V=100sin(200pi t +pi/4)

[John54321]

Homework Statement

I have the last 3 parts of this question to answer and I'm struggling with it and would appreciate any advice please or if what i have done is correct thanks.

Homework Equations

The Attempt at a Solution

Q1. The time to reach the greatest voltage rate of change

A. dv/dt =(2 x 10^4pi) cos (200pi t + pi/4)
maximum occurs when dv/dt = 0 and d^2v/dt^2 is negative
d^2v/dt^2 = -(4 x 10^6 pi^2) sin (200pi t + pi/4)
cos (200pi t + pi/4) = 0 when t = 0 and when t = 0 d^2v/dt^2 is negative

When t = 0 v = 100 as sin(200pi t +pi/t = 1

Q2. The value of the greatest voltage rate of change

dv/dt is maximum when cos (200pi t +pi/4) is either 1 or -1
which occurs when 200pi t + pi/4 = 0 or pi radians
t is either = 1/800 or 1/800 seconds i.e. t = -0.00125 or 0.00125 seconds

Q3. Using integral calculus, calculate the RMS value sketch a graph to visualise the process ?

 

[LCKurtz]

Homework Statement

I have the last 3 parts of this question to answer and I'm struggling with it and would appreciate any advice please or if what i have done is correct thanks.

Homework Equations

The Attempt at a Solution

Q1. The time to reach the greatest voltage rate of change

A. dv/dt =(2 x 10^4pi) cos (200pi t + pi/4)
maximum occurs when dv/dt = 0 and d^2v/dt^2 is negative
d^2v/dt^2 = -(4 x 10^6 pi^2) sin (200pi t + pi/4)
cos (200pi t + pi/4) = 0 when t = 0 and when t = 0 d^2v/dt^2 is negative

When $t=0$ you get $\cos(\frac \pi 4)\ne 0$.

When t = 0 v = 100 as sin(200pi t +pi/t = 1

You meant $\sin(200\pi t +\frac \pi 4)$ and that isn't $1$ when $t=0$.

Q2. The value of the greatest voltage rate of change

dv/dt is maximum when cos (200pi t +pi/4) is either 1 or -1
which occurs when 200pi t + pi/4 = 0 or pi radians
t is either = 1/800 or 1/800 seconds i.e. t = -0.00125 or 0.00125 seconds

Q3. Using integral calculus, calculate the RMS value sketch a graph to visualise the process ?

What is the integral you have to work?

 

[John54321]

The questions i have got are the same as I've written calculus integration. Thanks

 

[LCKurtz]

You haven't addressed the mistakes I pointed out. And you haven't shown us that you have at least looked up the integral you need to calculate, much less shown any effort at solving it.

 

[ChrisVer]

what is the question?
in both Q1 and Q2 you solve the same thing...(try to find the time of maximum)...in Q1 as it's pointed out you did it wrong, but in Q2 you got the right answer (if your voltage is as given).
Also are you looking for the rate of changes or the voltage itself? if you are looking for the rates of changes then you need to maximize dV/dt and not V...

 

[John54321]

Hi ChrisVer

I have the volatge when t is at 0 to be 71v
The voltage when t = 5 ms -71v

The question 10.6
The time to reach the gretaest voltage rate of change- note the greatest rate of change
- note that the greatest rate of change is when dv/dt is a maximum, ie d^2v/dt^2 = 0

10.7 The value of the greatest voltage rate of change

10.8 using integral calculus calculate the RMS value of the voltage sketch graph to visualise process