Final answer:
The time it takes for a capacitor to acquire 1/4 of its maximum charge can be determined using the RC time constant. In this case, it will take approximately 4.50 seconds. The current when the capacitor has acquired 1/4 of its maximum charge will be 0.833 A, which is not 1/4 of the maximum current.
Step-by-step explanation:
When a capacitor is charging through a resistor using a battery, the current decreases over time as the capacitor charges. The time it takes for the capacitor to charge to a certain fraction of its maximum charge can be determined using the RC time constant, given by the equation τ = R * C. In this case, the capacitance C is 1.50 F and the resistance R is 12.0 Ω, so the RC time constant is τ = 1.50 F * 12.0 Ω = 18.0 s.
For the capacitor to acquire 1/4 of its maximum charge, we can multiply the RC time constant by 1/4: τ * (1/4) = 18.0 s * (1/4) = 4.50 s. Therefore, it will take approximately 4.50 seconds for the capacitor to acquire 1/4 of its maximum charge.
As for the current, it will not be 1/4 of the maximum current. Instead, the current through the circuit at any given time can be calculated using Ohm's Law: I = V / R, where V is the voltage across the resistor and R is the resistance. Since the voltage across the resistor is equal to the battery voltage of 10.0 V, the current can be calculated as: I = 10.0 V / 12.0 Ω = 0.833 A. Therefore, the current when the capacitor has acquired 1/4 of its maximum charge will be 0.833 A, which is not 1/4 of the maximum current.