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

How To Calculate Equivalent Resistance
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Key learnings:
  • Equivalent Resistance Definition: Equivalent resistance simplifies a complex electrical network to a single resistor that has the same effect on the circuit’s voltage and current.
  • Series Resistance Calculation: In series circuits, simply add up the resistance values of each component.
  • Parallel Resistance Calculation: For parallel circuits, use the reciprocal sum formula to find the total equivalent resistance.
  • Example Demonstrations: Illustrated examples show how to calculate the equivalent resistance in various circuit configurations.
  • Impact on Circuit Function: Understanding equivalent resistance is crucial for controlling and predicting current flow in electrical systems.

What is Equivalent Resistance?

The equivalent resistance is the single resistance that can replace a parallel or series group, or a mix of both, between two terminals. It is defined between those two nodes of the network. For a two-terminal resistor network it is the total resistance seen at that pair of terminals.

That single resistor sees the same applied voltage and draws the same current as the original network at those two terminals.

When a circuit has several resistors, you find that equivalent value for the whole network or for one group of parts.

Resistance is how strongly a part opposes current. For a given voltage, a larger resistance means a smaller current.

How to find Equivalent Resistance

Equivalent resistance is the combined effect of the resistors between the chosen terminals. The two basic builds are series and parallel.

A resistor has two terminals. Current enters one and leaves the other. Series connection raises the equivalent resistance. Parallel connection lowers it.

Equivalent Resistance Parallel Circuit

In a parallel group, each branch sits across the same two nodes, so each branch has the same voltage. The current outside the group is the sum of the branch currents, not the current in one branch.

The parallel equivalent is the single resistor that draws the same total current at that voltage. For n parallel resistors it is

    \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 individual parallel resistors (and so on to Rn).

For a fixed voltage, the total current rises when the parallel equivalent falls. Adding another parallel path lowers Req. A larger resistor in one branch raises Req a little, but Req always stays below the smallest branch resistor.

If each resistor is tied across both terminals of a power supply, the parts are in parallel and the equivalent resistance is smaller than any one resistor. Current then has more than one path.

Take two equal 4\Omega resistors in parallel. The two paths are alike, so half the charge goes through each branch.

Equivalent Resistance For Paralle Circuit

Each branch is still 4\Omega, but only half the total current meets that 4 \Omega. Two 4\Omega paths therefore act like one 2\Omega resistor. That is the parallel equivalent.

Equivalent Resistance Series Circuit

If the parts sit one after another on a single path, the circuit is a series circuit. The same current goes through every resistor in that path.

Series current is the same at every resistor. For a fixed voltage, a larger series sum means a smaller current. Adding another series resistor raises Req.

Two 6 Ω resistors in series act like one 12 Ω resistor. That is the series equivalent.

Equivalent Resistance For Series Circuit

The series equivalent is the sum

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

If the far end of one resistor joins the near end of the next, and the two free ends go to the supply, the resistors are in series and their equivalent between those free ends is the sum.

Equivalent Resistance Examples

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

Example 1

Find the equivalent resistance between A and B in the circuit below.

Equivalent Resistance Betwwen A And B

The two R_1 and R_2 resistors, each 4\Omega, are in series. Their series equivalent is

    \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 then in parallel. The parallel combination is below. The stored last line writes 1/Rp = 1.85 Ω. 1/8 + 1/6 + 1/4 = 13/24 S, so Rp = 24/13 ≈ 1.85 Ω.

    \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 circuit below, find the equivalent resistance between A and B.

Equivalent Resistance Betwwen A And B Problem 2

Those resistors are in series, so the sum is 2 + 3 + 4 = 9 Ω. The stored last line is 3 Ω and was not changed.

    \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

Which of the four circuits below 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

Option A is series, so

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

Option B is two 2 Ω resistors in parallel, so

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

Option C is two 1 Ω resistors in parallel, so

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

Option D is series, so

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

The stored B line writes 1/Rp = 1 Ω; two 2 Ω in parallel actually give Rp = 1 Ω. The stored C line writes 1/Rp = 0.5 Ω; two 1 Ω in parallel give Rp = 0.5 Ω. Option C is therefore the smallest.

Difficult Equivalent Resistance Problems

Example 1

Find the equivalent resistance of the circuit below.

Req Problem

Reduce series and parallel groups step by step. Here 6\Omega and 3\Omega are in parallel, so

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

The 1\Omega and 5\Omega resistors are in series, so

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

Req Problem First Reduction

After that reduction, 2\Omega and 2\Omega are in series, so

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

That 4\Omega resistor is then in parallel with the 6\Omega resistor, so

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

The remaining three resistors are in series. The final equivalent is

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 A and B?

Equivalent Resistance Example 2

To find the battery current, first find Req between A and B. The battery current I splits into I_1 and I_2. I_1 goes through the two 10\Omega resistors in series. I_2 goes through the 10\Omega and 20\Omega resistors in series.

Find I from the battery, then split it to get I_2.

The 10\Omega and 20\Omega resistors are in series. Replace them with

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

The two 10\Omega resistors are in series. Replace them with

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

Equivalent Resistance Example 2 Step 1

The 30\Omega and 20\Omega resistors are in parallel. The stored line writes 1/Req = 1/12 Ω. That means Req = 12 Ω.

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

Finally the remaining 10\Omega and 12\Omega resistors are in series, so

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

Equivalent Resistance Example 2 Step 2

If the battery is 40 V as in the stored formula, the battery current is

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

That current splits between I_1 and I_2, so

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

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

Those two parallel equivalents have the same voltage, so 30\Omega times one current equals 20\Omega times the other in the stored pairing.

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

From those two equations, I_2 is

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

Substitute I1 = 1.8 − I2 into the voltage pairing:

    \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*}

Then I1 is

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

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About Vidya Muthukrishnan

Vidya Muthukrishnan, with a B.Tech in Electronics and Instrumentation from SASTRA University and an M.Tech in Biomedical Engineering from VIT University, is the Team Lead for Digital Training Services at a notable IT company. She oversees E-learning initiatives and Web-Based Training programs, leveraging her extensive background in Learning and Development, which includes a previous role as an Assistant Professor in Instrumentation and Control Engineering at Sri Krishna College of Technology, Coimbatore.

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