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HelloCthulhu

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- Homework Statement
- A 6V battery, 0.64mm diameter/14in length single/solid copper wire alligator clips how long and an 18cm disc parallel plate air capacitor are used in a circuit. What is the capacitance? How long will it take to charge the capacitor to 100% of its capacity?

- Relevant Equations
- A=πr2

capacitance

$$C=K\varepsilon_0\frac{A}{d}$$

Area of 18cm disc electrode = 0.0254469m2

d of separation = 0.01m

dielectric constant = 1

ε0 = 8.85×10−12 F/m−1

voltage respect to time

V(t) = VB(1-e[SUP]-t/RC[/SUP])

V(t) = voltage in respect to time

VB = battery voltage

-e (numerical constant) = 2.71828

-t = voltage in respect to time in sec

R = resistance of the wire

C = capacitance

RC = time constant = τ

resistance for the 14in long/ 0.64mm diameter wire found here:

https://www.cirris.com/learning-center/calculators/133-wire-resistance-calculator-table

$$C=K\varepsilon_0\frac{A}{d}=C=\frac{(8.85\times10^{-12})\times1\times 0.0254469}{0.01}=0.000022531F=22.531μF$$

6V*(1-2.71828

time constants

6*(1-2.71828

6*(1-2.71828

6*(1-2.71828

6*(0.982) = 5.89 sec

6*(0.993) = 5.95 sec

capacitance = 22.531μF

100% charge time = 6 sec

6V*(1-2.71828

^{-t/0.019Ω*0.0000225307uF})time constants

6*(1-2.71828

^{-1}) = 6*(0.632) = 3.79 sec6*(1-2.71828

^{-2}) = 6*(0.865) = 5.18 sec6*(1-2.71828

^{-3}) = 6*(0.950) = 5.70 sec6*(0.982) = 5.89 sec

6*(0.993) = 5.95 sec

capacitance = 22.531μF

100% charge time = 6 sec