Final answer:
The current in each resistor (a 5.0Ω, a 9.0Ω, and a 13.0Ω resistor connected in series with a 24.0V battery) is 0.89A, which is calculated using the total resistance and Ohm's Law.
Step-by-step explanation:
The question is asking to calculate the current through each resistor when a 5.0Ω resistor, a 9.0Ω resistor, and a 13.0Ω resistor are connected in series with a 24.0V battery. In a series circuit, the current flowing through each component is the same. To find the current, we first need to calculate the total resistance in the circuit, which is the sum of the resistances of all the resistors in series. The total resistance (Rtotal) is 5.0Ω + 9.0Ω + 13.0Ω = 27.0Ω.
Using Ohm's Law (V = IR), where V is the voltage, I is the current, and R is the resistance, we can solve for I (current) as follows:
I = V / Rtotal = 24.0V / 27.0Ω = 0.89A
Therefore, the current in each resistor is 0.89A, which corresponds to option (a).