Answer:
Time taken for the capacitor to charge to 0.75 of its maximum capacity = 2 × (Time take for the capacitor to charge to half of its capacity)
Explanation:
The charging of a capacitor/the build up of its voltage follows an exponential progression and is given by
V(t) = V₀ [1 - e⁻ᵏᵗ]
where k = (1/time constant)
when V(t) = V₀/2
(1/2) = 1 - e⁻ᵏᵗ
e⁻ᵏᵗ = 0.5
In e⁻ᵏᵗ = In 0.5 = - 0.693
-kt = - 0.693
kt = 0.693
t = (0.693/k)
Recall that k = (1/time constant)
Time to charge to half of max voltage = T(1/2)
T(1/2) = 0.693 (Time constant)
when V(t) = 0.75
0.75 = 1 - e⁻ᵏᵗ
e⁻ᵏᵗ = 0.25
In e⁻ᵏᵗ = In 0.25 = -1.386
-kt = - 1.386
kt = 1.386
t = 1.386(time constant) = 2 × 0.693(time constant)
Recall, T(1/2) = 0.693 (Time constant)
t = 2 × T(1/2)
Hope this Helps!!!
Answer:
the time taken for the capacitor to charge to 75% of its maximum is RC ln (1.33 )
Explanation:
applying an exponential progression
V = Vo ( 1 - [tex]e^{\frac{-t}{Rc} }[/tex] ) ---------- equation 1
note V = Vo/ 2
therefore equation 1 can as well be represented as
Vo / 2 = Vo ( 1 - [tex]e^{\frac{-t}{RC} }[/tex] ) ----------- equation 2
from equation 2 dividing both sides of the equation by 2
[tex]e^{\frac{-t}{RC} }[/tex] = 1 /2 therefore
[tex]\frac{t}{RC }[/tex] = ln 2
t ( 0.5 ) = RC ln 2
therefore
t ( 0.75 ) = RC ln(1.33)
R = resistance and C = capacitance
