Voltage Vs Current: What Are The Similarities & Differences?

If you're new to the physics of electricity, terms like ​voltage​ and ​amps​ might almost seem interchangeable based on the way they're used. But in reality, they are very different quantities, although they are closely linked by how they work together in an electrical circuit, as described by Ohm's law.

Really, "amps" are a measure of electrical current (which is measured in ​amperes​), and voltage is a term that means electric potential (measured in ​volts​), but unless you've learned the details, it's understandable that you could get the two confused with each other.

To understand the difference – and never get them mixed up again – you just need a basic primer on what they mean and how they relate to an electric circuit.

Voltage is another term for the electric potential difference between two points, and it can be simply defined as the electric potential energy per unit charge.

Just as gravitational potential is the potential energy an object has by virtue of its position within a gravitational field, electric potential is the potential energy a charged object has by virtue of its position in an electric field. Voltage specifically describes this per unit of electrical charge, and so it can be written:

\(V=\frac{E_{el}}{q}\)

Where ​V​ is the voltage, ​E​_el is the electrical potential energy and ​q​ is the electric charge. Since the unit for electrical potential energy is the joule (J) and the unit for electrical charge is the coulomb (C), the unit of voltage is the volt (V), where 1 V = 1 J/C, or in words, one volt is equal to one joule per coulomb.

This tells you that if you allow a charge of 1 coulomb to pass through a potential difference (i.e., a voltage) of 1 V, it will gain 1 J of energy, or conversely, it will take one joule of energy to move a coulomb of charge through a potential difference of 1 V. Voltage is also sometimes referred to as ​electromotive force​ (EMF).

The voltage difference (or potential difference) between two points, such as at either side of an element in an electric circuit, can be measured by connecting a voltmeter in parallel with the element you're interested in. As the name suggests, a voltmeter measures the voltage between two points on the circuit, but when you're using one, it must be connected ​in parallel​ to avoid interference with the voltage reading or damage to the device.

Electric current, which is sometimes referred to as the amperage (since it has the unit of the ampere), is the rate of flow of electric charge past a point in a circuit. The electric charge is carried by electrons, the negatively charged particles that surround the nucleus of an atom, so the amount of current really tells you the rate of flow of electrons. A simple mathematical definition of electrical current is:

\(I=\frac{q}{t}\)

Where ​I​ is the current (in amperes), ​q​ is the electric charge (in coulombs) and ​t​ is the time elapsed (in seconds). As this equation shows, the definition of an ampere (A) is 1 A = 1 C/s, or a flow of an electric charge of 1 coulomb per second. In terms of electrons, this is about 6.2 × 10¹⁸ electrons (about six billion billion) flowing past the reference point per second for a current flow of just 1 A.

Current can be measured in an electrical circuit by connecting an ammeter in series – meaning in the path of the main current – with the section of the circuit you want to measure the amount of current through.

If you're still struggling to understand the roles the voltage difference and electric current play within an electrical circuit, a widely used analogy between electricity and water should help to clarify things. Two different scenarios can be used to represent the voltage in an electric circuit: either a water pipe running down a hill, or a water tank filled up with an output spout at the bottom.

For the water pipe with one end at the top of a hill and the other end at the bottom, your intuition should tell you that water would flow through it faster if the hill was higher and slower if it the hill was lower. For the water tank example, if there were two water tanks filled to different levels, you would expect the more filled tank to release water from the outlet at a faster rate than the one filled to a lower level.

Whether it's the potential from the height of the hill (due to gravitational potential) or the potential created by the water pressure in the tank, both of these examples convey a key fact about voltage differences. The greater the potential, the faster the water (i.e., the current) will flow.

The flow of water itself is analogous to electric current. If you measured the water flowing past a single point on the pipe per second, this is like the current flow in a circuit, except with water in place of electrical charge in the form of electrons. So if all else is equal, a high voltage leads to a high current, and vice versa. The final part of the picture is resistance, which is analogous to the friction between the walls of the pipe and the water, or a physical obstruction placed in the pipe partially blocking the water flow.

\(\def\arraystretch{1.5}
\begin{array}{c:c}
\text{Similarities} & \text{Differences} \ \hline\hline
\text{Both pertain to electric circuits} & \text{Different units, voltage is measured in volts, where 1 V = 1 J/C} \
& \text{while current is measured in amperes, where 1 A = 1 C/s} \
\hline
\text{Both affect how much power is dissipated across a circuit element} & \text{Current is equally distributed in all components when in series}\
& \text{while voltage drop across components can differ}\
\hline
\text{Can both be in alternating polarity (e.g. alternating} & \text{Voltage drop is equal across all } \
\text{current or alternating voltage) or direct polarity } & \text{components connected in parallel, while current differs} \
\hline
\text{They are directly proportional to each other in line with Ohm's law} & \text{Voltage produces an electric field while current produces a magnetic field} \
\hline
& \text{Voltage causes current, while current is the effect of voltage} \
\hline
& \text{Current only flows when the circuit is complete, but voltage differences remain}
\end{array}\)

As the table shows, electric current and voltage have more differences than they do similarities, but there are some similarities too. The biggest difference between the two is the fact that they describe different quantities entirely, so once you understand the basics of what each one is, you're unlikely to get them confused with one another.

Voltage difference and electrical current are directly proportional to each other in line with Ohm's law, one of the most important equations in the physics of electric circuits. The equation relates the voltage (i.e., the potential difference created by the battery or other power source) to the current in the circuit and the resistance to the flow of current created by the components of the circuit.

Ohm's law states:

\(V = IR\)

Where ​V​ is the voltage, ​I​ is the electric current and ​R​ is the resistance (measured in ohms, Ω). For this reason, Ohm's law is sometimes referred to as the voltage, current and resistance equation. If you know any two quantities in this equation, you can re-arrange the equation to find the other quantity, which makes it useful in solving most electronics problems you'll encounter in physics class.

It's worth noting that Ohm's law isn't ​always​ valid, and as such it isn't a "true" law of physics, but a useful approximation for what are called ​ohmic​ materials. The linear relationship it implies between current and voltage doesn't hold for things such as a filament bulb, where the increase in temperature causes an increase in resistance and thus impacts the linear relationship. However, in most cases (and certainly most physics problems you'll be asked involving voltage and electric current) it can be used without issues.

Ohm's law is primarily used to relate voltage to current and resistance; however, there is an extension of the law that allows you to use the same quantities to calculate the electrical power dissipated in a circuit, where power ​P​ is the rate of energy transfer in watts (where 1 W = 1 J/s). The simplest form of this equation is:

\(P=IV\)

So in words, power equals current multiplied by voltage. Hence, this is a key area in which voltage difference and electric current are similar: They both share a directly proportional relationship with the power dissipated in a circuit. If you don't know the current, you can use a re-arrangement of Ohm's law (I = V / R) to express power as:

\(\begin{aligned}
P&=\frac{V}{R}× V \
&= \frac{V^2}{R}
\end{aligned}\)

Or using the standard form of Ohm's law, you can replace voltage and write:

\(P=I^2R\)

By re-arranging these equations, you can also express voltage, resistance or current in terms of power and another quantity.

Kirchhoff's laws are two of the other most important laws for electrical circuits, and they are particularly useful when you're analyzing a circuit with multiple components.

Kirchhoff's first law is sometimes called the current law, because it states that the total current flowing into a junction is equal to the current flowing out of it – essentially that charge is conserved.

Kirchhoff's second law is called the voltage law, and it states that for any closed loop in a circuit, the sum of all the voltages must equal zero. For the voltage law, you treat the battery as a positive voltage and treat the voltage drops across any component as a negative voltage.

In combination with Ohm's law, these two laws can be used to solve essentially any problem you're likely to encounter involving electric circuits.

Imagine you have a circuit involving a 12-V battery and two resistors, connected in series, with resistances of 30 Ω and 15 Ω. The total resistance for the circuit is given by the sum of these two resistances, so 30 Ω + 15 Ω = 45 Ω. Note that when resistors are arranged in parallel, the relationship involves reciprocals, but this isn't important for understanding the relationship between voltage difference and current, so this simple example will suffice for present purposes.

What is the electrical current flowing through the circuit? Try to apply Ohm's law yourself before reading on.

The following form of Ohm's law:

\(I=\frac{V}{R}\)

Allows you to calculate:

\(\begin{aligned}
I&=\frac{12 \text{ V}}{45 \text{ Ω}} \
&=0.27 \text{ A}
\end{aligned}\)

Now, knowing the current through the circuit, what is the voltage drop across the 15-Ω resistor? Ohm's law in the standard form can be used to address this question. Inserting the values of ​I​ = 0.27 A and ​R​ = 15 Ω gives:

\(\begin{aligned}
V &= IR \
&= 0.27 \text{ A} × 15 \text{ Ω} \
&= 4.05 \text{ V}
\end{aligned}\)

For the purposes of using Kirchhoff's laws, this will be a negative voltage (i.e., a voltage drop). As a final exercise, can you show that the total voltage around the closed loop will be equal to zero? Remember that the battery has a positive voltage, and all voltage drops are negative.