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

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.