RC discharge in basic electronics A 0.3 µF capacitor is initially at 2.7 V. It is discharged through a 10 kΩ resistor for 6.5 ms. What voltage remains across the capacitor after this time?

Introduction / Context:Capacitors in RC circuits discharge exponentially. This question tests your ability to apply the standard RC decay formula to find the remaining voltage after a given time constant interval.

Given Data / Assumptions:

• Capacitance C = 0.3 µF = 0.3 * 10^-6 F.
• Resistance R = 10 kΩ = 10,000 Ω.
• Initial voltage V0 = 2.7 V.
• Elapsed time t = 6.5 ms = 6.5 * 10^-3 s.

Concept / Approach:The capacitor voltage during discharge follows V(t) = V0 * exp( - t / (R * C) ). The product R * C is the time constant tau, which sets the exponential decay rate.

Step-by-Step Solution:

Verification / Alternative check:Since 6.5 ms is a little over two time constants (2 tau = 6 ms), a remaining fraction near exp( -2 ) ≈ 0.135 is expected; our more precise 0.1145 is reasonable because 6.5 ms > 6 ms.

Why Other Options Are Wrong:

• 0.0 V: Ideal exponential never reaches exactly zero at finite time.
• 2.7 V: That is the initial voltage; no discharge applied.
• 2.295 V: Implies very little decay; contradicts t > 2 tau.

Common Pitfalls:Mixing up units (µF, kΩ, ms), using linear instead of exponential decay, or forgetting to compute tau first.

Final Answer:0.309 V