Voltage on capacitor

  • Thread starter luigihs
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  • #1
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A 22 nF capacitor is initially charged to 10V. It is then discharged by connecting a 100k resistor across it. Approximately how long does it take for the voltage on the capacitor to fall by 10%?



v = v₀e^(–t/τ)
RC = τ = time constant



Im not sure if this the right equation what do you think guys?
 

Answers and Replies

  • #2
gneill
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The equation looks fine. How you make use of it is what's important :smile:
 
  • #3
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τ = 22nF x 100k = 2200µs = 2.2 ms

then
v/v₀ = 0.9 = e^(–t/2.2m)
–t/2.2m = ln 0.9 = –0.1054
t = 0.0002318 = 232 µs <-- answer!

is that alright?
 
Last edited:
  • #4
gneill
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Yes, that's fine.
 

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