How to Calculate Equivalent Resistance (Series & Parallel Circuit Examples)

How To Calculate Equivalent Resistance

What is Equivalent Resistance?

The equivalent resistance is defined as a point where the total resistance is measured in a parallel or series circuit (in either the whole circuit or in a part of the circuit). The equivalent resistance is defined between two terminals or nodes of the network. Equivalent resistance may sound complicated, but it’s just a technical way to say “total resistance”.

In the equivalent resistance of a network, a single resistor could substitute the complete network so that for a specific applied voltage and/or the equivalent current can be obtained similar to the one when used as a network.

When a circuit has more than one circuit component in it, there should be a way to calculate the total effective resistance of the entire circuit or for just one part of the circuit.

Before we discuss what equal resistance is, we can describe resistance. Resistance is a measure of how much a device or material can resist the movement of electricity through it. It is inversely related to current, higher resistance means reduced current flow; reduced resistance means higher current flow.

How to find Equivalent Resistance

The equivalent resistance represents the total effect of all resistors in the circuit. The equivalent resistance can be measured in either a series or parallel circuit.

Resistor comprises two junctions with which the current passes in and out of it. They are passive devices that utilize electricity. In order to improve the net resistance, the resistors must be wired in series and the resistors must be connected in parallel to reduce the resistance.

Equivalent Resistance Parallel Circuit

A parallel circuit is one in which elements are connected to different branches. In a parallel circuit, the voltage drop is the same for each parallel branch. The total current in each branch is equal to the current outside the branches.

The equivalent resistance of the circuit is the amount of resistance that a single resistor will require in order to equalise the total effect of the set of resistors present in the circuit. For parallel circuits, the equivalent resistance of a parallel circuit is given as

    \begin{align*}\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + …. + \frac{1}{R_n} \end{align*}

where R_1, R_2, and R_3 are the resistance values of the individual resistors that are connected in parallel.

The total amount of current will often vary inversely with the level of cumulative resistance. There is a direct relation between the resistance of the individual resistors and the total resistance of the resistance collection.

If all endpoints of the resistors are wired to both endpoints of the power supply, so the resistors are connected in parallel and their equivalent resistance declines between their endpoints. There is more than one direction to flow in parallel circuit current.

To investigate this relationship, let’s start with the simplest case of two resistors positioned in parallel branches, each of which has the same resistance value as 4\Omega. Since the circuit provides two equivalent paths for the charge transport, just one-half of the charge can choose to travel through the branch.

Equivalent Resistance For Paralle Circuit

Although each branch gives 4\Omega of resistance to any charge flowing through it, only one-half of all the charge flowing through the circuit may meet 4 \Omega of resistance of that branch. Thus, the presence of two 4\Omega resistors in parallel will be equal to one 2\Omega resistor in the circuit. This is the concept of equivalent resistance in a parallel circuit.

Equivalent Resistance Series Circuit

If all components are connected in series, the circuit is referred to as a series circuit. In a series circuit, each unit is connected in such a way that there is only one route through which the charge can travel through the external circuit. Every charge traveling through the external circuit loop would travel through each resistor in a sequential manner. In a series circuit, the current has only one path to flow.

Charge flows together over the external circuit at a rate that is the same everywhere. The current is not stronger at one place and weaker at another point. Inversely, the exact amount of current varies with the total resistance. There is a direct relationship between the resistance of the single resistors and the total resistance of all the resistors present in the circuit.

For example, when two 6-Ω resistors are connected in series, it would be equivalent to having one 12-Ω resistor in the circuit. This is the concept of equivalent resistance in a series circuit.

Equivalent Resistance For Series Circuit

For series circuits, the equivalent resistance of a series circuit is given as

    \begin{align*} R_s = R_1 + R_2 + R_3 + .... R_n\end{align*}

If the endpoint of one resistor is linearly connected to the endpoint of the neighboring resistor and the free end of one resistor and the free end of the other resistor are connected to the power supply. Then the two resistors are wired in series and their equal resistance increases between their endpoints.

Equivalent Resistance Examples

For the Combination of Resistors Shown, Find the Equivalent Resistance Between Points A and B.

Example 1

For the given circuit below, what is the equivalent resistance between points A and B?

Equivalent Resistance Betwwen A And B

The two resistors R_1 and R_2 with value 4\Omega are in series. So, their equivalent resistance value will be

    \begin{align*} R_s = R_1 + R_2 \end{align*}

    \begin{align*} R_s = 4\Omega + 4\Omega = 8\Omega \end{align*}

Equivalent Resistance Betwwen A And B Step 2

R_s , R_3 and R_4 are in parallel. The eqivalent resistance of the circuit.

    \begin{align*}\frac{1}{R_p} = \frac{1}{8\Omega} + \frac{1}{6\Omega} + \frac{1}{4\Omega} = \frac{13}{24}\Omega\end{align*}

    \begin{align*}\frac{1}{R_p} = 1.85 \Omega \end{align*}

Example 2

For the given circuit below, calculate the equivalent resistance between the end points A and B

Equivalent Resistance Betwwen A And B Problem 2

The expression for the equivalent resistance of the resistor connected in series is given as follows.

    \begin{align*} R_s = R_1 + R_2 +R_3\end{align*}

    \begin{align*} R_s = 2\Omega + 3\Omega +4\Omega\end{align*}

    \begin{align*} R_s = 3\Omega\end{align*}

Which Circuit Has the Smallest Equivalent Resistance

Example 1

From the below given circuits, identify the circuit that has the smallest equivalent resistance.

Smallest Resistance Problem Option A
Option A
Smallest Resistance Problem Option B
Option B
Smallest Resistance Problem Option C
Option C
Smallest Resistance Problem Option D
Option D

The first given is a series circuit. So, the equivalent resistance is given as

    \begin{align*} R_s = 2\Omega + 2\Omega\ = 4\Omega \end{align*}

The second given is a parallel circuit. So, the equivalent resistance is given as

    \begin{align*}\frac{1}{R_p} = \frac{1}{2\Omega} + \frac{1}{2\Omega} = 1\Omega\end{align*}

The second given is also a parallel circuit. So, the equivalent resistance is given as

    \begin{align*}\frac{1}{R_p} = \frac{1}{1\Omega} + \frac{1}{1\Omega} = 0.5\Omega\end{align*}

The fourth given is a series circuit. So, the equivalent resistance is given as

    \begin{align*} R_s = 1\Omega + 1\Omega\ = 2\Omega \end{align*}

So, from the above calculation it is seen that the third option has the smallest euqivalent resistance value.

Difficult Equivalent Resistance Problems

Example 1

Find the Equivalent Resistance of the given circuit.

Req Problem

To get the Equivalent Resistance we combine resistors in series and in parallel. Here, 6\Omega and 3\Omega are in parallel. So, the equivalent resistance is given as

    \begin{align*}\frac{6\times3}{6+3}=2\Omega \end{align*}

Also, the 1\Omega and 5\Omega resistors are in series. Hence the equivalent resistance will be given as,

    \begin{align*} 1\Omega + 5\Omega = 6\Omega\end{align*}

Req Problem First Reduction

After reduction, we now notice, 2\Omega and 2\Omega are in series, so the equivalent resistance

    \begin{align*} 2\Omega + 2\Omega = 4\Omega\end{align*}

This 4\Omega resistor is now in parallel with the 6\Omega resistor. So, their equivalent resistance will be given as

    \begin{align*}\frac{4\times 6}{4+6}=2.4\Omega \end{align*}

Now replacing the above circuit with appropriate values, the three resistors will be in series. So, the final equivalent resistance is given as

Req Problem Second Reduction

    \begin{align*} R_{eq} = 4\Omega + 2.4\Omega + 8\Omega = 14.4\Omega \end{align*}

Example 2

What is the equivalent resistance between points A and B?

Equivalent Resistance Example 2

To find the current through the battery we need to find the equivalent resistance of the circuit. The total current I is divided into I_1 and I_2. The current I_1 passes through two 10\Omega resistors as they are connected in series and have the same current. The current I_2 passes through 10\Omega and 20\Omega resistors as they have the same current.

We need to find the current I_2 by first calculating the current I that passes through the battery.

We see that 10\Omega and 20\Omega resistors are connected in series. We replace them with an equivalent resistor with a resistance of

    \begin{align*} R_{eq} = 10\Omega + 20\Omega = 30\Omega \end{align*}

Two 10\Omega resistors are connected in series. We replace them with an equivalent resistance of

    \begin{align*}R_{eq} = 10\Omega + 10\Omega = 20\Omega \end{align*}

Equivalent Resistance Example 2 Step 1

Now we have two resistors 30\Omega and 20\Omega connected in parallel. We can replace with an equivalent resistor.

    \begin{align*}\frac{1}{R_{eq}} =\frac{1}{30} + \frac{1}{20} = \frac{1}{12}\Omega \end{align*}

Finally, we have two resistors 10\Omega and 12\Omega connected in series. The equivalent resistance of these two resistors is

    \begin{align*}R_{eq} = 10\Omega + 12\Omega = 22\Omega \end{align*}

Equivalent Resistance Example 2 Step 2

Now we can find the current I through the battery. It is,

    \begin{align*} I = \frac{V}{R_{eq}} = \frac{40}{22} = 1.8 Ampere \end{align*}

This current is divided between two currents I_1 and I_2. So, the total current

    \begin{align*}I = I_1 + I_2\end{align*}

(1)   \begin{equation*}1.8 = I_1 + I_2\end{equation*}

The second equation, which relates the currents, is the condition that the voltage across the resistor 30\Omega is equal to the voltage across the resistor 20\Omega.

(2)   \begin{equation*}20\times I_1 = 30\times I_2\end{equation*}

From the above equations ((1) and (2) the current I_2 is found.

    \begin{align*}I_1= 1.8 - I_2\end{align*}

Then we substitute this relation into equation (2),

    \begin{align*}20(1.8 - I_2) = 30\times I_2 \end{align*}


    \begin{align*}36 = (20+30)I_2 \end{align*}


    \begin{align*}I_2 = \frac{36}{50} = 0.72A\end{align*}

So, now the current I_1 is given as

    \begin{align*}I_1= 1.8 - 0.72 = 1.08 A\end{align*}

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About Vidya Muthukrishnan

Vidya Muthukrishnan is currently employed as a Senior Engineer in a product based IT company. She has 5 years of professional teaching experience, previously Assistant Professor in the Department of Instrumentation and Control Engineering at the Sri Krishna College of Technology. She also has 1 year of industry experience with TCS, India. Vidya completed her B.Tech Electronics and Instrumentation from SASTRA University and M.Tech in Biomedical Engineering from VIT University Vellore.