Question

7) An AC generator has the following emf: ε-6," sin[or-(π /4)], where ε,-30V and ω-350 rad/s. The current produced in a connected circuit is: 1(1)-Im sin(at-(3π/4)] , where Im-б20 mA. (a) At what time after t =0 does the generator first reach a maximum? (B) At what time after t-0 does the current reach a maximum?

Answer 1

a)

From

From epsilon = epsilon_m Sin[omega t - pi/4] at E = E_m epsilon_m = epsilon_m Sin[omega t - pi/4] Sin[omega t - pi/4] = 1 omega t - pi/4 = sin^-1(1) = pi/2 Given W = 350 rad/s 350 t = pi/2 + pi/4 = 2.3562 t = 2.3562/350 t = 6.732 * 10^-3 s or 6.732 ms I(t) = I_m sin(omega t - 3 pi/4) I_m = I_m sin(omega t - 3 pi/4) Sin(omega t - 3 pi/4) = 1 omega t - 3 pi/4 = sin^-1(1) = pi/2 350 t = pi/2 + 3 pi/4 = 3.927 t = 3.927/350 t = 0.01122s or 11.22 ms

at E=E_m

$\varepsilon _{m} =\varepsilon _{m}Sin\left [ \omega t-\frac{\pi }{4} \right ]$

$Sin\left [ \omega t-\frac{\pi }{4} \right ]=1$

$\omega t - \frac{\pi }{4} =Sin^{-1}\left ( 1 \right ) = \frac{\pi }{2}$

Given W=350 rad/s

$350t= \frac{\pi }{2} + \frac{\pi }{4} = 2.3562$

t = 2.3562/350

t=6.732*10^-3 s or 6.732 ms

b)

$I(t) = I_{m}Sin\left ( \omega t - \frac{3\pi }{4} \right )$

$=>I_{m} = I_{m}Sin\left ( \omega t - \frac{3\pi }{4} \right )$

$Sin\left ( \omega t - \frac{3\pi }{4} \right ) = 1$

$\omega t - \frac{3\pi }{4} = Sin^{-1}(1) = \frac{\pi }{2}$

$350 t = \frac{\pi }{2} +\frac{3\pi }{4} = 3.927$

t=3.927/350

t=0.01122 s or 11.22 ms