A capacitor in a single-loop RC circuit is charged to 85% of its final potential difference in 2.4 s. What is the time constant for this circuit?
A) 1.5 s
B) 1.3 s
C) 1.7 s
D) 1.9 s
E) 2.9 s

Question

Quizplus · Accepted Answer

From the problem, we know that after charging for 2.4 s, the capacitor has reached 85% of its final potential difference. Let's define the time constant for this circuit as $	au$. Then we know that after one time constant, the capacitor should have charged to about 63% of its final potential difference.

Using the formula for the charging of a capacitor in an RC circuit:

$V_C = V_f \left(1 - e^{-t/	au}ight)$

where $V_C$ is the potential difference across the capacitor at time $t$, $V_f$ is the final potential difference across the capacitor, $e$ is the mathematical constant, and $	au$ is the time constant.

We can plug in the given values:

$0.85V_f = V_f \left(1 - e^{-2.4/	au}ight)$

Simplifying:

$0.15 = e^{-2.4/	au}$

Taking the natural logarithm of both sides:

$\ln(0.15) = -\frac{2.4}{	au}$

Solving for $	au$:

$	au = -\frac{2.4}{\ln(0.15)} \approx 1.3	ext{ s}$

Therefore, the best choice is B.

A Capacitor in a Single-Loop RC Circuit Is Charged to 85