Chapter 23
Kirchhoff’s Rules
Examine the circuit shown in Fig. 23.1. You’ll note that the techniques we used in the previous chapter are
not suited to analyze this circuit, due to the presence of the 3V battery in the middle of the circuit.
Instead, we must may use of another technique calledKirchhoff ’s rules. There are two of these rules:
- Kirchhoff ’s voltage rulestates that the sum of the voltage rises and drops around any complete loop in
the circuit equals zero. - Kirchhoff ’s current rulestates that at each junction in the circuit, the sum of the currents entering the
junction equals the sum of the currents leaving the junction.
23.1 Example Circuit
We’ll use the circuit shown in Fig. 23.1 as an example to illustrate how to apply Kirchhoff’s rules.
- Begin by identifying loops in the circuit. The circuit in Fig. 23.1 consists of three loops: the upper
loop, the lower loop, and the outer loop. We’ll need to choose any two of these three loops to work
with—let’s choose the upper and lower loops. - Next, we choose a direction in which to “evaluate” each loop. This can be either clockwise or coun-
terclockwise; the choice is completely arbitrary, and will not affect the final results. Let’s choose to
evaluate both the upper and lower loop in the clockwise direction, as indicated by the arrows in the
center of each loop. - Now identify the currents in the circuit. By inspection of the circuit, we can see that there are three
distinct currents: one in the upper branch, one in the middle branch, and one in the lower branch.
We’ll label these three currentsI 1 ,I 2 , andI 3 (respectively), and choose a direction for each current, as
shown in Fig. 23.1. It doesn’t matter whether we choose the directions for the currents correctly—we
just guess at each direction. If we guess the wrong direction for a current, then that current will come
out negative when we finish the analysis, so the real current flows opposite the direction we guessed. - The next step is to apply Kirchhoff’s voltage rule to the upper and lower loops of the circuit. Beginning
at any point in the loop, we move in the direction chosen in step 2, and write down the terms shown in
Fig. 23.2; the sum of these terms is then set to zero.
For the upper loop, beginning in the upper-left corner, we find:
I 1 R 1 I 1 R 2 C 3 VCI 2 R 3 C 6 VD 0 (23.1)