Final answer:
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.