GREPhysics.NET: GR9277 #86

GR9277 #86

Problem

GREPhysics.NET Official Solution

\prob{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?

$V_0$ and R
s and R
s and $V_0$
R and R'
The sensitivity of the oscilloscope

Lab Methods $\Rightarrow$ }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.)

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.

Reply to this comment

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.

Reply to this comment

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.

Reply to this comment

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.

Reply to this comment

Herminso
2009-09-22 14:33:45

Reply to this comment

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.

Reply to this comment

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.

Reply to this comment

Post A Comment!

You are replying to:

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!!

Bare Basic LaTeX Rosetta Stone
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
type this...	to get...
$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...

All-Solutions List | Latest Comments | Unanswered Questions (Cries for Help!) |::| About

This site is not affiliated with ETS, nor is it officially endorsed (yet) by ETS. This site is the work of an individual named Yosun Chang. The solutions, sans user comments and ETS questions, are © 2005 by Yosun Chang except where noted. All rights reserved. / The comments appending particular solutions are due to the respective users. / Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 1997 and © 2001 by ETS.

Bare Basic LaTeX Rosetta Stone