What if a resistor is short-circuited?

If a zero-ohm wire is connected across both terminals of a resistor, it is bypassed. Its effective resistance is zero and it should be removed from the calculation.

How do I handle a delta or wye configuration?

Standard series-parallel rules do not apply to bridge-like structures. You must apply a Y-$\Delta$ or $\Delta$-Y transformation to convert the sub-circuit before proceeding with step-by-step simplification.

What if the circuit has multiple voltage sources?

If you are finding the Thevenin equivalent resistance, turn off all independent sources first: replace independent voltage sources with short circuits ($0\,\Omega$) and independent current sources with open circuits (infinite resistance).

How to find equivalent resistance in a series-parallel circuit

Finding the equivalent resistance of a series-parallel circuit requires simplifying the network step-by-step into a single equivalent resistor. This method applies to any linear, purely resistive DC circuit without active components or complex bridge configurations requiring delta-wye transformations.

The setup

Identify the nodes and branches in the circuit diagram. Locate the power source or the two terminals across which the equivalent resistance is to be found. Begin analysis at the end of the circuit furthest from these terminals.

The steps

Identify a group of resistors that are strictly in series (same current, no branching) or strictly in parallel (connected across the same two nodes).
Calculate the equivalent resistance for that specific group. For series: $R_{eq} = R_1 + R_2 + \dots$ . For parallel: $1/R_{eq} = 1/R_1 + 1/R_2 + \dots$
Redraw the circuit, replacing the identified group with its single equivalent resistor.
Repeat steps 1-3 until only a single equivalent resistor remains between the terminals of interest.

Checking the result

Verify the bounding values for each step. The equivalent resistance of any parallel group must be strictly less than the smallest individual resistor in that group. The equivalent resistance of any series group must be strictly greater than the largest individual resistor in that group.

Common errors

A common mistake is treating resistors as being in series when a node exists between them that routes current to another branch. Another frequent error is forgetting to invert the sum of the reciprocals when calculating parallel resistance ( $R_{p} = (1/R_1 + 1/R_2)^{-1}$ ).

Worked example

Find the equivalent resistance between terminals A and B for a circuit where $R_1 = 4\,\Omega$ is in series with a parallel combination of $R_2 = 6\,\Omega$ and $R_3 = 3\,\Omega$ .

Step 1: Identify that $R_2$ and $R_3$ are in parallel because they share the same two nodes. Step 2: Calculate the equivalent resistance $R_{23}$ for this parallel pair: $\frac{1}{R_{23}} = \frac{1}{R_2} + \frac{1}{R_3}$ $\frac{1}{R_{23}} = \frac{1}{6} + \frac{1}{3} = \frac{3}{6}$ $R_{23} = 2\,\Omega$ Step 3: Redraw the circuit mentally or on paper. $R_1$ is now strictly in series with $R_{23}$ . Step 4: Calculate the total equivalent resistance $R_{eq}$ : $R_{eq} = R_1 + R_{23}$ $R_{eq} = 4 + 2 = 6\,\Omega$

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References: University Physics with Modern Physics, 15th Edition · OpenStax University Physics Volume 2, Chapter 10