In this practice problem, the book apparently got the value for the time constant to be 0.1ms. This means that it only considered the 100k resistor to be the Rth. Why is the 10k resistor ignored? Shouldn't the circuit we consider be the one after the switching event to obtain v(t) and v_o(t) for t>0? Hence, shouldn't the Rth be the parallel combination of the 100k and 10k resistor and the corresponding time constant be 9.09ms?
\$\begingroup\$
\$\endgroup\$
8
-
\$\begingroup\$ 0.1 ms? Did the textbook actually come up with that as a time constant? I may be missing something. (Like being a brick short of a full load.) \$\endgroup\$– jonkCommented Oct 23, 2021 at 21:47
-
\$\begingroup\$ The exponential term has a degree of -10t. I know that 10=1/0.1. Hence, -10t=-t/0.1 which has the form -t/tau. Hence, I inferred that tau=0.1ms \$\endgroup\$– FlashCommented Oct 23, 2021 at 21:52
-
\$\begingroup\$ So, how does the time constant of the following circuit depend upon your inferences from the input source behavior? I'm probably still missing something important. \$\endgroup\$– jonkCommented Oct 23, 2021 at 21:55
-
3\$\begingroup\$ As Jonk implied, the time constant, although independent of the input resistor, is 100k X 1 uF = 0.1 second not 0.1 ms. \$\endgroup\$– BarryCommented Oct 23, 2021 at 22:10
-
\$\begingroup\$ Oh, I may have overlooked the units there. But, I don't understand how the time constant is independent of the input resistor for t>0. The switch closes and the input resistor becomes part of the circuit? I tried replacing the capacitor with a test voltage of 1V source and finding the corresponding test current, I got 110 microamps. 1V/110microamps=9090.909ohms which is the same as saying 100k is parallel to the 10k. How is my solution wrong? \$\endgroup\$– FlashCommented Oct 24, 2021 at 5:57
|
Show 3 more comments
1 Answer
\$\begingroup\$
\$\endgroup\$
1
Why is the 10k resistor ignored?
It doesn't affect the time constant because the "-" terminal is a virtual ground, so its voltage doesn't vary.
However, it is not ignored, since its value affects the current into the input and therefore the gain. If you change its value, the general shape of the output doesn't change (i.e., the time constant remains the same), but its amplitude does change.
answered Oct 23, 2021 at 21:39
-
\$\begingroup\$ I don't understand how the time constant is independent of the input resistor for t>0. The switch closes and the input resistor becomes part of the circuit. I tried replacing the capacitor with a test voltage of 1V source and finding the corresponding test current, I got 110 microamps. Getting Rth, 1V/110microamps=9090.909ohms which is the same as saying 100k is parallel to the 10k. How is my solution wrong? \$\endgroup\$– FlashCommented Oct 24, 2021 at 6:00