Two resistors of values 7.0 and 15.0 Ω are connected in parallel.This combination in turn is hooked in series with a 3.8-Ω resistor and a 22.3-V battery.What is the current in the 3.8-Ω resistor?
A)21 A
B)2.6 A
C)0.86 A
D)5.9 A
E)190 A

Question

Quizplus · Accepted Answer

First, calculate the equivalent resistance of the 7.0 Ω and 15.0 Ω resistors in parallel: $R_{eq} = \frac{1}{\frac{1}{7} + \frac{1}{15}} = \frac{105}{22} \Omega$. Then, add this to the 3.8 Ω resistor to get the total resistance in the circuit: $R_{total} = \frac{105}{22} + 3.8 \approx 8.58 \Omega$. Finally, use Ohm's law ($V = IR$) to find the current: $I = \frac{V}{R_{total}} = \frac{22.3}{8.58} \approx 2.6 A$.

Two Resistors of Values 7