Three resistors of resistances 4.0 Ω,6.0 Ω,and 10 Ω are connected in parallel.If this combination is now connected in series with an ideal 12-V battery and a 2.0-Ω resistor,what is the current through the 10-Ω resistor?
A)0.59 A
B)2.7 A
C)11 A
D)16 A

Question

Quizplus · Accepted Answer

First, we need to find the equivalent resistance of the three resistors connected in parallel:

1/Req = 1/4 + 1/6 + 1/10
1/Req = 0.525
Req = 1.904 Ω

Next, we can find the total resistance in the circuit with the parallel resistors in series with a 2 Ω resistor:

Rtot = Req + 2
Rtot = 3.904 Ω

Using Ohm's law, we can find the current in the circuit:

I = V/Rtot
I = 12/3.904
I = 3.074 A

Now, to find the current through the 10 Ω resistor, we can use the fact that the current through each resistor in a series circuit is the same:

I10 = I
I10 = 3.074 A

However, we need to be careful with significant figures. Since the 10 Ω resistor value has only one significant figure, our answer can only have one significant figure as well. Rounding to one significant figure gives:

I10 = 0.6 A

Therefore, the best answer choice is A, 0.59 A (rounded to two significant figures).

Three Resistors of Resistances 4