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dsp73
3 years ago
11

The time to half-maximum voltage is how long it takes the capacitor to charge halfway. Based on your experimental results, how l

ong does it take (approximately) for the capacitor to charge to 75% of its maximum?
Engineering
2 answers:
BaLLatris [955]3 years ago
4 0

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 - e^{\frac{-t}{Rc} } ) ---------- equation 1

note V = Vo/ 2

therefore equation 1 can as well be represented as

Vo / 2 = Vo ( 1 - e^{\frac{-t}{RC} } ) ----------- equation 2

from equation 2 dividing both sides of the equation by 2

e^{\frac{-t}{RC} } = 1 /2 therefore

\frac{t}{RC } = ln 2

t ( 0.5 ) = RC ln 2

therefore

t ( 0.75 ) = RC ln(1.33)

R = resistance and C = capacitance

satela [25.4K]3 years ago
3 0

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!!!

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2 years ago
A consolidation test was performed on a sample of fine-grained soil sample taken from a depth such that the vertical effective s
Scorpion4ik [409]

Answer:

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Explanation:

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where

'H' is the initial depth of the layer

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Applying the given values we get

\Delta H=\frac{8\times 0.3}{1+0.87}log(\frac{154+28}{154})=1.04

3 0
3 years ago
A continuous random variable, X, whose probability density function is given by f(x) = ( λe−λx , if x ≥ 0 0, otherwise is said t
Ganezh [65]

Answer:

a) F(x) = \lambda \int_0^{\infty} e^{-\lambda x} dx= -e^{-\lambda x} \Big|_0^{\infty} = 1- e^{-\lambda x} \

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Explanation:

Previous concepts

The cumulative distribution function (CDF) F(x),"describes the probability that a random variableX with a given probability distribution will be found at a value less than or equal to x".

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Part a

Let X the random variable of interest. We know on this case that X\sim Exp(\lambda)

And we know the probability denisty function for x given by:

f(x) = \lambda e^{-\lambda x} , x\geq 0

In order to find the cdf we need to do the following integral:

F(x) = \lambda \int_0^{\infty} e^{-\lambda x} dx= -e^{-\lambda x} \Big|_0^{\infty} = 1- e^{-\lambda x} \

Part b

Assuming that X \sim Exp(\lambda =0.1), then the density function is given by:

f(x) = 0.1 e^{-0.1 x} dx , x\geq 0

And for this case we want this probability:

P(10 < X

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P(10 < X

4 0
3 years ago
A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.
Galina-37 [17]

Answer:

elongation of the brass rod is 0.01956 mm

Explanation:

given data

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diameter = 4.50 mm

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load = 610 N

to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = \frac{PL}{AE}    ................1

here δ is change in length and P is applied load  and A id cross section area and E is Young's modulus and L is length

so all value in equation 1

δ = \frac{PL}{AE}  

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7 0
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Yuliya22 [10]

Answer:

the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

Explanation:

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= 37.88 psi

Therefore, the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

5 0
3 years ago
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