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Resistors and Capacitors in the Same Circuit!!. R-C Circuits. Previous assumptions and how they are changing. So far we have assumed resistance (R), electromotive force or source voltage ( ε ), potential (V), current (I), and power (P) are constant.

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Presentation Transcript

Previous assumptions and

how they are changing.

So far we have assumed resistance (R), electromotive force or source voltage (ε), potential (V), current (I), and power (P) are constant.

When charging or discharging a capacitor I, V, P change with time, we will use lower-case i, v, and p for the instantaneous values of current, voltage, and power.


ε

R - C in Series

Initial conditions

S

An ideal source (r=0)

a

b

c

R

C

At time t=0 we will close the switch. The instantaneous current at this time is i=0 and the instantaneous charge on the capacitor is q = 0 because at this instant there has been no chance for charge to build on the capacitor…yet!


R - C in Series

Charging the Capacitor

  • Remember:

  • Current through a series

  • circuit is the same

  • through all parts.

  • Total voltage input is

  • equal to the sum of the

  • voltage drops.

ε

S

i=0

+q

-q

a

b

c

R

C

At time t=0, Vbc = 0,

So Vab = ε

Therefore, at t=0,

As capacitor C charges,

Vbc , so Vab ,

Which causes I , so…


R - C in Series

Charging the Capacitor

ε

S

After a long time C will be fully charged and i = 0, so Vab = 0 and Vbc = ε

i=0

+q

-q

a

b

c

R

C

At any time t, instantaneous voltages across the resistor and capacitor are…

During charging…

let q be the charge on the capacitor,

and i be the current at any time t

Choose “+” current for a “+” charge on the left capacitor plate…

Vab = iR and Vbc =q/C


R - C in Series

Using Kirchoff’s Law

ε

S

i=0

+q

-q

Then solve for i…

a

b

c

R

C

Remember: At t=0, q=0 so initial current I0 = ε/R.

As q , q/RC , capacitor charge approaches final Qf and i to zero.

When i=0, … so…


R - C in Series

Now for the Calculus!

ε

S

i=0

+q

-q

Rearrange equation to put all terms with q on one side and everything else on the other…

a

b

c

R

C

Now integrate both sides…

Use “u-substitution”

to evaluate the integral…

Eliminate the “ln”…


R - C in Series

Results of the Calculus!

ε

S

i=0

+q

-q

a

At any time t during the charging, the instantaneous charge is…

b

c

R

C

To find the instantaneous current, remember that i = dq/dt


R - C in Series

The Graphs

i

Io

Io/2

q

Io/e

Qf

t

RC

Qf/e

Qf/2

RC


What is so special about time rc
What is so special about time = RC?

When time t=RC (yes…the product of resistance and capacitance)…

Or about 37% of the original current

Or about 63% of the original voltage

The product of R and C is called the time constant of the circuit. The time constant is a measure of how fast the capacitor charges. The symbol for time constant is τ(greek letter “tau”). It is calculated as RC because a that time the exponent of the “e” function becomes “-1” which gives us the equations above.

At t=10τ, I = 0.000045I0 or current is approximately zero.

NOTE:

If the time constant is small, the capacitor will charge quickly.

If the resistance is small, current is small, so the capacitor charges more quickly.

If the time constant is large, it will take more time for the capacitor to charge.


R-C Circuits

Discharging the capacitor

Remove battery to discharge capacitor

S

Assume the capacitor, C, is fully charged, Q0

+Q0

-Q0

a

b

c

R

C

Initial Conditions at time t=0….

At t = 0, close the switch…

and


R-C Circuits

Finding the instantaneous charge during capacitor discharge

S

i

Assume the capacitor, C, is fully charged, Q0

i

+q

-q

a

b

c

R

C

The time-varying current, i, …

Now integrate and solve for q(t)


R-C Circuits

Finding the instantaneous current during capacitor discharge

S

q

The time-varying charge, q, …

i

Qo

i

+q

-q

Qo/2

Qo/e

a

b

c

R

C

t

t

RC

RC

Io/e

Io/2

Io

i


R-C Circuits

Energy & Power

During charging…

The power of the battery, resistor and capacitor are as follows…

So…the total power of the circuit is…

Where ½ the energy is stored in the capacitor and the other ½ is dissipated by the resistor.


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