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?

Answer 1

The current in the 10.0-Ω resistor is approximately 5.33 A.

Step-by-step explanation:

To find the current in the 10.0-Ω resistor, we first need to calculate the equivalent resistance of the three resistors connected in parallel. The formula for calculating the equivalent resistance of parallel resistors is:

Req = 1 / (1/R1 + 1/R2 + 1/R3)

Plugging in the given resistance values, we have:

Req = 1 / (1/4.0 + 1/6.0 + 1/10.0)

Simplifying, we get Req ≈ 1.875 Ω

Now that we have the equivalent resistance, we can use Ohm's Law to find the current in the 10.0-Ω resistor:

I = V / R

Since the series circuit has a 12-V battery and a 2.0-Ω resistor, the voltage across the parallel combination of resistors is 12 - 2.0 = 10 V

Now, substituting the values into the equation, we have:

I = 10 V / 1.875 Ω

Calculating, we get I ≈ 5.33 A

Therefore, the current in the 10.0-Ω resistor is approximately 5.33 A.