Kirchhoff’s Rules

• Analyze a complex circuit using Kirchhoff’s rules, using the conventions for determining the correct signs of various terms.

• Kirchhoff’s first rule—the junction rule. The sum of all currents entering a junction must equal the sum of all currents leaving the junction.
• Kirchhoff’s second rule—the loop rule. The algebraic sum of changes in potential around any closed circuit path (loop) must be zero.

Kirchhoff’s rules for circuit analysis are applications of conservation laws to circuits. The first rule is the application of conservation of charge, while the second rule is the application of conservation of energy. Conservation laws, even used in a specific application, such as circuit analysis, are so basic as to form the foundation of that application.

Find the currents flowing in the circuit in [link].

Strategy

This circuit is sufficiently complex that the currents cannot be found using Ohm’s law and the series-parallel techniques—it is necessary to use Kirchhoff’s rules. Currents have been labeled $I_1$

, $I_2$

, and $I_3$

in the figure and assumptions have been made about their directions. Locations on the diagram have been labeled with letters a through h. In the solution we will apply the junction and loop rules, seeking three independent equations to allow us to solve for the three unknown currents.

Solution

We begin by applying Kirchhoff’s first or junction rule at point a. This gives

since $I_1$

flows into the junction, while $I_2$

and $I_3$

flow out. Applying the junction rule at e produces exactly the same equation, so that no new information is obtained. This is a single equation with three unknowns—three independent equations are needed, and so the loop rule must be applied.

Now we consider the loop abcdea. Going from a to b, we traverse $R_2$

in the same (assumed) direction of the current $I_2$

, and so the change in potential is $-I_2{R}_2$

. Then going from b to c, we go from $–$

to +, so that the change in potential is $+{\text{emf}}_1$

. Traversing the internal resistance $r_1$

from c to d gives $-I_2{r}_1$

. Completing the loop by going from d to a again traverses a resistor in the same direction as its current, giving a change in potential of $-I_1{R}_1$

.

The loop rule states that the changes in potential sum to zero. Thus,

Substituting values from the circuit diagram for the resistances and emf, and canceling the ampere unit gives

Now applying the loop rule to aefgha (we could have chosen abcdefgha as well) similarly gives

Note that the signs are reversed compared with the other loop, because elements are traversed in the opposite direction. With values entered, this becomes

These three equations are sufficient to solve for the three unknown currents. First, solve the second equation for $I_2$

:

Now solve the third equation for $I_3$

:

Substituting these two new equations into the first one allows us to find a value for $I_1$

:

Combining terms gives

Substituting this value for $I_1$

back into the fourth equation gives

The minus sign means $I_2$

flows in the direction opposite to that assumed in [link].

Finally, substituting the value for $I_1$

into the fifth equation gives

Discussion

Just as a check, we note that indeed $I_1=I_2+I_3$

. The results could also have been checked by entering all of the values into the equation for the abcdefgha loop.

1. Make certain there is a clear circuit diagram on which you can label all known and unknown resistances, emfs, and currents. If a current is unknown, you must assign it a direction. This is necessary for determining the signs of potential changes. If you assign the direction incorrectly, the current will be found to have a negative value—no harm done.
2. Apply the junction rule to any junction in the circuit. Each time the junction rule is applied, you should get an equation with a current that does not appear in a previous application—if not, then the equation is redundant.
3. Apply the loop rule to as many loops as needed to solve for the unknowns in the problem. (There must be as many independent equations as unknowns.) To apply the loop rule, you must choose a direction to go around the loop. Then carefully and consistently determine the signs of the potential changes for each element using the four bulleted points discussed above in conjunction with [link].
4. Solve the simultaneous equations for the unknowns. This may involve many algebraic steps, requiring careful checking and rechecking.
5. Check to see whether the answers are reasonable and consistent. The numbers should be of the correct order of magnitude, neither exceedingly large nor vanishingly small. The signs should be reasonable—for example, no resistance should be negative. Check to see that the values obtained satisfy the various equations obtained from applying the rules. The currents should satisfy the junction rule, for example.

Glossary

Kirchhoff’s rules

a set of two rules, based on conservation of charge and energy, governing current and changes in potential in an electric circuit

junction rule

Kirchhoff’s first rule, which applies the conservation of charge to a junction; current is the flow of charge; thus, whatever charge flows into the junction must flow out; the rule can be stated $I_1=I_2+I_3$

loop rule

Kirchhoff’s second rule, which states that in a closed loop, whatever energy is supplied by emf must be transferred into other forms by devices in the loop, since there are no other ways in which energy can be transferred into or out of the circuit. Thus, the emf equals the sum of the $\text{IR}$

(voltage) drops in the loop and can be stated:

$\text{emf}=\text{Ir}+{\text{IR}}_1+{\text{IR}}_2$

conservation laws

require that energy and charge be conserved in a system