RC charging time: A 4.7 MΩ resistor is in series with a 0.015 µF capacitor across a 12 V source. Estimate how long it takes for the capacitor to be effectively fully charged (use the standard 5τ rule).

Introduction / Context:
Capacitors in RC networks charge exponentially. Engineers often use the 5τ guideline (five time constants) to approximate the time for a capacitor to be essentially fully charged (over 99%).

Given Data / Assumptions:

• R = 4.7 MΩ.
• C = 0.015 µF.
• Series RC driven by a 12 V DC source.
• Use t_full ≈ 5 * tau.

Concept / Approach:
Time constant tau = R * C. Convert units carefully before multiplying: megaohms to ohms, microfarads to farads. Then multiply tau by 5 to approximate full charge time.

Step-by-Step Solution:
R = 4.7 MΩ = 4.7 * 10^6 ΩC = 0.015 µF = 15 * 10^-9 Ftau = R * C = 4.7 * 10^6 * 15 * 10^-9tau = 70.5 * 10^-3 s = 70.5 mst_full ≈ 5 * tau = 5 * 70.5 ms = 352.5 msClosest option is 352 ms.

Verification / Alternative check:
At t = 5τ, the capacitor reaches about 99.3% of the final voltage, which is the standard engineering rule-of-thumb for “fully” charged in practice.

Why Other Options Are Wrong:

• 70.5 ms: This is only 1τ, not near full charge.
• 35 ms: Less than 1τ; underestimates by a factor of about 10.
• 3.5 s: Overestimates by an order of magnitude.

Common Pitfalls:
Using 3τ instead of 5τ when the question explicitly asks for fully charged, or mis-converting µF and MΩ. Always convert to base units before multiplying.

Final Answer:
352 ms