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
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
I meant 0.012 sec for part A
(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
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?
τ=RC=(4/3)•9•10^-6=
=1.333•9•10^-6=0.000012=
=1.2^10^-5 s.
To answer the three parts of the question:
A) To calculate the time constant for the RC circuit, you need the values of the resistors (R1 and R2) and capacitors (C1 and C2) in the circuit. From the information provided, the capacitors have values of 3 uF and 6 uF, and the resistors have values of 4 k-ohm and 2 k-ohm.
The formula for the time constant in an RC circuit is: TC = (C1+C2) * (R1 * R2) / (R1 + R2)
Substituting the given values, we get:
TC = (3 uF + 6 uF) * (4 k-ohm * 2 k-ohm) / (4 k-ohm + 2 k-ohm)
Simplifying this expression gives:
TC = 12 ms
Note that ms stands for milliseconds, not seconds. Therefore, the time constant for the RC circuit is 12 milliseconds (ms), not 12 seconds.
B) To calculate the time required for the voltage across the capacitor to reach 6 V, you can use the formula for the charge on a charging capacitor, which is given by: q = q0 * (1 - e^(-t/TC))
In this case, the initial charge (q0) is 12 V. We need to find the time (t) when the voltage across the capacitor (q) reaches 6 V.
Rearranging the formula:
6 V = 12 V * (1 - e^(-t/TC))
Divide both sides by 12 V:
1/2 = (1 - e^(-t/TC))
Rearranging further:
e^(-t/TC) = 1 - 1/2
e^(-t/TC) = 1/2
Taking the natural logarithm (ln) of both sides:
-t/TC = ln(1/2)
t/TC = -ln(1/2)
t/TC = ln(2)
t = TC * ln(2)
Substituting the value of the time constant (TC = 12 ms) into the equation, we get:
t = 12 ms * ln(2)
Using a calculator, this is approximately:
t = 8.316 ms
So, the time required for the voltage across the capacitor to reach 6 V is approximately 8.316 milliseconds (ms).
C) To find the net charge stored in the capacitor after 10 nanoseconds (ns), you can use the formula for the charge on a charging capacitor, which is q = q0 * (1 - e^(-t/TC)).
In this case, we are given the time (t) as 10 ns (which is equivalent to 10^-8 seconds), but we don't have the required values of q0 or TC.
To find the net charge stored in the capacitor, you need to know the initial charge (q0) or the voltage across the capacitor at time t = 0, as well as the value of the time constant (TC). Unfortunately, we don't have that information in this scenario, so we cannot calculate the net charge stored in the capacitor after 10 ns.