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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

MIT 6.002 Circuits and Electronics, Spring 2007
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@chaand.i check dis.
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thats correct aparna! now can u put a transient anlalysis graph pic here?
ok 1 min

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Other answers:

aparna,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?
this is v min
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@jason10 , i am attaching my transiant analysis , kindly now tell me the answer for last two parts from these graphs.
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capture 3 n4 r my values n i got it right
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@jason10 , i have attached my file , can u tell me the answer for last two parts ????
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@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
@aparna.b , what is the stop time we need to set ? is that 5ms always ?
I checked two points of tran and my vmin value became bigger than vmax, it's ok?
it ur wish. we can use even 10ms. but time time u choose must be very small n in ms
@aparna.b , my vmin = 42.22m and vmax = 5.925 .. is it ook ?
@fatdrfrog vmin will be smaller than vmax. i think u calculated as V. its not V its mV
@chaand.i i think they r ryt.
but @aparna.b , i am getting confused that how vmin can be 42.22m ?? :/
vmin is the min voltage across resistor. n if u point ur marker at min value u get dat
y is it that sum r getting it ryt n sum r getting it wrong?
i dunno. but wat method i did is right.
@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.
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V0=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
@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.
for question 6

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