# Kirchoff’s Voltage Law

### What does Kirchoff’s Voltage Law state?

KVL states that the voltage rises around a loop is equal to the voltage drops.

$\sum_{}^{} V_{rises} = \sum_{}^{} V_{drops}$

Consider Figure 3.  The voltage source supplies 12 V to the two resistors.  This is a voltage rise.  The resistors, therefore, must drop the voltage 12 V to match the 12 V produced by the supply.  When determining KVL loop equations,

1)     Determine a starting point (a voltage source is typically a good place).

2)    Choose a direction (clockwise or counterclockwise).

3)    Label each component with a plus on one side and a minus on the other.

Obviously, below the voltage source, the voltage is 0 V and above it the voltage is 12 V.  The 12 V can be seen as a voltage rise or drop depending on which way you move around the loop.  The source is interpreted as a voltage rise when moving clockwise (minus to plus) and a voltage drop when moving counterclockwise (plus to minus).

### A few Kirchoff’s Voltage Law examples

Example: Find the voltage across resistor R1.

Solution: To find this voltage, we will start at the voltage source and move clockwise.  When moving clockwise, the voltage source is the only voltage rise (minus to plus) and the two resistors are both voltage drops (plus to minus).  Therefore the equation can be written as

$\sum_{}^{} V_{rises} = \sum_{}^{} V_{drops}$

$12 = V_{R1} + 3$

$V_{R1} = 9V$

It does not matter whether you go clockwise or counterclockwise.  If we had gone counterclockwise, the voltage source would have become the voltage drop and the voltage across the two resistors the voltages rises.  Be careful of the orientation of the voltage across components.  If VR1 was labeled with the plus on the right and the minus on the left, the voltage calculated would be -9 V.

Example: Find VR1

Solution: In nearly all the problems you’ll see, the orientation of the voltages is already labeled for you (orientation, as in which side has the plus and which has the minus).  You only need to worry about using the given orientation properly within your KVL equation.

In this case we will start from V1 and move clockwise.  V1 and V3 are the only voltage rises.  V2 is opposing the other two sources so it is a voltage drop, as are VR1 and VR2.  Our equation becomes:

$\sum_{}^{} V_{rises} = \sum_{}^{} V_{drops}$ $V_{1} + V_{3} = V_{R1} + V_{2} + V_{R2}$ $20 + 20 = V_{R1} + 30 + 3$

$V_{R1} = 7 V$ (Always label your values!)

So far, we have only worked with single loops, but you should recognize that there can be multiple loops in one circuit.  There can even be loops within loops.  For every loop, Kirchoff’s voltage law still applies.  Three loops are shown in Figure 7, but there are others that aren’t marked.  As you work with more complicated circuits, you should learn to identify any loops in a circuit and be able to apply KVL to each loop.  Written by Ryan Eatinger (reatinge@ksu.edu).  Thank you!