Question

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

Answer 1

Answer:

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 $e^\frac{-t}{\tau}$

Q = Q₀ / 2 , t = ?

Q₀ / 2  =  Q₀ $e^\frac{-t}{\tau}$

1 / 2 =  $e^\frac{-t}{\tau}$

2 = $e^\frac{t}{\tau}$

ln2 = $\frac{t}{\tau}$

t = τ x ln2

= 49.5  x .693

= 34.3

= 34.3 s.