Kirchhoff’s First Rule

Kirchhoff’s first rule (the junction rule) is an application of the conservation of charge to a junction; it is illustrated in [link]. Current is the flow of charge, and charge is conserved; thus, whatever charge flows into the junction must flow out. Kirchhoff’s first rule requires that $I_1=I_2+I_3$ (see figure). Equations like this can and will be used to analyze circuits and to solve circuit problems.

Kirchhoff’s Second Rule

Kirchhoff’s second rule (the loop rule) is an application of conservation of energy. The loop rule is stated in terms of potential, $V$, rather than potential energy, but the two are related since ${\text{PE}}_{\text{elec}}=\text{qV}$. Recall that emf is the potential difference of a source when no current is flowing. 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. [link] illustrates the changes in potential in a simple series circuit loop.

Kirchhoff’s second rule requires $\text{emf}-\text{Ir}-{\text{IR}}_1-{\text{IR}}_2=0$. Rearranged, this is $\text{emf}=\text{Ir}+{\text{IR}}_1+{\text{IR}}_2$, which means the emf equals the sum of the $\text{IR}$ (voltage) drops in the loop.

Applying Kirchhoff’s Rules

By applying Kirchhoff’s rules, we generate equations that allow us to find the unknowns in circuits. The unknowns may be currents, emfs, or resistances. Each time a rule is applied, an equation is produced. If there are as many independent equations as unknowns, then the problem can be solved. There are two decisions you must make when applying Kirchhoff’s rules. These decisions determine the signs of various quantities in the equations you obtain from applying the rules.

1. When applying Kirchhoff’s first rule, the junction rule, you must label the current in each branch and decide in what direction it is going. For example, in [link], [link], and [link], currents are labeled $I_1$, $I_2$, $I_3$, and $I$, and arrows indicate their directions. There is no risk here, for if you choose the wrong direction, the current will be of the correct magnitude but negative.
2. When applying Kirchhoff’s second rule, the loop rule, you must identify a closed loop and decide in which direction to go around it, clockwise or counterclockwise. For example, in [link] the loop was traversed in the same direction as the current (clockwise). Again, there is no risk; going around the circuit in the opposite direction reverses the sign of every term in the equation, which is like multiplying both sides of the equation by $\mathrm{–1.}$

[link] and the following points will help you get the plus or minus signs right when applying the loop rule. Note that the resistors and emfs are traversed by going from a to b. In many circuits, it will be necessary to construct more than one loop. In traversing each loop, one needs to be consistent for the sign of the change in potential. (See [link].)

• When a resistor is traversed in the same direction as the current, the change in potential is $-\text{IR}$. (See [link].)
• When a resistor is traversed in the direction opposite to the current, the change in potential is $+\text{IR}$. (See [link].)
• When an emf is traversed from $–$ to + (the same direction it moves positive charge), the change in potential is +emf. (See [link].)
• When an emf is traversed from + to $–$ (opposite to the direction it moves positive charge), the change in potential is $-$emf. (See [link].)

Calculating Current: Using Kirchhoff’s Rules

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

$I_1=I_2+I_3,$

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,

$-I_2{R}_2+{\text{emf}}_1-I_2{r}_1-I_1{R}_1=-I_2(R_2+r_1)+{\text{emf}}_1-I_1{R}_1=0.$

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

$-{3I}_2+\text{18}-{6I}_1=0.$

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

$+I_1{R}_1+I_3{R}_3+I_3{r}_2-{\text{emf}}_2\text{= +}{I}_1{R}_1+I_3\left(R_3+r_2\right)-{\text{emf}}_2=0.$

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

$+{6I}_1+{2I}_3-\text{45}=0.$

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

$I_2=6-{2I}_1.$

Now solve the third equation for $I_3$:

$I_3=\text{22}\text{.}5-{3I}_1.$

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

$I_1=I_2+I_3=(6-{2I}_1)+(\text{22}\text{.}5-{3I}_1)=\text{28}\text{.}5-{5I}_1.$

Combining terms gives

${6I}_1=\text{28}\text{.}\mathrm{5,\; and}$

$I_1=4\text{.}\text{75 A}.$

Substituting this value for $I_1$ back into the fourth equation gives

$I_2=6-{2I}_1=6-9.50$

$I_2=-3\text{.}\text{50 A}.$

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

$I_3=\text{22.5}-{3I}_1=\text{22.5}-\text{14}\text{.}\text{25}$

$I_3=8\text{.}\text{25 A}.$

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.

The material in this section is correct in theory. We should be able to verify it by making measurements of current and voltage. In fact, some of the devices used to make such measurements are straightforward applications of the principles covered so far and are explored in the next modules. As we shall see, a very basic, even profound, fact results—making a measurement alters the quantity being measured.