Series, Parallel and Junction Currents

Nodes, Branches and Junctions

First, let’s begin by defining what those terms mean. Let’s begin with a branch.

A branch is simply a two terminal electronic component in a circuit. It is connected to two nodes using a wire coming from each terminal. Each coloured line represents one branch on the above screenshot. The same current flows through both wires (both sides of the component):

Next up we have a node…

A node is a common connection of two or more circuit elements. Node has the same voltage at any point. Current flows into a node and spreads out into the branches. Current flowing into a node must equal the current flowing out of a node.

Let’s have a look at a junction now…

Junction is a point on a node at which it branches out. Similarly to a node, current flowing into a junction must equal to current flowing out of it.

Series Circuit Current

In series circuits, current has only one path through which it can travel. This current will be the same at every point and will be dependant on the voltage source and resistance in the circuit.

Resistors in series combine to create a single total value of resistance:

$R_{Series}=R_{1}+R_{2}$

Therefore the current according to Ohm’s law is:

$I=\frac{V_{S}}{R_{Series}}=\frac{5V}{6\Omega}=0.83333A=833.33mA$

We can find the volt drop across a resistor by multiplying the total current flowing through the circuit by the resistance value of that resistor:

$V_{R1}=I\times R_{1}=0.83333A\times 2\Omega = 1.667V$

Which means that we can find the current flowing through this circuit by rearranging the formula a little:

$I=\frac{V_{R1}}{R1} = \frac{1.667V}{2\Omega}=0.83333A=833.33mA$

Parallel Circuit Current

Components are connected in parallel when they share the same two nodes together. In parallel circuits, current splits among the branches:

$R_{1},R_{2},R_{3},V_{S}$ are in parallel

Currents across each one of the resistors ( $R_{1},R_{2},R_{3}$ ) are as follows:

$I_{1}=\frac{20V}{2\Omega}=10A$

$I_{2}=\frac{20V}{3\Omega}=6.67A$

$I_{3}=\frac{20V}{6\Omega}=3.33A$

In order to get the current flowing from the source (or in other words current flowing into the node) we need to sum those individual currents we just calculated:

$I_{Total}=I_{1}+I_{2}+I_{3}=10A+6.67A+3.33A=20A$

We can check by combining all parallel resistors into one equivalent resistor in series with the supply:

$\frac{1}{R_{Parallel}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$

$\frac{1}{R_{Parallel}}=\frac{1}{2\Omega}+\frac{1}{3\Omega}+\frac{1}{6\Omega}$

$\frac{1}{R_{Parallel}}=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\frac{6}{6}=\frac{1}{1}$

$\frac{R_{Parallel}}{1}=\frac{1}{1}$

$R_{Parallel}=1\Omega$

And then divide the supply voltage by the equivalent series resistance:

$I=\frac{V_{S}}{R_{Parallel}}=\frac{20V}{1\Omega}=20A$

Current going into the node must equal the current that goes out of the node. Current goes in and out of branches.

Junction Current

Take the same circuit as above and focus on point A:

Current flowing into a junction must equal current flowing out of it.

In this example (at point A) we see one current flowing in and two flowing out.

From previous calculations, we know that the current flowing into the node is equal to 20A. We also know that the current flowing through the 2Ohm resistor is equal to 10A. The remaining current must be therefore:

$I_{In}=I_{Out}$

$20A=10A+I_{J3} //-10A$

$I_{J3}=10A$

This 10A current then flows into the next branch and splits in the same manner.

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