Maple TA 20-2 Quiz

View previous topic View next topic Go down

Maple TA 20-2 Quiz

Post  sck105 on Sat Mar 07, 2009 12:26 pm

Hey everyone anyone figured out how to do q2 and q3 on this quiz?

sck105
Guest


Back to top Go down

Q3

Post  almost on Sat Mar 07, 2009 1:32 pm

still working on #2 but here's #3

Imax=2*pi*f*C*Vmax, then plug in converting C *10^-6
so C=19.73
Vmax=70.5
Imax=.2608

.2608=2*pi*f*(19.73*10^-6)*70.5
f=29.84
answer is negative

almost
Guest


Back to top Go down

Q1+4

Post  almost on Sat Mar 07, 2009 1:37 pm

while im here might as well put these up too

1)
Irms=(Imax/sqrt(2))
given Irms, solve for Imax
answer is negative


4)
first find Vmax by
Vrms=(Vm/sqrt(2))

then
Vmax=2*pi*f*L*Imax
solve for Imax, its a straight plug in problem L is in Henrys
answer is positive

almost
Guest


Back to top Go down

#2

Post  gina on Sat Mar 07, 2009 3:09 pm

voltage given/square root of 2 = V_rms

(V_rms/resistance given)*V_rms = answer

answer is positive

gina
Guest


Back to top Go down

Re: Maple TA 20-2 Quiz

Post  u on Sat Mar 07, 2009 7:35 pm

What are we supposed to convert Henry to?

u
Guest


Back to top Go down

Re: Maple TA 20-2 Quiz

Post  Kathleen on Sat Mar 07, 2009 9:36 pm

nothing, you dont have to convert henrys

Kathleen

Posts : 42
Join date : 2008-12-18

View user profile

Back to top Go down

Re: Maple TA 20-2 Quiz

Post  yosalsjk on Sun Mar 08, 2009 8:27 pm

i love jimberlain

yosalsjk
Guest


Back to top Go down

question2

Post  guest786 on Sun Mar 08, 2009 11:53 pm

Hey i am confused about the equation given.

voltage given/square root of 2 = V_rms

(V_rms/resistance given)*V_rms = answer


the second equation why do you have to divide and multiply by V_rms? Is there an error in this equation because it isnt working for me.

guest786
Guest


Back to top Go down

Re: Maple TA 20-2 Quiz

Post  Sponsored content


Sponsored content


Back to top Go down

View previous topic View next topic Back to top

- Similar topics

 
Permissions in this forum:
You cannot reply to topics in this forum