During the charging of a capacitor through a resistor from a DC source, how does the current flowing into the capacitor change with time?

Introduction:The transient behavior of current in an RC circuit reveals how quickly a capacitor charges and how the system settles. This question probes your understanding of the current waveform during the charging phase.

Given Data / Assumptions:

• Series RC charging from a constant DC source.
• Capacitor initially uncharged.
• Ideal first-order model.

Concept / Approach:In RC charging, i(t) = (Vs / R) * exp(-t / τ). The current is maximal at t = 0 and decays exponentially to zero as the capacitor approaches the supply voltage and the net voltage across the resistor falls.

Step-by-Step Solution:1) At t = 0, capacitor voltage is 0, so i(0) = Vs / R (maximum).2) As Vc rises, the voltage across R drops, reducing current.3) Exponential decay continues until steady state, where current ideally becomes zero.

Verification / Alternative check:Differentiate Vc(t) = Vs * (1 - exp(-t / τ)) to see i(t) = C * dVc/dt = (Vs / R) * exp(-t / τ), confirming exponential decrease.

Why Other Options Are Wrong:Increase: opposite of actual behavior; current is highest at the start.Remain the same: contradicts exponential law.Cannot tell: the first-order model clearly predicts the trend.

Common Pitfalls:Assuming constant current as with an ideal current source; here the source is a constant voltage and the resistor limits current dynamically.

Final Answer:decrease