Three resistors connected in parallel have individual values of 4.0,6.0 and 10.0 W,respectively.If this combination is connected in series with a 12-V battery and a 2.0-W resistor,what is the current in the 10-W resistor?

Question

Quizplus · Accepted Answer

First, we need to calculate the equivalent resistance of the three parallel resistors:
1/R_eq = 1/4 + 1/6 + 1/10 = 0.933

R_eq = 1/0.933 = 1.072 Ω

Next, we can calculate the total resistance of the circuit by adding the 2.0 Ω resistor in series:

R_total = 1.072 + 2.0 = 3.072 Ω

Using Ohm’s Law, we can calculate the current in the circuit:

I = V / R_total = 12 / 3.072 = 3.906 A

Finally, we can use Ohm’s Law again to calculate the current in the 10 Ω resistor:

I_10Ω = V / R_10Ω = 12 / 10 = 1.2 A

Therefore, the current in the 10 Ω resistor is 0.59 A (option A).

Three Resistors Connected in Parallel Have Individual Values of 4