Question

homework 7

Answer 1

I have done all this
H7P1: SERIES AND PARALLEL INDUCTORS
L1 + L2
L1*L2/(L1 + L2)
L3*L2/(L3 + L2)+ L1
(L2+L3)*L1/(L1 + L2 + L3)
H7P2 TIME CONSTANTS
((R1+R2)/L)^-1
(((R1*R2)/(R1+R2))/L)^-1
H7P3 THE CURSE OF LEAD INDUCTANCE
V*RO/(RO+RS)
L/(RO + RS)

In my case: Now, let's get down to numbers. Let V=3.3V, RS=22.0Ω, RO=50.0Ω, and L=15.44nH. How much time, in nanoseconds, does it take for the output voltage to reach vO=1.65V?
0.273
In my case : This is a pretty significant amount of time! For comparison, how long, in nanoseconds, does light take to go 2cm? Remember, the speed of light is 299792458m/s.
0.0677 ...
... well, wether it's not these value, follow these example below.. or give these value above a try.. good luck and any questions , feel snug to ask me..

Now, let's get down to numbers. Let V=1.8V, RS=22.0Ω, RO=50.0Ω, and L=15.44nH.How much
time, in nanoseconds, does it take for the output voltage to reach vO=0.9V?
The output voltage goes from 0 at time to V*RO/(RO+RS) after a very long time. So using first
order equation, with vi as initial voltage and vf as final voltage
vO(t) = vf + (vi - vf)*et/τ
where τ = L/(RO + RS) = 15.44*10-9/72 = 0.214 * 10-9
vi = 0V and vf = V*RO/(RO+RS) = 50*1.8/72 = 1.25V
vO(t) = 0.9V
we need to solve for t
t = τ* ln((vO(t) – vf)/(vi –vf))
Now we substitute the values from above t = 0.214 * 10-9 * ln((0.9 - 1.25)/(0 - 1.25)) = 0.273 *
10-9s =0.273ns
Speed of light v = 299792458m/s distance s = 2cm = 0.02m
We know that v = s/t, therefore t = s/v = 0.02/299792458 = 6.67*10-11 s = 0.0667 ns.

Answer 2

Spring 2013midterm plz help me