AP Physics 2 · Question of the Day

AP Physics 2 Question of the Day

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Friday, November 6, 2026

A $6\ \text{V}$ ideal battery connects to $R_1=1\ \Omega$ in series, then reaches junction $J$ where it splits into two branches that rejoin at $K$ . Branch 1 has $R_2=2\ \Omega$ ; Branch 2 has $R_3=6\ \Omega$ . Which statement correctly compares the currents through $R_2$ and $R_1$ ?

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Question of the Day

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The current through $R_2$ is greater than the current through $R_1$ .
The current through $R_2$ equals the current through $R_1$ .
The current through $R_2$ is less than the current through $R_1$ . (correct answer)
The current through $R_2$ is zero because current splits and is consumed.

Explanation: This problem examines current relationships in compound DC circuits. In this configuration, R₁ carries the total circuit current before it splits at junction J into two parallel branches containing R₂ and R₃. By Kirchhoff's current law, the current through R₁ equals the sum of currents through R₂ and R₃, making the current through R₁ greater than the current through either parallel branch alone. Since current divides between parallel branches, the current through R₂ must be less than the current through R₁. Choice D incorrectly suggests current is "consumed" when it splits, misunderstanding that current is conserved at junctions. When analyzing compound circuits, trace the current path and remember that series elements carry the same current while parallel branches divide the total current.