Q.A circular loop has a resistance of 40 \Omega. Two points P and Q of the loop, which are one quarter of the circumference apart, are connected to a 24 V battery, having an internal resistance of 0.5 \Omega. What is the current flowing through the battery? (A) 0.5 A (B) 1 A (C) 2 A (D) 3 A

[!TLDR] The two arcs between P and Q (10 \Omega and 30 \Omega) are in parallel, giving 7.5 \Omega; adding the 0.5 \Omega internal resistanc …

The circular loop has total resistance 40 \Omega, and P, Q are one quarter of the circumference apart, so the SHORTER arc (a quarter of the loop) has resistance $40/4=10\ \Omega$ and the LONGER arc (the remaining three-quarters) has resistance $3\times10=30\ \Omega$. Since both arcs connect the same two points P and Q, they are effectively in PARALLEL as seen by the external battery:

$R_{PQ} = \frac{10\times30}{10+30} = \frac{300}{40} = 7.5\ \Omega$ …

Split the loop into its two arcs by the quarter/three-quarter ratio, combine them as resistors in parallel (since both ends are shared), then add the b …

Treating the two arcs as being in series instead of parallel, or forgetting to add the battery's own inte …