In the circuit below, at time t = 0, the switch is closed, causing the capacitors to charge. The voltage across the battery is 12 V

Question

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

Answer 1

I meant 0.012 sec for part A

Answer 2

(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

Answer 3

I 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?

Answer 4

τ=RC=(4/3)•9•10^-6=
=1.333•9•10^-6=0.000012=
=1.2^10^-5 s.