A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
answers !!!!
qs 6
a) and b) L/(rw+rl)
c and d) go to sandbox, make the circuit, take a f=1/(2*T) and a 5ms trans and measure vmax and vmin
anonymous
 4 years ago
answers !!!! qs 6 a) and b) L/(rw+rl) c and d) go to sandbox, make the circuit, take a f=1/(2*T) and a 5ms trans and measure vmax and vmin

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thats correct aparna! now can u put a transient anlalysis graph pic here?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0aparna,do 1 thing,attach ur file showing transient analysis and then tell , where is vmax and where is vmin, i mean at what time do you measure them?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@jason10 , i am attaching my transiant analysis , kindly now tell me the answer for last two parts from these graphs.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0capture 3 n4 r my values n i got it right

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@jason10 , i have attached my file , can u tell me the answer for last two parts ????

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@chaand.i for vmin u need to place the pointer a the min value i.e at 2ms(ur snap) n for max u need to place it at max peak value

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@aparna.b , what is the stop time we need to set ? is that 5ms always ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I checked two points of tran and my vmin value became bigger than vmax, it's ok?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it ur wish. we can use even 10ms. but time time u choose must be very small n in ms

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@aparna.b , my vmin = 42.22m and vmax = 5.925 .. is it ook ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@fatdrfrog vmin will be smaller than vmax. i think u calculated as V. its not V its mV

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@chaand.i i think they r ryt.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but @aparna.b , i am getting confused that how vmin can be 42.22m ?? :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0vmin is the min voltage across resistor. n if u point ur marker at min value u get dat

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0y is it that sum r getting it ryt n sum r getting it wrong?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i dunno. but wat method i did is right.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@jason10 am I right please check i am left with last check. i got part c as 4.32 and part d as 0.67 please tell me the answer i got 1st two green ticks, last two left.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0V0=5V, T=1ms, Rw=1Ω, Lw=1mH and RL=1Ω. vmax = ???? vmin = 0.67 4.308 and 4.328 are wrong, and I have only one check

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@chaand.i I have the following values :V0=10V, T=0.5ms, Rw=1Ω, Lw=0.5mH and RL=4Ω.. i have a=0.1, b=0.1, c= 7.942. still having problems with d.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.