Kirchhoff’s Voltage Law

The Voltage Law of Gustav Kirchhoff is the second of his essential laws that we can apply for circuit analysis. His voltage law asserts that the algebraic sum of all the voltages around any closed loop in a circuit is zero for a closed-loop series path. This is because a circuit loop has a closed conducting line, which means that no energy is lost.

To put it another way, the algebraic sum of ALL potential differences around the loop must equal zero, as follows: V = 0. It’s worth noting that the phrase “algebraic sum” refers to accounting for the polarities and signs of the sources as well as voltage dips along the loop.

The Conservation of Energy is a notion proposed by Kirchhoff that states that traveling through a closed-loop, or circuit, will return you to where you started in the circuit and hence to the same beginning potential with no voltage loss. As a result, any voltage drops encountered along the way must be equal to any voltage sources encountered.

When applying Kirchhoff’s voltage law to a given circuit element, it’s critical to pay close attention to the algebraic signs, (+ and -), of voltage drops across elements and source EMFs, or our calculations will be incorrect.

But, before diving into Kirchhoff’s voltage law (KVL), it’s important to understand the voltage drop across a single element, such as a resistor.

Exceptions to Kirchhoff’s Voltage Law

Kirchhoff’s Voltage Law is fascinating because we’ve just argued so forcefully for why it must be “clearly” true…

However, you might be surprised to hear that the KVL is really untrue in the underlying physics of Maxwell’s equations! Faraday’s induction law is as follows:

This states that the rate of change of magnetic flux across the loop’s enclosing surface is equal to the voltage induced in the loop. So, if there is no time-varying magnetic flux through a loop, the voltage around it is 0.

In our discussion of Electrons at Rest, we mentioned this issue. To recapitulate, we must assume that Kirchhoff’s Voltage Law holds true in our Lumped Element Model, but we make some tweaks here and there.

For example, Every inductor and transformer have time-varying magnetic flux, but we only incorporate this in the circuit element’s model. An inductor’s voltage is the same as the right-hand part in Faraday’s law, but instead of treating it as a correction to the KVL, we treat it as a voltage source in and of itself.

However, if there are external time-varying magnetic fields, we may have to be concerned. In electronics, this can be a cause of interference. This is one of the reasons why huge electronic systems with loops within can be a problem, and one of the reasons why ground loops can create spurious voltages in our system: they form a large surface for time-varying magnetic flux to cause spurious voltages. However, if we want to, we can generally simulate this effect as a separate voltage source.

For the time being, you should assume Kirchhoff’s Voltage Law is correct in your electronics studies. Simply remember this information in case you need to work with time-varying magnetic fields in the future!

A Single Circuit Element

Just for simplicity, we’ll assume that the current, I, flows in the same direction as the positive charge flow, which is known as conventional current flow.

The current flow through the resistor here is from point A to point B, or from a positive to a negative terminal. Because we are moving in the same direction as current, there will be a drop in potential across the resistive element, resulting in a -IR voltage drop.

If the current flowed in the opposite direction from point B to point A, the potential across the resistive element would grow as we moved from a – to a + potential, resulting in a +I*R voltage drop.

As a result, in order to accurately apply Kirchhoff’s voltage law to a circuit, we must first understand the polarity, and as we can see, the sign of the voltage drop across the resistive element is dependent on the direction of the current flowing through it. As a general rule, moving in the same direction as current across an element will lose potential, whereas moving in the direction of an emf source will gain potential.

Current flow around a closed circuit can be considered to be clockwise or anticlockwise, and either direction can be chosen. The conclusion will still be correct and valid if the chosen direction differs from the real direction of current flow, but the algebraic answer will have a minus sign.

To better grasp this concept, consider a single circuit loop to check if Kirchhoff’s Voltage Law holds true.

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.
2. It must provide a complete path for current flow from any location to the loop’s center 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.

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.

ΣE_source = Σ I R

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.

Using Kirchhoff’s Voltage Law:

ΣE_source = Σ I R

50 – 70 = 30I + 10I

-20 = 40I

I = -20/40 = -0.5 amps

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

See Also: 4-20 mA Transmitter Wiring Types: 2-Wire, 3-Wire, 4-Wire