Given Information:
Inductance = L = 5 mH = 0.005 H
Time = t = 2 seconds
Required Information:
Current at t = 2 seconds = i(t) = ?
Energy at t = 2 seconds = W = ?
Answer:
Current at t = 2 seconds = i(t) = 735.75 A
Energy at t = 2 seconds = W = 1353.32 J
Explanation:
The voltage across an inductor is given as
The current flowing through the inductor is given by
Where L is the inductance and i(0) is the initial current in the inductor which we will assume to be zero since it is not given.
![i(t) = \frac{1}{0.005} \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \,+ 0\\\\i(t) = 200 \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \\\\i(t) = 200 \: [ {5\: (t + \frac{e^{-0.5t}}{0.5})]_0^t \\i(t) = 200\times5\: \: [ { (t + 2e^{-0.5t} + 2 )] \\](https://tex.z-dn.net/?f=i%28t%29%20%3D%20%5Cfrac%7B1%7D%7B0.005%7D%20%5Cint_0%5Et%20%5Cmathrm%7B5%281-e%5E%7B-0.5t%7D%7D%29%20%5C%2C%5Cmathrm%7Bd%7Dt%20%5C%2C%2B%200%5C%5C%5C%5Ci%28t%29%20%3D%20200%20%5Cint_0%5Et%20%5Cmathrm%7B5%281-e%5E%7B-0.5t%7D%7D%29%20%5C%2C%5Cmathrm%7Bd%7Dt%20%5C%5C%5C%5Ci%28t%29%20%3D%20200%20%5C%3A%20%5B%20%7B5%5C%3A%20%28t%20%2B%20%5Cfrac%7Be%5E%7B-0.5t%7D%7D%7B0.5%7D%29%5D_0%5Et%20%5C%5Ci%28t%29%20%3D%20200%5Ctimes5%5C%3A%20%5C%3A%20%5B%20%7B%20%28t%20%2B%202e%5E%7B-0.5t%7D%20%2B%202%20%29%5D%20%5C%5C)
So the current at t = 2 seconds is
The energy stored in the inductor at t = 2 seconds is
Explanation:
There are two ways to find out the equivalent impulse response of the system.
1. Convolution in time domain
2. Simple multiplication in Laplace domain
2nd method is efficient, easy and is less time consuming.
Step 1: Take the Laplace transform of the given three impulse response functions to convert time domain signals into s-domain
Step 2: Once we get signals in s-domain, multiply them algebraically to get the equivalent s-domain response.
Step 3: Take inverse Laplace transform of the equivalent impulse response to convert from s-domain into time domain.
Solution using Matlab:
Step 1: Take Laplace Transform
Ys1 = 1/(s + 1)
Ys2 = 1/s - exp(-s/2)/s
Ys3 = exp(-3*s)
Step 2: Multiplication in s-domain
Y = (exp(-(7*s)/2)*(exp(s/2) - 1))/(s*(s + 1))
Step 3: Inverse Laplace Transform (Final Solution in Time Domain)
h = heaviside(t - 7/2)*(exp(7/2 - t) - 1) - heaviside(t - 3)*(exp(3 - t) - 1)
Answer:
Throat diameter 
=28.60 mm
Explanation:
Bore diameter 
⇒
Manometric deflection x=235 mm
Flow rate Q=240 Lt/min⇒ Q=.004
Coefficient of discharge 
=0.8
We know that discharge through venturi meter
=13.6 for Hg and 
=1 for water.
h=2.961 m
Now by putting the all value in
⇒=28.60 mm
So throat diameter =28.60 mm
a bricklayer, a bricklayer builds houses repairs walls and chimneys etc.
Answer:
Explanation:
The pressure drop is directly proportional to the length of the pipe. Then, the new pressure drop is two time the previous one.
