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 Figure 4.24. 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 $I_1=I_2+{}_{}$ (see figure). Equations like this can and will be used to analyze circuits and to solve circuit problems.

Figure 4.24 The junction rule. The diagram shows an example of Kirchhoff’s first rule where the sum of the currents into a junction equals the sum of the currents out of a junction. In this case, the current going into the junction splits and comes out as two currents so that $I_1=I_2+I_3\text{.}$ Here, $I_1$ must be 11 A since $I_2$ is 7 A and $I_3$ is 4 A.

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 ${\text{PE}}_{\text{elec}}=\text{qV}\text{.}$ 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. Figure 4.25 illustrates the changes in potential in a simple series circuit loop.

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

Figure 4.25 The loop rule. An example of Kirchhoff’s second rule where the sum of the changes in potential around a closed loop must be zero. (a) In this standard schematic of a simple series circuit, the emf supplies 18 V, which is reduced to zero by the resistances, with 1 V across the internal resistance and 12 V and 5 V across the two load resistances for a total of 18 V. (b) This perspective view represents the potential as something like a roller coaster, where charge is raised in potential by the emf and lowered by the resistances. (Note that the script E stands for emf.)

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 Figure 4.23, Figure 4.24, and Figure 4.25, currents are labeled $I_1\text{,}$ $I_2\text{,}$ $I_3\text{,}$ 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 Figure 4.25, 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 $\mathrm{–1.}$

Figure 4.26 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 Example 4.5).

Figure 4.26 Each of these resistors and voltage sources is traversed from a to b. The potential changes are shown beneath each element and are explained in the text. (Note that the script E stands for emf.)

• When a resistor is traversed in the same direction as the current, the change in potential is $-\text{IR}$ (see Figure 4.26).
• When a resistor is traversed in the direction opposite to the current, the change in potential is $+\text{IR}$ (see Figure 4.26).
• When an emf is traversed from $–$ to + (the same direction it moves positive charge), the change in potential is +emf (see Figure 4.26).
• When an emf is traversed from + to $–$ (opposite to the direction it moves positive charge), the change in potential is $-$emf (see Figure 4.26).

Example 4.5 Calculating Current: Using Kirchhoff’s Rules

Find the currents flowing in the circuit in Figure 4.27.

Figure 4.27 This circuit is similar to that in Figure 4.23, but the resistances and emfs are specified. (Each emf is denoted by script E.) The currents in each branch are labeled and assumed to move in the directions shown. This example uses Kirchhoff’s rules to find the currents.

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 $I_1\text{,}$ $I_2\text{,}$ 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

4.54 $$I_1=I_2+I_3$$

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, 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 $I_2\text{,}$ so the change in potential is $-I_2{R}_2\text{.}$ Then going from b to c, we go from $–$ to +, so the change in potential is $+{\text{emf}}_1\text{.}$ Traversing the internal resistance $r_1$ from c to d gives $-I_2{r}_1\text{.}$ 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 $-I_1{R}_1\text{.}$

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

4.55 $$-I_2{R}_2+{\text{emf}}_1-I_2{r}_1-I_1{R}_1=-I_2(R_2+r_1)+{\text{emf}}_1-I_1{R}_1=0.$$

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

4.56 $$-{3I}_2+\text{18}-{6I}_1=0.$$

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

4.57 $$+I_1{R}_1+I_3{R}_3+I_3{r}_2-{\text{emf}}_2\text{= +}{I}_1{R}_1+I_3\left(R_3+r_2\right)-{\text{emf}}_2=0.$$

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

4.58 $$+{6I}_1+{2I}_3-\text{45}=0.$$

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

4.59 $$I_2=6-{2I}_1$$

Now solve the third equation for $I_3\text{.}$

4.60 $$I_3=\text{22}\text{.}5-{3I}_1$$

Substituting these two new equations into the first one allows us to find a value for $I_1\text{.}$

4.61 $$I_1=I_2+I_3=(6-{2I}_1)+(\text{22}\text{.}5-{3I}_1)=\text{28}\text{.}5-{5I}_1$$

Combining terms gives

4.62 $${6I}_1=\text{28}\text{.}\mathrm{5\; and}$$

4.63 $$I_1=4\text{.}\text{75 A}.$$

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

4.64 $$I_2=6-{2I}_1=6-9.50$$

4.65 $$I_2=-3\text{.}\text{50 A}.$$

The minus sign means that $I_2$ flows in the direction opposite to that assumed in Figure 4.27.

Finally, substituting the value for $I_1$ into the fifth equation gives

4.66 $$I_3=\text{22.5}-{3I}_1=\text{22.5}-\text{14}\text{.}\text{25}$$

4.67 $$I_3=8\text{.}\text{25 A}.$$

Discussion

Just as a check, we note that indeed $I_1=I_2+I_3\text{.}$ 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.