In the circuit shown R1=5.0 k, R2=10 k, and V = 12.0 V. The capacitor is initially uncharged. After the switch has been closed for 1.30 μs the potential difference across the capacitor is 5.0 V. (a) Calculate the time constant of the circuit. (b) Find the value of C. (c) Sketch the current, charge, and potential difference across the capacitor as a function of time.

2.41x10-6 s

0.16 nF

Someday, when I have time, I will make this into a nice diagram!

In the circuit shown R1=5.0 k, R2=10 k, and V = 12.0 V. The capacitor is initially uncharged. After the switch has been closed for 1.30 μs the potential difference across the capacitor is 5.0 V. (a) Calculate the time constant of the circuit..We can’t use = RC because we don’t know C.

We are told the capacitor is charging, and given information about potential difference, so we derive an equation for V(t).

t-RC

finalq t =Q 1-e

t-RC

0C V t =C V 1-e

t-RC

0V t =V 1-e

In the circuit shown R1=5.0 k, R2=10 k, and V = 12.0 V. The capacitor is initially uncharged. After the switch has been closed for 1.30 μs the potential difference across the capacitor is 5.0 V. (a) Calculate the time constant of the circuit.

T-RC

0V T =V 1-e

Let T = 1.30x10-6 s (for simplicity of writing equations).

T-RC

0

V T= 1-e

V

T-RC

0

V Te = 1-

V

In the circuit shown R1=5.0 k, R2=10 k, and V = 12.0 V. The capacitor is initially uncharged. After the switch has been closed for 1.30 μs the potential difference across the capacitor is 5.0 V. (a) Calculate the time constant of the circuit..

0

V TT- =ln 1-RC V

0

T- =RC=

V Tln 1-

V

-6-61.3 10

=- =2.41 10 s5

ln 1-12

Take natural log of both sides of last equation on previous slide.

In the circuit shown R1=5.0 k, R2=10 k, and V = 12.0 V. The capacitor is initially uncharged. After the switch has been closed for 1.30 μs the potential difference across the capacitor is 5.0 V. (b) Find the value of C.

-6=2.41 10 s = RC

-6

31 2

2.41 10C= = =

R R +R 15 10

-9C=0.161 10 F=0.161 nF

In the circuit shown R1=5.0 k, R2=10 k, and V = 12.0 V. The capacitor is initially uncharged. (c) Sketch the current, charge, and potential difference across the capacitor as a function of time.

Charging capacitor (you aren’t required to derive these equations):

t-RC

finalq t =Q 1-e

ε t-RCI t = e

R

ε

t-RCV t = 1-e

Charging Capacitor

0

0.002

0.004

0.006

0.008

0.01

0 0.2 0.4 0.6 0.8 1

t (s)

q (

C)

Charging Capacitor

0

0.01

0.02

0.03

0.04

0.05

0 0.2 0.4 0.6 0.8 1

t (s)I (

A)

Charging Capacitor

0

0.002

0.004

0.006

0.008

0.01

0 0.2 0.4 0.6 0.8 1

t (s)

q (

C)

time

time

time

charg

ecu

rrent

volt

age

In the circuit shown R1=5.0 k, R2=10 k, and V = 12.0 V. The capacitor is allowed to fully charge and the battery removed from the circuit. How long does it take for the voltage across the capacitor to drop to ¼ of its fully-charged value?Capacitor has been charged to Qfinal = CV0 and is now discharging.

t-RC

0q(t)=Q e

t-RC

0CV(t)=CV e

t-RC

0V(t)=V e

t-RC

0

V(t)= e

V

In the circuit shown R1=5.0 k, R2=10 k, and V = 12.0 V. The capacitor is allowed to fully charge and the battery removed from the circuit. How long does it take for the voltage across the capacitor to drop to ¼ of its fully-charged value?

0

V(t) tln = -

V RC

0 0

V(t) V(t)t=-RC ln = - ln

V V

0-6

0

V14t=- ln = - 2.41 10 ln

V 4

Take natural log of both sides of last equation on previous slide.

-6t=3.34 10 = 3.34 s