Resistors in combination: Three resistors having resistances of 4.0 Ω, 6.0 Ω, and 10.0 Ω are connected in parallel. If the combination is connected in series with an ideal 12-V battery and a 2.0-Ω resistor, what is the current through the 10.0-Ω resistor?

Question

Quizplus · Accepted Answer

First, calculate the equivalent resistance of the parallel combination:
1/R = 1/4 + 1/6 + 1/10 = 0.65
R = 1/0.65 = 1.54 Ω
Next, calculate the total resistance of the entire circuit (including the 2.0-Ω resistor):
R_total = 1.54 + 2.0 + 4.0 + 6.0 + 10.0 = 23.54 Ω
Using Ohm's law, we can calculate the current through the circuit:
I = V/R_total = 12/23.54 = 0.51 A
Now, we can use the current divider rule to determine the current through the 10.0-Ω resistor. The current through each branch of the circuit is proportional to the resistance of that branch. Thus, the current through the 10.0-Ω resistor is:
I_10Ω = (10/23.54) * I = 0.21 A
Alternatively, we could use the fact that the current through each resistor in a parallel combination is the same. Therefore, the current through the 10.0-Ω resistor is:
I_10Ω = I_parallel = (1/4 + 1/6 + 1/10)*I = 0.59 A
The closest answer choice is A, which is the correct answer.

Resistors in Combination: Three Resistors Having Resistances of 4