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October 2, 2014

October 2, 2014

Posted by **Dan** on Friday, July 27, 2012 at 12:28pm.

The circuit:

flickr.

com/photos/

68751220@N03

/6848519773/

A)Calculate the time constant for the RC circuit

B)Calculate the time required for the voltage across the capacitor to reach 6 V

C)What will be the net charge stored in the capacitor after 10 nanoseconds?

What I did:

A) Time constant= [(C1+C2)*(R1R2)]/(R1+R2)

= [(3uF+6uF)*(4k-ohm*2k-ohm)]/(4k-ohm+2k-ohm)

= 12 ms= 12 sec?

TC= time constant

B) q= q0[1-e^(-t/TC)]

t= 8.316 sec?

C) I don't know what to do and can't find an equation that I think would work

- physics -
**Dan**, Friday, July 27, 2012 at 12:32pmI meant 0.012 sec for part A

- physics -
**Elena**, Friday, July 27, 2012 at 3:35pm(a) 1/R=1/R1+1/R2=

=1/4+1/2=3/4 =>

R=4/3 Ohm.

C=C1+C2 =3+6=9 μF

τ=RC=(4/3)•9•10^-6=1.2^10^-5 s.

(b)

V(C)=V(o) •e^(- t/RC)=>

V(C)/V(o)= e^(- t/RC)=>

6/12=1/2 = e^(- t/RC)=>

ln(1/2)= -t/1.2^10^-5,

t=8.32•10^-6 s.

Q(max) = C•U=

=9•10^-6•12=108•10^-6 Coulombs=

=108μC

(a) 1/R=1/R1+1/R2=

=1/4+1/2=3/4 =>

R=4/3 Ohm.

C=C1+C2 =3+6=9 μF

τ=RC=(4/3)•9•10^-6=1.2^10^-5 s.

(b)

V(C)=V(o) •e^(- t/RC)=>

V(C)/V(o)= e^(- t/RC)=>

6/12=1/2 = e^(- t/RC)=>

ln(1/2)= -t/1.2^10^-5,

t=8.32•10^-6 s.

Q(max) = C•U=

=9•10^-6•12=108•10^-6 Coulombs=

=108μC

(a) 1/R=1/R1+1/R2=

=1/4+1/2=3/4 =>

R=4/3 Ohm.

C=C1+C2 =3+6=9 μF

τ=RC=(4/3)•9•10^-6=1.2^10^-5 s.

(b)

V(C)=V(o) •e^(- t/RC)=>

V(C)/V(o)= e^(- t/RC)=>

6/12=1/2 = e^(- t/RC)=>

ln(1/2)= -t/1.2^10^-5,

t=8.32•10^-6 s.

Q(max) = C•U=

=9•10^-6•12=108•10^-6 Coulombs=

=108μC

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capdis.html

- physics -
**Dan**, Friday, July 27, 2012 at 4:47pmI plugged the #'s into that site you supplied, but I still get my answer of 0.012 seconds, not the 1.2E-5 seconds you put?

- physics -
**Elena**, Saturday, July 28, 2012 at 2:49pmτ=RC=(4/3)•9•10^-6=

=1.333•9•10^-6=0.000012=

=1.2^10^-5 s.

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