Kirchoff's Loop Law

Two different examples are given below. The first is explained in detail. The second is more concise. Both examples involve the circuit shown in the diagram at right. Voltage values are given for batteries A and B as well as capacitor C. We will also assume that the top plate of capacitor C is positively charged since it is connected directly to the positive terminal of battery B. The goal of the two examples is to determine the voltages across capacitors D and E.

Detailed Example

Kirchoff's Loop Law states that the voltage rises and drops around any closed loop must sum to zero. A closed loop is a path around the circuit that starts and stops at the same location. The circuit diagram that is shown at right is identical to the one above except that a particular closed loop has been highlighted. The loop is colored green except for capacitor E, which is colored red. Capacitor E is colored differently to remind you that you are trying to determine the voltage across it.

To apply Kirchoff’s Loop Law, imagine ‘walking’ around the closed loop in one direction or the other. It doesn’t matter whether you walk clockwise or counterclockwise around the loop. It also doesn’t matter where in the loop you start because you will stop where you started. For the sake of this example, imagine that you start next to the negative terminal of battery B and walk clockwise around the highlighted loop. Before returning to your starting point you will walk past battery B, then capacitor C, then capacitor E. Hence, according to Kirchoff’s Loop Law:

VB + VC + VE = 0

where VB, VC and VE represent the voltages across B, C and E respectively. For this equation to be true, some of these voltages must be rises (i.e., positive) and some must be drops (i.e., negative).

Rules for determining whether a voltage is a rise or a drop:

Battery Rule
Capacitor Rule

Recall that Kirchoff’s Loop Law applied to the closed loop in the circuit above resulted in the equation:

VB + VC + VE = 0

In addition, we determined that:

VB = +12V and VC = -14V.

Hence:

+12V + -14V + VE = 0.

Solving for VE gives VE = +2V.

Notice that the value of VE is positive. This implies that it is the left plate of capacitor E that is positively charged.

Concise Example

The circuit at right is identical to the circuit in the previous example, but this time a different closed loop is highlighted. As before, most of the loop is colored green except for one red capacitor. The red capacitor is colored differently to remind you that you are trying to determine the voltage across it.

VA + VB + VC + VD = 0.