Question

The voltage, v, (volts) in an alternating current at any time, t, (seconds) is given by: V = 100 sin (200πt + π/4)
Using differential calculus, calculate: 1. The time to reach the greatest voltage rate of change 2. The value of the greatest voltage rate of change 3. Using integral Calclulus, calculate the RMS value of the voltage

Answer 1

differentiate dV/dt and equate to zero

Answer 2

So first off, what is the voltage rate of change? That's the derivative of the voltage with respect to time.

Now where is that at it's maximum? Well when it's at its maximum, it will have a slope of 0, so take the derivative again and set it equal to zero, and you can solve for time t.

Since you're looking for the time to reach it the first time, you choose the value of t which is the smallest and non-negative, unless you have a time machine.

So you differentiate twice, not once like cwrw is saying.

Answer 3

lol time machine !!
yes u r right kalnui - however i was only indicating how to start the problem

Answer 4

Err, sorry I might have misread what you wrote. Anyways, I'm gone so good luck guys!

Answer 5

So far I have calculated the amplitude =100V, Frequency = 100Hz and Phase Angle = 45degrees (don't know how to get the little zero for degrees on keyboard.)
The voltage when t=0 = v=100Sin(0+0.785) = 100Sin(45Degrees) = 71V
The Voltage when t=5ms = 100Sin(225degress) = -71V
The time when voltage first reaches 50V.

Answer 6

When V=50V
then 50V= 100Sin(200*pi*t+0.785)
50/100=Sin(200*pi*t+0.785)
(200*pi*t+0.785)=arcsin 50/100
=30degrees or 0.524rad

200*pi*t= 0.524+0.785 = 1.309

So when V=50v
Time t = 1.309/200*pi = 2.08ms

Answer 7

v=100Sin(200*pi*t+pi/4)

Using differential calculus, calculate:
The time for the the voltage to first reach a maximum:
Maximum volatge = 100V

Hence 100=100Sin(200*pi*t+0.785)
1=Sin(200*pi*t+0.785)
arcsin1=200*pi*t+0.785
=90 degress or 1.5708rad
200*pi*t=1.5708 - 0.785 = 0.7858

t=0.7858/200*pi = 1.25ms

Answer 8

The time to reach the greatest voltage rate of change

dv/dt is a maximum when Cos(200πt+π/4) is either 1 or -1 which occurs when 200πt+π/4=0 or π radians

t is either =(-1)/800 secs or 1/800 secs

t=-0.00125 secs or t=0.00125 secs

Answer 9

Is this correct please as I have the same question