rc circuit analysis

8/2/2019 RC Circuit Analysis

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9/29/04 UWisconsin, Physics 104, Fall 2003 1

RC Circuit Analysis

Circuit with resistance and capacitance

Charging and discharging


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Problem-Solving Strategy

Kirchhoffs Rules Draw the circuit diagram and assign labels

and symbols to all known and unknownquantities. Assign directions to the currents.

Apply the junction rule to any junction in thecircuit

Apply the loop rule to as many loops as areneeded to solve for the unknowns

Solve the equations simultaneously for theunknown quantities.


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Example

R1

R2 R3

I1 I3

I2

+

-

+

+

+

Loop 1: 1+I1R1 I2R2 = 0

1. Label all currents (Choose any direction)

2. Label +/- for all elements (Current goes + - )3. Choose loop and direction (Your choice!)

4. Write down voltage drops (First sign you hit is sign to use!)

-

-

-

Loop 2:

1

5. Write down node equation

Node: I1 +I2 = I3

2

3 Equations, 3 unknowns the rest is algebra!

In the circuit below your are given E1, E2, R1, R2 and R3.

Find I1, I2 and I3.

Loop 1

Loop 2

+-

+I2R2 +I3R3 +2= 0


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What is the resistance

of a capacitor? Answer depends on time!

Uncharged

Initially voltage drop between the plates is zero

The charges flow in and build up on the plates

Resistance starts off from zero, R0 = V / I = 0 / I = 0

Fully charged

Charge on the plates builds to a maximum No more current flow

Voltage reaches a maximum

Resistance approaches infinity, R

= V / I = V / 0 =


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Combine R+C Circuits

Gives time dependence

Current is not constant, I(t)

Charge on the capacitor is not constant, q(t)

Used for timing Pacemaker

Intermittent windshield wipers

Models nervous system Olympic Trivia: How soon after starting gun

can you run w/o getting a False Start?

0.1 seconds


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RC Circuits: Charging

Kirchoffs Voltage Rule:

-V+I(t)R+q(t)/C = 0

Just after S1 closing: q=q0= 0-V + I0R=0 I0 = V /R

Long time after: I=0

-V + q/C =0 q

=V C

Intermediate (more complex)

q(t) = q(1-e-t/RC)

I(t) = I0e-t/RC

C

V

R

S1

S2

+

+

+

I-

-

-

The switches are originally open and the capacitoris uncharged. Then switch S1 is closed.

t

q

RC 2RC

0

q

Q=CV


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Charging in an RC Circuit

The charge on the capacitor

varies with time

q = Q(1 e -t/RC)

The time constant, =RC

The time constant

represents the time required

for the charge to increase

from zero to 63.2% of its

maximum


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Discharging in an RC Circuit

When a chargedcapacitor is placed inthe circuit, it can bedischarged q = Qe -t/RC

The charge decreasesexponentially

At t = = RC, thecharge decreases to0.368 Qmax In other words, in one

time constant, thecapacitor loses 63.2% ofits initial charge


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Preflight: Short Term

2R

CV R

S2

Both switches are initially open, and the capacitor isuncharged. What is the current through the batteryjust after switch S1 is closed?

1) Ib = 0

2) Ib = V/(3R)

3) Ib = V/(2R)

4) Ib = V/R

S1KVL: -V + I(2R) + q/C = 0

q = 0

I = V/(2R)

Ib

+

-

+

+

-

-9%42%

22%

27%

0% 10% 20% 30% 40% 50%

Pretty Sure

Not Quite Sure

Just Guessing


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Preflight: Long Time

2R

CV R

S2

Both switches are initially open, and thecapacitor is uncharged. What is the current

through the battery after switch 1 has been

closed a long time?

1) Ib = 02) Ib = V/(3R)

3) Ib = V/(2R)

4) Ib = V/R

S1

Ib

+

-

+

+

-

-

Long time current through capacitor is zero

Ib=0 because the battery and capacitor are in series.

KVL: -V + 0 + q

/C = 0 q

= VC

11%

38%

22%

29%

0% 10% 20% 30% 40%

Pretty Sure

Not Quite Sure

Just Guessing


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Preflight: Discharge

2R

CV R

S2

After being closed for a long time, the switch 1 isopened, while switch 2 is closed simultaneously.

What is the current through the resistor, R?1) IR = 0

2) IR = V/(3R)3) IR = V/(2R)

4) IR = V/R

S1

Ib

+

-

+

+

-

-

Long time capacitor is fully charged to V

Switch 1 open, switch 2 closed means C discharges

through R.

KVR: V + IRR = 0

q= VC

23%

18%

26%

34%

0% 10% 20% 30% 40%

IR

Pretty Sure

Not Quite Sure

Just Guessing


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Charging: Intermediate Times

V

2R

C R

S2

S1

Ib

+

-

+

+

-

-

R= 10 , V= 50 Volts, C= 100F

Calculate the charge on the capacitor 310-3 seconds

after switch 1 is closed.

q(t) = q(1-e-t/RC)

= q(1-e-310

-3/(2010010

-6)))

= q

(0.78)

Recall q= VC

= (50)(100x10-6) (0.78)

= 3.9 x10-3 Coulombs


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Electrical Safety

Electric shock can result in fatal burns

Electric shock can cause the muscles of vital organs

(such as the heart) to malfunction

The degree of damage depends on

the magnitude of the current the length of time it acts

the part of the body through which it passes


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Effects of Various Currents

5 mA or less

can cause a sensation of shock

generally little or no damage

10 mA

hand muscles contract may be unable to let go a of live wire

100 mA

if passes through the body for 1 second or less, can be fatal


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Ground Wire

Electrical equipment

manufacturers use

electrical cords that

have a third wire,

called a ground

Prevents shocks


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Ground Fault Interrupts (GFI)

Special power outlets

Used in hazardous areas

Designed to protect people from

electrical shock Senses currents (of about 5 mA or

greater) leaking to ground

Shuts off the current when above this

level


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Electrical Signals in Neurons

Specialized cells in the body, called neurons,form a complex network that receives,processes, and transmits information fromone part of the body to another

Three classes of neurons Sensory neurons

Receive stimuli from sensory organs that monitor theexternal and internal environment of the body

Motor neurons

Carry messages that control the muscle cells

Interneurons Transmit information from one neuron to another


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Diagram of a Neuron

rc circuit analysis

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