A capacitor is charged from an 8 V DC source through a resistor (first-order RC). What voltage will appear across the capacitor exactly at t = 2τ (two time constants) during the charging process?

Introduction:
First-order RC charging follows an exponential law. Knowing how to compute the capacitor voltage at a specified multiple of the time constant is essential for timing, settling, and design margin decisions.

Given Data / Assumptions:

• Supply voltage Vs = 8 V.
• Time instant t = 2τ.
• Standard RC charging starting from 0 V initial condition.

Concept / Approach:
The charging law is Vc(t) = Vs * (1 - exp(-t / τ)). Substitute t = 2τ to compute the fraction of the final value reached at two time constants.

Step-by-Step Solution:
1) Start with Vc(t) = 8 * (1 - exp(-t / τ)).2) At t = 2τ, Vc = 8 * (1 - exp(-2)).3) exp(-2) ≈ 0.1353.4) 1 - 0.1353 = 0.8647.5) Vc ≈ 8 * 0.8647 ≈ 6.9176 V, which rounds to about 6.9 V, commonly reported as 7 V.

Verification / Alternative check:
Check limits: at t = 0, Vc = 0; as t increases, Vc approaches 8 V. Two time constants giving roughly 86% aligns with standard RC charging tables.

Why Other Options Are Wrong:
5 volts: corresponds to about 62.5% of 8 V, closer to t ≈ 1τ.3 volts: too low; would be early in the transient.1 volt: far too early; not consistent with 2τ.

Common Pitfalls:
Forgetting the exponential form and attempting linear interpolation; RC charging is not linear with time.

Final Answer:
7 volts