# Capacitance from a curve

Only just started ac currents today and this was on my first quiz, I got 8/10 for it but got this question wrong.

## Homework Statement

A 100 ohm resistor is connected in series with a capacitor and a power supply, and the supply voltage is brought up to 200 mV.
The graph shows- for three different capacitors- the decay of the voltage across the capacitor after the voltage source is "shorted out" .

From the graph, estimate the capacitance in the circuit which gave the red decay curve.

V=IR
t=RC
t=l/C

## The Attempt at a Solution

Really had no idea what to do on this question.
V=IR --> R=V/I --> R=40/100 --> 0.4ohms
c=t/R --> c=0.0001/0.4 --> 2.5E-4 --> 25MicroFarads.

I know this is completely wrong but I got in trouble for not showing working last time.

Any help would be appreciated.

#### Attachments

• TimeConstants.jpeg
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Simon Bridge
Homework Helper
$\tau = RC$ is called the mean-life of the capacitor/resistor circuit.

R=100Ω
τ= from the graph
C = what you want to find

The vurve is:
$V(t)=V_0e^{-t/\tau}$

$ln|V(t)| = ln|V_0| - t/\tau$
... so a plot of $ln|V|$ against time will have a slope $1/\tau$

OR: the mean time is how long the capacitor would have discharged in if it was not a curve ... estimate the initial slope of the graph and draw the line.
OR: use the half-life to derive the mean-life (read the half-life off the graph)
OR: read the time the capacitor takes to decay to 2/3 maximum (off the graph) - that's about the mean life.

http://en.wikipedia.org/wiki/Capacitor#DC_circuits

Last edited:
Awesome thanks for that. Didn't know what to search cause I didn't know what it was called but this helps a lot.
Thank you

Simon Bridge