Final answer:
The current when a 1.70 µF capacitor has acquired 1/4 of its maximum charge through a 10.0 Ω resistor using a 10.0 V battery is not 1/4 of the maximum current; instead, it must be calculated by finding the time at which the charge is 1/4 of maximum and then applying the exponential charging equation.
Step-by-step explanation:
When a 1.70 µF capacitor is charging through a 10.0 Ω resistor, using a 10.0 V battery, the current when the capacitor has acquired 1/4 of its maximum charge will not be 1/4 of the maximum current.
The charging of a capacitor through a resistor follows an exponential curve given by I(t) = (V/R)e-t/RC,
where V is the battery voltage, R is the resistance, C is the capacitance, and t is the time.
The initial current (I0) can be found using Ohm's law:
I0 = V/R.
When the capacitor has acquired 1/4 of its maximum charge Qmax, which is C⋅V, the current is not simply 1/4 of I0 because as the capacitor charges, the voltage across it increases and the voltage across the resistor (and hence the current) decreases. To find the current at the time when the capacitor has 1/4 Qmax, we first find the time at which this occurs, then we use the exponential curve equation to find the current at that time.
Complete question is as follows :
Question: 26.44) A 1.70 capacitor is charging through a 10.0 resistor using a 10.0
26.44) A 1.70 - µF
capacitor is charging through a 10.0 Ω
resistor using a 10.0 -Vbattery.
What will be the current when the capacitor has acquired 1/4 of its maximum charge? Will it be 1/4 of the maximum current?