Calculate the time taken for a capacitor to lose half of its charge when the Capacitance is 330μF and the Resistance is 150kΩ

Question

asked May 27, 2021 130k views

1 Answer

Karlerik · Answer 1 · 2021-06-03T05:26:51+0000

Answer:

Step-by-step explanation:

decay constant τ = CR , C is capacitance , R is resistance .

= 330 x 10⁻⁶ x 150x 10³

= 49.5 s .

τ = 49.5 s.

Relation of decay of charge

Q = Q₀
$e^(-t)/(\tau)$

Q = Q₀ / 2 , t = ?

Q₀ / 2 = Q₀
$e^(-t)/(\tau)$

1 / 2 =
$e^(-t)/(\tau)$

2 =
$e^(t)/(\tau)$

ln2 =
$(t)/(\tau)$

t = τ x ln2

= 49.5 x .693

= 34.3

= 34.3 s.