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  GR9277 #86
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\prob{86}
9277_86

THe circuit shown above is used to measure the size of the capacitance C. The y-coordinate of the spot on the oscilloscope screen is proportional to the potential difference across R, and the x-coordinate of the spot is swept at a constant speed s. The switch is closed and then opened. One can then calculate C from the shape and the size of the curve on the screen plus a knowledge of which of the following?


  1. $V_0$ and R
  2. s and R
  3. s and $V_0$
  4. R and R'
  5. The sensitivity of the oscilloscope

Lab Methods}Oscilloscopes

The discharge of the capacitor after it has been charged to V_0 is just V(t) = V_0(1-e^{-\omega t}), where \omega = 1/(RC). One can find C by knowing R and the sweep rate, which is related to t.
(Solution due to David Latchman.)

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
Herminso
2009-09-22 14:33:45
The discharge of the capacitor after it has been charged to V_0 is just

V=V_0e^{-t/RC} \Rightarrow C=-\frac{t}{R}\ln{(V_0/V)},

thus after the switch is opened, since R is in series with the capacitor we need to find t, which we can determine by how fast the trace sweeps, s. We need to find R, will be given. Thus the answer is (B). The ration V_o/V can be read off the vertical parts of the scope.
Alternate Solution - Unverified
Comments
ehsana
2013-10-07 17:36:19
Voltage through a discharging capacitor should be V(t)=V_0 e^{-t/RC}.

V(t)=V_0(1- e^{-t/RC}), as it's written in the solution, is the voltage through a charging capacitor.
Typo Alert!
astro_girl
2013-09-13 06:30:48
The voltage across the capacitor and acrossR, when the switch is closed, is not V_o but V_o\frac{R}{R'+R}. When the switch is closed the capacitor branch is open , but current keeps flowing through R. This doesn't affect the answer though. NEC
flyboy621
2010-10-22 21:23:15
You can do this one by elimination.

When the switch is opened, R' is out of the picture, which rules out (D). (E) is too vague to be meaningful, so that choice must be wrong. Finally, it should be obvious that V_0 will affect only the scale of the scope display and not the time-behavior of it. Thus (A) and (C) are ruled out. (B) is the only choice left.
NEC
Herminso
2009-09-22 14:33:45
The discharge of the capacitor after it has been charged to V_0 is just

V=V_0e^{-t/RC} \Rightarrow C=-\frac{t}{R}\ln{(V_0/V)},

thus after the switch is opened, since R is in series with the capacitor we need to find t, which we can determine by how fast the trace sweeps, s. We need to find R, will be given. Thus the answer is (B). The ration V_o/V can be read off the vertical parts of the scope.
Alternate Solution - Unverified
oSciL8
2009-03-29 19:04:58
Wait a minute - I thought the time constant (tau = 1/RC) was dependent on the resistor in series with the capacitor? This would mean that one would need to know the sweep rate and R', not R...would someone explain?


Thanks for this website btw - I'll see how helpful its been in a week!!
gt2009
2009-06-21 04:13:49
After the switch is opened, R is in series with the capacitor.
Answered Question!
hamood
2007-04-05 14:42:14
Capacitor discharging is V = Vo \e(-t/CR)
Capacitor charging is V(t) = Vo(1-\e(-t/CR))
grace
2010-11-10 22:50:28
I agree with you.
NEC

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Capacitor discharging is V = Vo \e(-t/CR) Capacitor charging is V(t) = Vo(1-\e(-t/CR))

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