The time to half-maximum voltage is how long it takes the capacitor to charge halfway. Based on your experimental results, how long does it take (approximately) for the capacitor to charge to 75% of its maximum?

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!!!

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 - ) ---------- equation 1

note V = Vo/ 2

therefore equation 1 can as well be represented as

Vo / 2 = Vo ( 1 - ) ----------- equation 2

from equation 2 dividing both sides of the equation by 2

= 1 /2 therefore

= ln 2

t ( 0.5 ) = RC ln 2

therefore

t ( 0.75 ) = RC ln(1.33)

R = resistance and C = capacitance

answer:

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