A $6.0 \Omega$ resistor and a $3.0 \Omega$ resistor are connected in parallel to a $1.5 \mathrm{~V}$ battery of negligible internal resistance. What is the current in the $3.0 \Omega$ resistor?
A) $0.75 \mathrm{~A}$
B) $0.25 \mathrm{~A}$
C) $2.0 \mathrm{~A}$
D) $0.17 \mathrm{~A}$
E) $0.50 \mathrm{~A}$

Question

Quizplus · Accepted Answer

The total resistance in parallel is given by $1/R_{total} = 1/R_1 + 1/R_2$, which calculates to $1/R_{total} = 1/6 + 1/3 = 1/2$, so $R_{total} = 2 \Omega$. The total current from the battery is $I_{total} = V/R_{total} = 1.5V / 2\Omega = 0.75A$. The current through the $3.0 \Omega$ resistor can be found using the voltage across it, which is the same as the battery's voltage (1.5V), so $I = V/R = 1.5V / 3\Omega = 0.5A$.

A 6.0Ω6.0 \Omega6.0Ω Resistor and A 3.0Ω3.0 \Omega3.0Ω Resistor Are Connected in Parallel to A

A $6.0 \Omega$ Resistor and A $3.0 \Omega$ Resistor Are Connected in Parallel to A