RC charging at t = 0+: an uncharged capacitor and a 1 kΩ resistor are placed in series with a switch and a 6 V battery. Immediately after closing the switch, what is the voltage across the capacitor?

Introduction / Context:
Transient analysis of RC circuits starts with initial conditions. Knowing the instantaneous voltage across a capacitor at the moment a switch closes is a foundational concept for timing circuits, debouncing, and analog signal steps.

Given Data / Assumptions:

• An initially uncharged capacitor (Vc(0^-) = 0 V).
• A series resistor of 1 kΩ and an ideal 6 V DC source.
• Switch closes at t = 0, no prior pre-charge.

Concept / Approach:
A capacitor’s voltage cannot change instantaneously; it changes according to the exponential charging law Vc(t) = Vs * (1 - e^(-t/RC)). At the instant just after closing (t = 0+), Vc(0+) equals the initial value Vc(0^-) which is 0 V for an uncharged capacitor.

Step-by-Step Solution:

Verification / Alternative check:
Use the charging equation: Vc(t) = 6 * (1 - e^(-t/(R*C))). At t = 0, e^0 = 1, so Vc(0) = 6 * (1 - 1) = 0 V. Current initially is I(0+) = 6 / R = 6 mA, then decays as the capacitor charges.

Why Other Options Are Wrong:

• 6 V: That is the long-term value as t → ∞, not at t = 0+.
• 3 V or 2 V: Arbitrary fractions; the initial value is determined by continuity, not averaging.
• Depends on capacitance value: The initial voltage is zero regardless of C; C changes the time constant, not the initial condition.

Common Pitfalls:

• Assuming immediate charge to source voltage; real capacitors cannot change voltage instantaneously.
• Confusing initial current (which can be large) with initial capacitor voltage.

Final Answer:
0 V