H Hancko New member Joined Feb 18, 2021 Messages 16 Apr 11, 2021 #1 Dear Tutors, Given equations: i = 5cos(60(pi)t) p = vi v = (6+10(integration of idt) I have encountered the following problem considering integration and am unable to complete this sum: This is my attempt:
Dear Tutors, Given equations: i = 5cos(60(pi)t) p = vi v = (6+10(integration of idt) I have encountered the following problem considering integration and am unable to complete this sum: This is my attempt:
skeeter Elite Member Joined Dec 15, 2005 Messages 3,204 Apr 11, 2021 #2 [MATH]i = 5\cos(60\pi t)[/MATH] [MATH]v = 6 + \frac{5}{6\pi} \sin(60 \pi t)[/MATH] I get [MATH]p=iv \approx 18.3 \, W [/MATH] at t = 5 msec.
[MATH]i = 5\cos(60\pi t)[/MATH] [MATH]v = 6 + \frac{5}{6\pi} \sin(60 \pi t)[/MATH] I get [MATH]p=iv \approx 18.3 \, W [/MATH] at t = 5 msec.
H Hancko New member Joined Feb 18, 2021 Messages 16 Apr 11, 2021 #3 skeeter said: [MATH]i = 5\cos(60\pi t)[/MATH] [MATH]v = 6 + \frac{5}{6\pi} \sin(60 \pi t)[/MATH] I get [MATH]p=iv \approx 18.3 \, W [/MATH] at t = 5 msec. Click to expand... I do not understand how you got 5/6(pi)
skeeter said: [MATH]i = 5\cos(60\pi t)[/MATH] [MATH]v = 6 + \frac{5}{6\pi} \sin(60 \pi t)[/MATH] I get [MATH]p=iv \approx 18.3 \, W [/MATH] at t = 5 msec. Click to expand... I do not understand how you got 5/6(pi)
skeeter Elite Member Joined Dec 15, 2005 Messages 3,204 Apr 11, 2021 #4 the antiderivative of [MATH]\cos(60 \pi t)[/MATH] is [MATH]\frac{\sin(60 \pi t)}{60\pi}[/MATH] [MATH]6 + 10 \int 5\cos(60 \pi t) \, dt = 6 + 50 \int \cos(60 \pi t) \, dt = 6 + \frac{50\sin(60 \pi t)}{60 \pi}[/MATH] reduce the fraction
the antiderivative of [MATH]\cos(60 \pi t)[/MATH] is [MATH]\frac{\sin(60 \pi t)}{60\pi}[/MATH] [MATH]6 + 10 \int 5\cos(60 \pi t) \, dt = 6 + 50 \int \cos(60 \pi t) \, dt = 6 + \frac{50\sin(60 \pi t)}{60 \pi}[/MATH] reduce the fraction
