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?
A) 0.59 A
B) 2.7 A
C) 6.4 A
D) 11.2 A
E) 16 A

Question

Quizplus · Accepted Answer

When resistors are connected in parallel, the total equivalent resistance is given by:

1/R = 1/R1 + 1/R2 + 1/R3
1/R = 1/4 + 1/6 + 1/10
1/R = 15/120
R = 8 Ω

The current passing through this parallel combination when connected to a 12-V battery can be calculated using Ohm's law:

I = V/R
I = 12/8
I = 1.5 A

Now, when this combination is connected in series with a 2-Ω resistor, the total equivalent resistance is:

R = 8 + 2
R = 10 Ω

The current passing through the resistance of 10 Ω can be calculated using Ohm's law:

I = V/R
I = 12/10
I = 1.2 A

Therefore, the current through the 10.0-Ω resistor is 1.2 A (choice A).

Three Resistors Having Resistances of 4