You have a 81.9 81.9 mF capacitor initially charged to a potential difference of 11.1 11.1 V. You discharge the capacitor throug

Question

Archy [21]

4 years ago

15

You have a 81.9 81.9 mF capacitor initially charged to a potential difference of 11.1 11.1 V. You discharge the capacitor throug

h a 3.19 3.19 Ω resistor. What is the time constant?

Physics

2 answers:

-BARSIC- [3] · Answer 1 · 2021-12-11T16:51:02+03:00

Answer:

The time constant is 0.26 s

Explanation:

From ohm's law,

Current (I) = potential difference ÷ resistance = 11.1 ÷ 3.19 = 3.48 A

Quantity of electricity (Q) = capacitance × potential difference = 81.9×10×^-3 × 11.1 = 0.90909 C

time constant (t) = Q/I = 0.90909/3.48 = 0.26 s

lukranit [14] · Answer 2 · 2021-12-11T18:57:20+03:00

<h2>Answer:</h2>

0.261s

<h2>Explanation:</h2>

For an R-C discharging circuit, the time constant τ, is the product of the capacitance (C) of the capacitor and the resistance (R) of the resistor. i.e

τ = R x C ------------------(i)

From the question;

R = 3.19Ω

C = 81.9mF = 81.9 x 10⁻³F

Substitute these values into equation (i) as follows;

τ = 81.9 x 10⁻³ x 3.19

τ = 261.261 x 10⁻³

τ = 0.261s

Therefore, the time constant is 0.261s