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Six 30 ohm resistors are wired into three branches connected in parallel across a 12 V battery. The middle branch has two of the resistors wired one after the other. What power is dissipated in the middle branch?
A4.8 W
B2.4 W
C1.2 W
D0.4 W
Answer & Solution
Correct answer: B. 2.4 W
1. The two 30 ohm resistors in the middle branch are wired one after the other, so their resistance adds: R_middle = 30 ohm + 30 ohm = 60 ohm.
2. Since the branch is connected directly across the battery, the full battery voltage of 12 V appears across it.
3. Use P = V^2 / R_middle = (12 V)^2 / 60 ohm.
4. P_middle = 144 / 60 W = 2.4 W.
5. 4.8 W would result from using only a single 30 ohm resistor instead of the two combined.
_Source: OpenStax Physics (CC BY 4.0), Ch 19 "Electrical Circuits", section Power through a Branch of a Circuit_
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