# Capacitance from a curve

1. Nov 1, 2011

### KingBigness

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

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
V=IR
t=RC
t=l/C

3. 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.

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2. Nov 1, 2011

### Simon Bridge

$\tau = RC$ 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: Nov 1, 2011
3. Nov 1, 2011

### KingBigness

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

4. Nov 1, 2011

### Simon Bridge

Searching "capacitor" would have given you wikipedia in the top ten. "capacitor physics" would be better and "discharging capacitor" even better.

Reading wikipedia would have given you better search terms like "exponential decay".