Series-Parallel Branch Current: Public Worked Question

Question

A $12\,\text{V}$ battery drives a circuit consisting of a $3.0\,\Omega$ resistor in series with two $6.0\,\Omega$ resistors that are connected in parallel with each other. What is the current through one of the $6.0\,\Omega$ resistors?

1. a.$0.8\,\text{A}$
2. b.$1.0\,\text{A}$
3. c.$2.0\,\text{A}$
4. d.$4.0\,\text{A}$

Full reasoning

1. 1Combine the parallel pair first: two equal $6.0\,\Omega$ resistors give $6.0/2 = 3.0\,\Omega$.
2. 2The total resistance is $R_{eq} = 3.0 + 3.0 = 6.0\,\Omega$, so the battery current is $I = 12/6.0 = 2.0\,\text{A}$.
3. 3The voltage across the parallel pair is $V = (2.0\,\text{A})(3.0\,\Omega) = 6.0\,\text{V}$.
4. 4Each branch then carries $I = 6.0\,\text{V}/6.0\,\Omega = 1.0\,\text{A}$, consistent with the $2.0\,\text{A}$ splitting equally by symmetry.

Why each choice is right or wrong

Related formula

Related topic and practice

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Source and public-release record

Authored for this public explainer registry. It has no database question ID and is not copied from a protected or official exam bank.

1. University Physics Volume 2, Section 10.2: Resistors in Series and Parallel — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.