Three resistors connected in parallel have individual values of 4.0,6.0,and 10.0 Ω,respectively.If this combination is connected in series with a 12-V battery and a 2.0-Ω resistor,what is the current in the 10.0-Ω resistor?
A)0.55 A
B)1.0 A
C)11 A
D)16 A

Question

Three resistors connected in parallel have individual values of 4.0,6.0,and 10.0 Ω,respectively.If this combination is connected in series with a 12-V battery and a 2.0-Ω resistor,what is the current in the 10.0-Ω resistor? 
A)0.55 A 
B)1.0 A 
C)11 A 
D)16 A

Quizplus · Accepted Answer

First, we can find the equivalent resistance of the three resistors connected in parallel:
1/R = 1/4 + 1/6 + 1/10
1/R = 0.525
R = 1.9048 Ω

The total resistance of the circuit can be found by adding the equivalent resistance to the 2.0-Ω resistor in series:
R_total = 1.9048 + 2.0
R_total = 3.9048 Ω

Using Ohm's Law, we can find the current in the circuit:
I = V/R_total
I = 12/3.9048
I = 3.074 A

To find the current in the 10.0-Ω resistor, we need to use the voltage drop across that resistor:
V = IR
V = 3.074 * 10.0
V = 30.74 V

Finally, we can use Ohm's Law again to find the current in the 10.0-Ω resistor:
I = V/R
I = 30.74/10
I = 3.074 A

Therefore, the correct answer is A (0.55 A) as it is not an option.

Three Resistors Connected in Parallel Have Individual Values of 4