Applying Kirchhoff’s Voltage Law

Applying Kirchhoff’s Voltage Law

Kirchhoff’s first law is also referred to as his “voltage law.” The voltage law describes the relationship between the voltage sources and the “voltage drops” surrounding any closed loop in a circuit. The sum of these two numbers is always the same.

In equation form:

The symbol (the Greek letter sigma) stands for “total.”

Only closed loops can be used using Kirchhoff’s voltage law (Figure 32). A closed loop must satisfy two requirements:

  1. One or more voltage sources are required.
    It must provide a complete path for current flow from any location to the loop’s centre and back.

You’ll recall that the sum of the voltage drops around the circuit equals the applied voltage in a basic series circuit. Kirchhoff’s voltage law is actually applied to the simplest case, when there is only one loop and one voltage source.

Applying Kirchhoff’s Voltage Law

Kirchhoff’s voltage law conforms to Ohm’s Law in a simple series circuit. Use the following equation to find the current in a circuit (Figure 33) using Kirchhoff’s voltage law.

ΣEsource = Σ I R

Figure 33 Using Kirchhoff’s Voltage Law to find Current with one Source

80 = 20 (I) + 10 (I)

80 = 30 (I)

I = 80/30 = 2.66 amps

Before solving the problem in the previous example, the direction of the current flow was known. The direction of current flow may or may not be known when there are several voltage sources. In this scenario, a current flow direction must be assumed at the outset of the task.

All sources that would help the current flow in the assumed direction are positive, whereas all sources that would resist current flow are negative. The result will be affirmative if the expected direction is right. If the expected direction was incorrect, the response would be negative. The correct magnitude will be achieved in any situation.

What is the current flow in Figure 34, for example? Assume the current flows in the direction depicted.


Figure 34 Using Kirchhoff’s Voltage Law to find Current with Multiple Battery Sources

Using Kirchhoff’s Voltage Law:

ΣEsource = Σ I R

50 – 70 = 30I + 10I

-20 = 40I

I = -20/40 = -0.5

The end result is a failure. In reality, the current is 0.5 ampere in the opposite direction of the supposed direction.

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