Three resistors of 4.0, 6.0, and 10.0 Ω are connected in parallel. If the combination is connected in series with a 12.0-V battery and a 2.0-Ω resistor, what is the current through the 10.0-Ω resistor?
A)0.59 A
B)2.7 A
C)6.4 A
D)11.2 A
E)16.0 A

Question

Quizplus · Accepted Answer

The equivalent resistance of the parallel resistors is 1/(1/4 + 1/6 + 1/10) = 1.71 Ω. The total resistance in the circuit is 1.71 + 2.0 = 3.71 Ω. The total current from the battery is 12.0 V / 3.71 Ω = 3.23 A. The current through the 10.0-Ω resistor is (3.23 A) * (1.71 Ω / 10.0 Ω) = 0.59 A.

Three Resistors of 4