by **Guest** » Sat Jun 27, 2015 6:24 am

The maximum value [tex]\sin[/tex] attains is 1. So the maximum voltage attained by [tex]V = V_0\sin(\omega t+\alpha)[/tex] is simply [tex]V_0[/tex] (assuming [tex]V_0[/tex] is positive which for the purpose of this question we are free to do). This means [tex]V_0=312[/tex] volts.

(If we assume [tex]V_0[/tex] is negative then [tex]V_0=-312[/tex] volts and the value of [tex]\alpha[/tex] we calculate later should be replaced by [tex]\alpha+180^\circ[/tex]).

The maximum peaks in [tex]\sin[/tex] occur at intervals of [tex]360^\circ[/tex] (or [tex]2\pi[/tex] radians). Let [tex]t_1[/tex] be the time of the first peak and let [tex]t_2[/tex] be the time of the second peak. We know that [tex](\omega t_2 +\alpha) - (\omega t_1+\alpha) = 360^\circ[/tex] (assuming [tex]\omega[/tex] is positive which for the purpose of this question we are free to do) which simplifies to [tex]\omega(t_2-t_1)=360^\circ[/tex] which in turn tells us that [tex]\omega = 360^\circ/(t_2-t_1)[/tex]. The question tells us that [tex]t_2-t_1=0.02[/tex] seconds, so that means [tex]\omega = 18000[/tex] degrees/second (or [tex]100\pi[/tex] radians/second).

(If we assume [tex]\omega[/tex] is negative then [tex]\omega = -18000[/tex] degrees/second.)

At [tex]t=0[/tex] the question tells us that [tex]V=200[/tex] volts. Substituting this into [tex]V = V_0\sin(\omega t+\alpha)[/tex] gives [tex]200 = 312\sin\alpha[/tex]. So [tex]\sin\alpha = 200/312 = 25/39[/tex] and therefore [tex]\alpha = \sin^{-1} (25/39) = 39.86...^\circ[/tex] or [tex]140.13...^\circ[/tex] plus or minus multiples of [tex]360^\circ[/tex] which for the purposes of this question make no difference, (or equivalently 0.6958... or 2.4457... radians plus or minus multiples of [tex]2\pi[/tex]).

Unfortunately the question is badly written and there is no way to determine which of the two solutions for [tex]\alpha[/tex] we should take.

"What will the voltage be after 1.2 seconds?"

Substituting this into the formula gives [tex]V = 312\sin (18000\times 1.2+\alpha) = 312\sin(216000+\alpha) = 312\sin(60\times 360+\alpha) = 312\sin\alpha = 312\times \frac{25}{39} = 200[/tex] volts.

In 1.2 seconds the voltage goes through 60 full oscillations to end up back where it started which is at 200 volts.

"What are the first 3 times that the voltage will be 240 volts?"

This depends on which of the two solutions for [tex]\alpha[/tex] we take.

[tex]240 = 312\sin(18000t+\alpha)[/tex]

which implies [tex]18000t+\alpha = \sin^{-1}(240/312) = 50.28...+360n[/tex] or [tex]129.71...+360n[/tex] for some integer [tex]n[/tex].

This rearranges to [tex]t = (50.28-\alpha)/18000+0.02n[/tex] or [tex](129.71-\alpha)/18000+0.02n[/tex].

If [tex]\alpha = 39.86...[/tex] we get

[tex]t = ..., -0.0194213, 0.000578695, 0.020578695, 0.040578695, ...[/tex] or [tex]..., -0.0150085 ,0.00499148, 0.02499148, 0.04499148, ...[/tex]

So the first three times are: [tex]0.000578695, 0.00499148, 0.020578695[/tex].

If [tex]\alpha = 140.13...[/tex] we get

[tex]t = ..., -0.00499148, 0.0150085, 0.0350085, 0.0550085, ...[/tex] or [tex]..., -0.000578695, 0.0194213, 0.0394213, 0.0594213 ...[/tex]

So the first three times are: [tex]0.0150085, 0.0194213, 0.0350085[/tex].

Hope this helped,

R. Baber.