omega$ , $9.0 = I (1.0 + 2.0)$ , so $I = 3.0\,\mathrm{A}$ .

Signs, and a method that always works

Going round the loop in your chosen direction…	Count it as
through a cell from its − terminal to its + terminal	$+ E$ (energy gained)
through a cell from + to −	$- E$
through a resistor in the same direction as the current you assumed	$+ I R$ (energy given up)
through a resistor against the current you assumed	$- I R$

Draw the circuit. Give each branch a current and a direction (a guess is fine). Use the junction rule straight away to reduce the number of unknowns.
Choose enough independent loops (one per unknown left over), each with a direction of travel.
Write $\sum E = \sum IR$ for each loop, using the sign rules.
Solve the simultaneous equations. A negative answer means the current flows opposite to the direction you assumed; its size is still right.
Check with a loop you did not use.

Worked example 1: two cells in one loop

A 12 V battery charges a 6.0 V battery: the two are connected positive to positive through a $2.5\,\Omega$ resistor. Their internal resistances are $0.30\,\Omega$ and $0.20\,\Omega$ . Find the charging current.

Going round the loop in the direction the 12 V battery drives current, we go − to + through the 12 V battery ( $+ 12$ ) and + to − through the 6.0 V battery ( $- 6.0$ ):

12 - 6.0 = I(2.5 + 0.30 + 0.20) \;\Rightarrow\; I = \frac{6.0}{3.0} = 2.0\,\mathrm{A}

The current enters the 6.0 V battery at its positive terminal, which is what charging means: energy is being stored in it.

A two-loop circuit

Figure 4.2: Two sources, three branches: a job for Kirchhoff's laws.

Worked example 2: solving Figure 4.2

Find the current in each branch (ignore internal resistance).

Junction P: $I_{3} = I_{1} + I_{2}$ .

Loop 1 (left loop, clockwise: up through $E_{1}$ , along $R_{1}$ , down $R_{3}$ ): $12 = 4.0 I_{1} + 6.0 I_{3} = 4.0 I_{1} + 6.0 (I_{1} + I_{2})$ , so

10I_1 + 6.0I_2 = 12 \qquad (1)

Loop 2 (right loop, anticlockwise: up through $E_{2}$ , along $R_{2}$ , down $R_{3}$ ): $6.0 = 2.0 I_{2} + 6.0 I_{3} = 2.0 I_{2} + 6.0 (I_{1} + I_{2})$ , so

6.0I_1 + 8.0I_2 = 6.0 \qquad (2)

Multiply (1) by 3 and (2) by 5 and subtract: $(30 - 30) I_{1} + (18 - 40) I_{2} = 36 - 30$ , so $I_2 = -\frac{6}{22} = -0.27\,\mathrm{A}$ . Then from (1), $I_1 = \frac{12 - 6.0(-0.273)}{10} = 1.36\,\mathrm{A}$ , and $I_3 = 1.36 - 0.27 = 1.09\,\mathrm{A}$ .

I_1 = 1.36\,\mathrm{A}, \qquad I_2 = -0.27\,\mathrm{A}, \qquad I_3 = 1.09\,\mathrm{A}

$I_{2}$ is negative: the current in the right branch flows away from P, entering $E_{2}$ at its positive terminal, so the 12 V source is charging the 6.0 V one. Check with the outer loop (clockwise round the whole circuit): $12 - 6.0 = 4.0 \times 1.36 - 2.0 \times (-0.27) = 5.45 + 0.55 = 6.0\,\mathrm{V}$ . ✓

Experiment 4.1: testing Kirchhoff's laws

Apparatus: two cells or low-voltage supplies of different e.m.f. (for example 3.0 V and 1.5 V), three resistors (about $10\,\Omega$ to $47\,\Omega$ ), three ammeters (or one moved between branches), a voltmeter, leads.

Build a circuit with the shape of Figure 4.2. Measure the current in each branch, noting its direction from the meter's polarity. Check that the currents into the middle junction add up to the current out.
Measure the p.d. across each component. Go round each loop and check that the p.d.s across the resistors add up to the sum of the e.m.f.s, taking signs into account.
Compare the measured currents with those calculated with Kirchhoff's laws from the measured resistances and e.m.f.s.

Small differences come from the cells' internal resistances and the meters' own resistances.

Warning

Safety. Use only low-voltage cells or supplies. Connecting cells of different e.m.f. directly in parallel, or short-circuiting a cell, can make wires and cells dangerously hot: always include resistors in every branch.

Warning

Common mistakes. (1) Mixing sign conventions within one loop. (2) Writing a loop equation for a path that is not closed. (3) Writing more loop equations than unknowns, or two equations that are not independent; use the third loop only as a check. (4) "Correcting" a negative answer by changing its sign without changing the direction of the current.

Tip

Use the junction rule first, in the diagram itself (label the third current $I_{1} + I_{2}$ ). Fewer unknowns means fewer equations and fewer chances to slip.

Solving circuit problems with Kirchhoff's rules (Physics Ninja). Watch how currents are labelled, loops chosen and the simultaneous equations solved for a circuit with more than one source.

✦

Practice

State Kirchhoff's two laws and the conservation law on which each is based.
At a junction, 3.2 A and 1.8 A flow in, and 2.5 A flows out along one wire. Find the current in the only other wire, and its direction.
A 9.0 V battery (internal resistance $0.50\,\Omega$ ) and a 3.0 V battery (internal resistance $0.50\,\Omega$ ) are connected in series, opposing each other, with a $5.0\,\Omega$ resistor. Find the current.
In a circuit like Figure 4.2, $E_1 = 10\,\mathrm{V}$ , $R_1 = 2.0\,\Omega$ , $E_2 = 5.0\,\mathrm{V}$ , $R_2 = 1.0\,\Omega$ and $R_3 = 4.0\,\Omega$ . Find the three currents.
For question 4, find the p.d. across $R_{3}$ and check it with both outer branches.
Show that Kirchhoff's laws give the rule $R = R_{1} + R_{2}$ for resistors in series and $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$ for resistors in parallel.
A student solves a two-loop circuit and finds all three currents positive, but the current she calculated in the middle branch is larger than the sum of the other two. Evaluate her answer.

Show answersHide answers

Junction law: current in = current out at any junction (conservation of charge). Loop law: sum of e.m.f.s = sum of $I R$ round any closed loop (conservation of energy).
$3.2 + 1.8 = 2.5 + I$ , so $I = 2.5\,\mathrm{A}$ , flowing out of the junction.
$I = \frac{9.0 - 3.0}{5.0 + 0.50 + 0.50} = 1.0\,\mathrm{A}$ .
Loop 1: $10 = 2.0I_1 + 4.0(I_1 + I_2) \Rightarrow 6.0I_1 + 4.0I_2 = 10$ . Loop 2: $5.0 = 1.0I_2 + 4.0(I_1 + I_2) \Rightarrow 4.0I_1 + 5.0I_2 = 5.0$ . Solving: $I_1 = \frac{15}{7} = 2.14\,\mathrm{A}$ , $I_2 = -\frac{5}{7} = -0.71\,\mathrm{A}$ (so the 5.0 V source is being charged), $I_3 = \frac{10}{7} = 1.43\,\mathrm{A}$ .
$V_3 = 4.0 \times 1.43 = 5.7\,\mathrm{V}$ . Left branch: $10 - 2.0 \times 2.14 = 5.7\,\mathrm{V}$ . Right branch: $5.0 - 1.0 \times (-0.71) = 5.7\,\mathrm{V}$ . ✓
Series: one current $I$ ; loop law $V = I R_{1} + I R_{2} = I (R_{1} + R_{2})$ , so $R = R_{1} + R_{2}$ . Parallel: same p.d. $V$ across each (loop law round each branch); junction law $I = \frac{V}{R_1} + \frac{V}{R_2}$ , so $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$ .
The answer must be wrong. If all three currents are defined as flowing into or out of the junction as in Figure 4.2, the junction law requires $I_{3} = I_{1} + I_{2}$ exactly; a larger $I_{3}$ breaks conservation of charge. She has made an arithmetic or sign error, which a check with the third loop would reveal.

Quick check

Key points

Junction law: $\sum I_{\text{in}} = \sum I_{\text{out}}$ (conservation of charge).
Loop law: $\sum E = \sum IR$ round any closed loop (conservation of energy), with consistent signs.
Method: label currents (use the junction law first), choose independent loops, solve, and check with another loop. A negative current flows the other way.

Sources and credits

21.3: Kirchhoff's Rules, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Photo: Gustav Robert Kirchhoff, Library of Congress, Wikimedia Commons, public domain.
Video: Solving Circuit Problems using Kirchhoff's Rules, Physics Ninja, embedded from YouTube.

The diagrams, the method, the experiment, the worked examples and the practice exercises are written by Ganzaa.

1When series and parallel are not enough

2Kirchhoff's first law: the junction rule

3Kirchhoff's second law: the loop rule

4Signs, and a method that always works

5A two-loop circuit

6Experiment 4.1: testing Kirchhoff's laws