Final answer:
The maximum current flows immediately after the switch is closed and declines as the capacitor charges. After a significant period, the voltage across each capacitor equals the battery emf, and the charge equals the product of capacitance and voltage. The time constant depends on the equivalent capacitance and resistance in circuit.
Step-by-step explanation:
Understanding Capacitor Charging in Circuits
When a switch in an electrical circuit containing a capacitor and resistors is closed, current begins to flow, causing the initially uncharged capacitor to start charging. The initial current is at a maximum as the voltage across the capacitor starts from zero. Over time, the voltage across the capacitor increases, leading to a decrease in current, following the equation V = emf(1 - e-t/RC) where V is voltage, emf is the electromotive force of the battery, t is time, R is resistance, and C is capacitance.
Immediately after the switch is closed, the ammeter will read the maximum current, but this current will decline as the capacitor charges. After a long time (several RC time constants), the capacitor is considered to be fully charged, the current falls to zero, and the voltage across the capacitor equals the battery's emf. At this steady state, the voltage across each resistor will be zero if they are in parallel with the capacitor, and the maximum charge stored in the capacitor is given by Q = CV, where Q is charge and V is the final voltage across the capacitor.
If capacitors are connected in parallel, as in the case when three 20-mF capacitors are utilized, the equivalent capacitance of the capacitors is the sum of the individual capacitances. The RC time constant, which influences how quickly the current declines to a certain percentage of its initial value, depends on this equivalent capacitance and the resistance through which the capacitors are charging.