How to calculate voltage drop

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Electrical devices have revolutionized the way humans live and interact with each other. This revolution was possible thanks to the discovery of electricity. A key factor when describing electric circuits is the electric potential difference, or voltage, which refers to the physical work that an electric field can eventually perform on charged particles, for example, those flowing through the internal circuits of the device you are using to read this. 

In order to understand how the energy available to move charges is used in a circuit, we need to understand the concept of voltage drop. This basic concept refers to the electric energy that is used by electrical elements, such as resistors, capacitors, or inductors. 

Although voltage drop is an essential part of electric circuits, it can also affect a device’s efficiency when energy is dissipated by undesired effects, such as voltage being consumed just to transport charges along a very long wire. 

Let’s take a trip together through the world of basic electrostatics to understand in detail where voltage drops come from and how to measure it. 

How to calculate voltage drop

To calculate the voltage drop (V) across the most common electrical elements, use the following expressions: 

Electrical elementSymbolVoltage drop

Where I is the current flowing through the element, R the resistance, C the capacitance, and L the inductance. 

What is a voltage drop?

To understand what voltage drop is, we need to understand first the concept of potential energy. You have probably heard about it when dealing with a very specific kind: gravitational potential energy. 

When you hold a ball at some distance above the ground, its velocity with respect to you is zero. It does not move from your hand as long as you hold it tight. Once you release the ball, it will fall to the ground due the force exerted on it by the Earth’s gravity.

When the ball finally hits the ground, it will have a certain speed, which means its kinetic energy would have gone from zero to a certain positive value. Now, since energy cannot be created or destroyed, the resulting kinetic energy in the ball must have come from somewhere. 

Since gravity is a type of force that does not require contact to act on a body, it is usually described as a field that surrounds bodies near to its source. So, all bodies on our planet are submerged in its gravitational field and will be subjected to a certain gravitational force depending on their location with respect to Earth’s center of gravity. 

Let’s go back to our previous example. When you hold the ball above the ground, the Earth’s gravitational field pulls it down, so you must exert an equal force in the opposite direction, meaning upwards, to hold it in place. In this situation, we can think of the ball as having the potential to gain speed by falling to the ground. 

This potential is represented by the gravitational potential energy, which is a function of a body’s vertical position with respect to a reference level. In general, the existence of potential energy implies that a physical system —such as a ball falling to the ground— has the possibility to release other types of energies.

Potential energy is a concept that is most usually related to its gravitational counterpart. Nevertheless, it also exists in other forms. For example, when you compress a spring, it will exert a force in the opposite direction until you release it. So, in a way, the spring stores energy until it is released, and its initial shape is recovered. This is called elastic potential energy

The key type of potential energy that will allow us to understand voltage drop is called the electric potential energy. To explain it, let’s imagine an isolated positive point charge, which we will call A. Since our imaginary particle possesses charge, it will create an electric field around it pointing outwards according to physicists’ conventions. 

Now, imagine a second positive point charge, B, fixed within a few centimeters of A. According to basic electrostatics, since B is a charge submerged in the electric field produced by A, it will interact with it. Since the field points out of A and given that a positive charge will experience a force in the same direction of the external electric field, the result is that B will be pulled away from A. Conversely, since B creates an electric field on its own, A will interact with it and will be pulled away from it as well. In short, both charges will tend to separate, as the following image shows: 

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Now imagine you fix A at a certain position and want to move B closer to it. To achieve this, you will need to apply a force on B that counteracts the electrostatic force exerted on it by A’s electric field (FA-B). Since the force you apply generates a displacement of B, you will be performing work on it. This can be expressed as: 


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Where W is the total work performed by the force F (applied by you) to cause a displacement d of the particle B

Now B is located closer to A. This means, the electrostatic force experienced by B due to A’s electric field is bigger than at the beginning. If you are not sure about why this is, go ahead and read about Coulomb’s law. Since the force is greater, so is B’s potential to accelerate away from A once you release it. 

The opposite situation is basically the same: if you hold B close to A and release it, the electric field will exert a force on it, which will accelerate it away from A. This time, the electric field performs work on B. If you stop B at the same spot where it initially was when you started moving it towards A, the work performed by the electric field would equal the one performed by you in the opposite direction. This is because both forces (yours and the electric field’s) are equal —but opposite in direction— and generate the same displacement. 

The following image shows this scenario: B starts located close to A, where a certain electrostatic force is generated on it. When it is released, it displaces away from A and the electrostatic force decreases. 


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In a way, this is equivalent to our previous example of dropping a ball: here, the electric field in which B is submerged provides it with the potential to increase its velocity in a certain direction. This is called the electric potential energy because it arises in connection to an electric field, and it is measured in joules. Its exact mathematical expression will depend on the specific characteristics of the electric field.

Now, as you can see from this thought experiment, there is a relation between the potential energy gained by a point charge and the work either you or the electric field need to perform on it to locate it at the spot where you want to determine it. 

In summary, the electric potential energy varies depending on the position of an electric charge within an electric field and on the field itself. At any given spot a charged particle will have a certain electric potential energy. This quantity can increase or decrease as the particle is displaced inside the field. Since work is required to generate said change, and since energy cannot be created or destroyed, both quantities must be equal, meaning: 


Where U represents the electric potential energy at a given spot. Here, the work needed to move a charged particle from point 1 to point 2 equals the electric potential energy change of the particle. Keep in mind that, for this example, we have assumed the electric field within which charges move is static, meaning, it does not change in time. 

So far, we have associated the work performed by an electric field on a charged particle to the change in its electric potential energy. The actual value of both variables depends on the electric field, the location of the particle within it, and on its charge. The last one is important because the force experienced by a charged particle in an electric field, which can eventually perform work on it, is directly proportional to its charge. 

A way to generalize this scenario is to express the electric potential energy by unit charge at a certain location. This way we would be able to obtain the electric potential energy for any charged particle located at that spot. This is called the electric potential and is defined as: 


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Where V is the electric potential, U the electric potential energy at the desired location within the electric field and q the electric charge. Replacing equation 3 in equation 2, we get: 


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This equation can be expressed like this: the work per unit charge performed on a particle to displace it from one point to another equals the electric potential difference between both locations. Notice that we require two points in space to make sense of this definition. This means that the electric potential will always require a reference. 

Now, voltage is defined as the electric potential difference between two points in space, as shown on the right-hand side of equation 4. Since the units for the electric potential energy are joules and the units for the electric charge are coulombs, the units for voltage are joule/coulomb. 1 volt equals 1 joule of work performed per each coulomb of charge

Whenever we see a voltage, for example, on electric batteries, it refers to the amount of work it can perform per unit charge. So, when you introduce it into an electrical device, the battery will be able to push charges through its circuits by performing a certain amount of work on them. 

Finally, the energy that a certain voltage can release as movement of charges within an electric field can be altered by a lot of factors. For example, if charges flowing through an electric circuit have to travel through a spot where it is hard for them to move, more voltage will be needed per charge to reach the desired final displacement. This is precisely what happens in resistors. 

Since a portion of the available voltage is used when reaching a resistor to force charges through, we say there is a voltage drop at the resistor. In general, a voltage drop is a decrease in the electric potential along the path charges follow inside an electric circuit. 

Types of voltage drops and how to measure them

In electric circuits voltage is usually used to “push” charges across electric components such as resistors, capacitors, and inductors. Since the nature of each type of component is different, the amount of voltage that drops across them also differs. 

Since the movement of charges is expressed as a current, voltage drop calculations for each of the electric components mentioned above usually involve it. The table shown at the beginning of this article is a summary of the most common mathematical expressions to calculate the voltage drop across resistors, capacitors or inductors. 

If you want to measure the voltage drop using a multimeter, follow these steps: 

  1. Set the measuring device to the appropriate measurement type and range. Usually, multimeters allow to set up the measurement for very small voltages, in the range of microvolts, or for big potential differences in the range of tens of volts. 
  2. Place one of the tips of the multimeter at the entrance of the electric element you want to determine the voltage drop for, for example, a resistor. 
  3. Place the other tip at the exit of the electric element under measurement. This will create a circuit parallel to the one under study, passing through the multimeter. 
  4. Read the potential difference value on the multimeter’s display, which equals the voltage drop at the selected element. 

Since voltage in parallel circuits is equal on both branches, the multimeter will be able to measure the potential difference between the entrance and exit points, and display the resulting voltage drop. 

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Wires used in electric circuits are conductors. This means they allow charges to move readily across them. Nevertheless, they are not perfect conductors. This means, some voltage can drop in a wire, especially when it is very long. This is critical for many large-scale applications since this phenomenon can limit their efficiency. 

Use the following voltage drop calculator by to find the total voltage drop for a wire of specific material, gauge, length, and operation voltage and current. 

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