Question

V = 100 sin (200πt + π/4)
the time to reach the greatest voltage rate of change

Answer 1

That's cool. Now where's the Q?

Answer 2

when is Sin theta maximum? for what angle?

Answer 3

Oh, voltage rate of change...not voltage...then u ll need to differentiate first

Answer 4

@telltoamit, I suspect your suspicions on the question is correct.

Answer 5

question are correct.*

Answer 6

yea, u ll jus get a Cos term

Answer 7

is this correct
dv/dt = (2 x 10^4 π) cos (200 π t + π/4)

Answer 8

absolutely

Answer 9

@tomtom, please write your questions as a question in the English language next time. It makes you more easily understood.

Answer 10

what next
after differentiation

Answer 11

for what angle is Cos theta maximum?

Answer 12

45?

Answer 13

Nah. 0. Cos zero = 1

Answer 14

is this my answer
dv/dt = (2 x 104 π) cos (200 π t + π/4)

maximum occurs when dv/dt = 0 and d2v/dt2 is negative

d2v/dt2 = -(4 x 106 π2) sin (200 π t + π/4)

cos (200 π t + π/4) = 0 when t = 0 and when t = 0 d2v/dt2 is negative

when t = 0, v = 100 as sin (200 π + π/t) = 1
dv/dt is a maximum when cos (200 π t + π/4) is either 1 or -1
which occurs when 200 π t + π/4 = 0 or π radians

t is either = -1/800 or 1/800 seconds

t = -0.00125 or 0.00125 seconds

Answer 15

No. U hv to find when is dv/dt maximum and not when v is maximum.

Answer 16

maximum occurs when dv/dt = 0
so 0 = (2 x 10^4 π) cos (200 π t + π/4)

Answer 17

No, the greatest voltage rate of change
maximum for dv/dt occurs when d2v/dt2 = 0

Answer 18

d2v/dt2 = (4 x 10^6 π2) sin (200 π t + π/4) = 0