Definition of a Simple Electrical Series Circuit

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Getting to grips with the basics of electronics means understanding circuits, how they work and how to calculate things like the total resistance around different types of circuits. Real-world circuits can get complicated, but you can understand them with the basic knowledge you pick up from simpler, idealized circuits.

The two main types of circuits are series and parallel. In a series circuit, all of the components (such as resistors) are arranged in a line, with a single loop of wire making up the circuit. A parallel circuit splits off into multiple paths with one or more components on each. Calculating series circuits is easy, but it’s important to understand the differences and how to work with both types.

The Basics of Electrical Circuits

Electricity only flows in circuits. In other words, it needs a complete loop in order for something to work. If you break that loop with a switch, the power stops flowing, and your light (for example) will turn off. A simple circuit definition is a closed loop of a conductor that electrons can travel around, usually consisting of a power source (a battery, for example) and an electrical component or device (like a resistor or a light bulb) and conducting wire.

You’ll need to get to grips with some basic terminology to understand how circuits work, but you’ll be familiar with most of the terms from day-to-day life.

A “voltage difference” is a term for the difference in electrical potential energy between two places, per unit charge. Batteries work by creating a difference in potential between their two terminals, which allows a current to flow from one to the other when they’re connected in a circuit. The potential at one point is technically the voltage, but differences in voltage are the important thing in practice. A 5-volt battery has a potential difference of 5 volts between the two terminals, and 1 volt = 1 joule per coulomb.

Connecting a conductor (such as a wire) to both terminals of a battery creates a circuit, with an electric current flowing around it. The current is measured in amps, which means coulombs (of charge) per second.

Any conductor will have electrical “resistance,” which means the material’s opposition to the flow of current. Resistance is measured in ohms (Ω), and a conductor with 1 ohm of resistance connected across a voltage of 1 volt would allow a current of 1 amp to flow.

The relationship between these is encapsulated by Ohm’s law:


In words, “voltage equals current multiplied by resistance.”

Series vs. Parallel Circuits

The two main types of circuits are distinguished by how components are arranged in them.

A simple series circuit definition is, “A circuit with the components arranged in a straight line, so all of the current flows through each component in turn.” If you made a basic loop circuit with a battery connected to two resistors, and then have a connection running back to the battery, the two resistors would be in series. So the current would go from the positive terminal of the battery (by convention you treat current as if it emerges from the positive end) to the first resistor, from that to the second resistor and then back to the battery.

A parallel circuit is different. A circuit with two resistors in parallel would split into two tracks, with a resistor on each. When the current reaches a junction, the same amount of current that enters the junction has to leave the junction as well. This is called the conservation of charge, or specifically for electronics, Kirchhoff’s current law. If the two paths have equal resistance, an equal current will flow down them, so if 6 amps of current reaches a junction with equal resistance on both paths, 3 amps will flow down each one. The paths then rejoin before reconnecting to the battery to complete the circuit.

Calculating Resistance for a Series Circuit

Calculating the total resistance from multiple resistors emphasizes the distinction between series vs. parallel circuits. For a series circuit, the total resistance (​Rtotal) is just the sum of the individual resistances, so:

R_{total}=R_1 + R_2 + R_3 + ...

The fact that it’s a series circuit means the total resistance on the path is just the sum of the individual resistances on it.

For a practice problem, imagine a series circuit with three resistances: ​R1 = 2 Ω, ​R2 = 4 Ω and ​R3 = 6 Ω. Calculate the total resistance in the circuit.

This is simply the sum of the individual resistances, so the solution is:

\begin{aligned} R_{total}&=R_1 + R_2 + R_3 \\ &=2 \; \Omega \; + 4 \; \Omega \; +6 \; \Omega \\ &=12 \; \Omega \end{aligned}

Calculating Resistance for a Parallel Circuit

For parallel circuits, the calculation of ​Rtotal is a bit more complicated. The formula is:

{1 \above{2pt}R_{total}} = {1 \above{2pt}R_1} + {1 \above{2pt}R_2} + {1 \above{2pt}R_3}

Remember that this formula gives you the reciprocal of the resistance (i.e., one divided by the resistance). So you need to divide one by the answer to get the total resistance.

Imagine those same three resistors from before were arranged in parallel instead. The total resistance would be given by:

\begin{aligned} {1 \above{2pt}R_{total}} &= {1 \above{2pt}R_1} + {1 \above{2pt}R_2} + {1 \above{2pt}R_3}\\ &= {1 \above{2pt}2 \; Ω} + {1 \above{2pt}4 \; Ω} + {1 \above{2pt}6\; Ω}\\ &= {6 \above{2pt}12 \; Ω} + {3 \above{2pt}12 \; Ω} + {2 \above{2pt}12 \; Ω}\\ &= {11 \above{2pt}12Ω}\\ &= 0.917 \; Ω^{-1} \end{aligned}

But this is 1 / ​Rtotal, so the answer is:

\begin{aligned} \ R_{total} &= {1 \above{2pt}0.917 \; Ω^{-1}}\\ &= 1.09 \; \Omega \end{aligned}

How to Solve a Series and Parallel Combination Circuit

You can break down all circuits into combinations of series and parallel circuits. A branch of a parallel circuit might have three components in series, and a circuit could be composed of a series of three parallel, branching sections in a row.

Solving problems like this just means breaking down the circuit into sections and working them out in turn. Consider a simple example, where there are three branches on a parallel circuit, but one of those branches has a series of three resistors attached.

The trick to solving the problem is to incorporate the series resistance calculation into the bigger one for the whole circuit. For a parallel circuit, you have to use the expression:

{1 \above{2pt}R_{total}} = {1 \above{2pt}R_1} + {1 \above{2pt}R_2} + {1 \above{2pt}R_3}

But the first branch, ​R1, is actually made of three different resistors in series. So if you focus on this first, you know that:

R_1=R_4 + R_5 + R_6

Imagine that ​R4 = 12 Ω, ​R5 = 5 Ω and ​R6 = 3 Ω. The total resistance is:

\begin{aligned} R_1&=R_4 + R_5 + R_6 \\ &= 12 \; \Omega \; + 5 \; \Omega \; + 3 \; \Omega \\ &= 20 \; \Omega \end{aligned}

With this result for the first branch, you can go onto the main problem. With a single resistor on each of the remaining paths, say that ​R2 = 40 Ω and ​R3 = 10 Ω. You can now calculate:

\begin{aligned} {1 \above{2pt}R_{total}} &= {1 \above{2pt}R_1} + {1 \above{2pt}R_2} + {1 \above{2pt}R_3}\\ &= {1 \above{2pt}20 \; Ω} + {1 \above{2pt}40 \; Ω} + {1 \above{2pt}10\; Ω}\\ &= {2 \above{2pt}40 \; Ω} + {1 \above{2pt}40 \; Ω} + {4 \above{2pt}40 \; Ω}\\ &= {7 \above{2pt}40 \; Ω}\\ &= 0.175 \; Ω^{-1} \end{aligned}

So that means:

\begin{aligned} \ R_{total} &= {1 \above{2pt}0.175 \; Ω^{-1}}\\ &= 5.7 \; \Omega \end{aligned}

Other Calculations

Resistance is much easier to calculate on a series circuit than a parallel circuit, but that isn’t always the case. The equations for capacitance (​C​) in series and parallel circuits basically work the opposite way around. For a series circuit, you have an equation for the reciprocal of capacitance, so you calculate the total capacitance (​Ctotal) with:

{1 \above{2pt}C_{total}} = {1 \above{2pt}C_1} + {1 \above{2pt}C_2} + {1 \above{2pt}C_3} + ....

And then you have to divide one by this result to find ​Ctotal.

For a parallel circuit you have a simpler equation:

C_{total} = C_1 + C_2 + C_3 + ....

However, the basic approach to solving problems with series vs. parallel circuits is the same.

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