How long after the switch is closed does the voltage across

AI Thread Summary
The discussion focuses on calculating the time it takes for the voltage across a resistor to drop to 20V after a switch is closed in a circuit. The relevant equations include the time constant Tc = RC and the charge equation Q = Qm(1-e-t/RC). An initial calculation led to an incorrect answer due to using the wrong formula. After correcting the approach to I = (V0/R)*e-t/RC, the correct time was determined to be approximately 91.886 microseconds. The importance of using the correct voltage values in calculations was emphasized throughout the discussion.
beeteep
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Homework Statement


Consider the circuit
2s5yyd0.jpg


How long after the switch is closed does the voltage across the resistor drop to Vf = 20V. Answer in units of s.

Homework Equations


Tc = RC
Q=Qm(1-e-t/RC)

The Attempt at a Solution


[/B]
RC = .0003026

11 = 31(1-e-t/RC)

.3548 = 1-e-t/RC

1.3548 = e-t/RC

ln(1.3548) = -t/RC

.3037*.0003026 = -t

91.886x10-6 s
Answers I've also come to that have been incorrect:

Previous Tries

Try#1: .945

Try#2: .000313554 (using 20 instead of 11 for Q)

Try#3: 91.886e-6
 

Attachments

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Where in your calculations does the 20V appear?
 
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rude man said:
Where in your calculations does the 20V appear?
20 = 31(1-e-t/RC)
 
beeteep said:
20 = 31(1-e-t/RC)

Or what I think you may be asking

(31V-20V) = 31(1-e-t/RC)
 
Looks like I was using the wrong formula.

I changed to I = (V0/R)*e-t/RC and was able to get the correct answer.
 
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