lect_9_f03

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Overview

Discuss Test 1

Review

Kirchoff's Rules for Circuits

Resistors in Series & Parallel

RC Circuits

Text Reference: Chapter 27, 28.1-4

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UNIT: Ampere = A = C/s

If there is a potential difference between two

points then, if there is a conducting path, freecharge will flow from the higher to the lowerpotential. The amount of charge which flowsper unit time is defined as the current I, i.e.

current is charge flow per unit time.

Current- a Definition

dqI

dt

i.e. no longerelectrostatics.

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Resistors:

Purpose is to limit current drawn in a circuit. Resistors are

basically bad conductors. Actually all conductors have some

resistance to the flow of charge.

Devices

Resistance

Resistance is defined to be theratio of the applied voltage to thecurrent passing through.

V

I I

R

UNIT: OHM =

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Resistivity

LA

E

I

Property of bulk matter

related to resistance :

The flow of charge is easier with a larger

cross sectional area, it is harder if L is

large.

eg, for a copper wire, ~ 10-8 -m, 1mm radius, 1 m long, then R .01

The constant of proportionality is called the

resistivity r

The resistivity depends on the details of the atomic structure whichmakes up the resistor (see chapter 27 in text)

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Ohm's Law

Demo:

Vary applied voltage V.

Measure current I

Does ratio (V/I)remain

constant??

V

I

slope = R

V

II

R

Only true for ideal resisitor!

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Lecture 11, CQ 1 Two cylindrical resistors, R1 and R2, are made of

identical material. R2has twice the length ofR1 but half

the radius ofR1.

These resistors are then connected to a battery Vas shown:

VI1

I2

What is the relation between I1, the currentflowing in R1, and I2, the current flowing in R2?

(a) I1 < I2 (b) I1 = I2 (c) I1 > I2

The resistivity of both resistors is the same . Therefore the resistances are related as:

RL

A

L

A

L

AR2

2

2

1

1

1

11

2

48 8 r r r

( / )

The resistors have the same voltage across them; therefore

I

V

R

V

RI2

2 1

18

1

8

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Kirchoff's First Rule"Loop Rule" orKirchoffs Voltage Law (KVL)

"When any closed circuit loop is traversed, the algebraicsum of the changes in potential must equal zero."

This is just a restatement of what you already know: that thepotential difference is independent of path!

Vnloop

0KVL:

We will follow the convention that voltage drops enter with a + sign andvoltage gains enter with a - sign in this equation.

RULES OF THE ROAD:

R1 R2IMoveclockwisearound

circuit:

R1 R2I

IR1 IR2 0

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Kirchoff's Second Rule"Junction Rule" orKirchoffs Current Law (KCL)

In deriving the formula for the equivalent resistance of2 resistors in parallel, we applied Kirchoff's SecondRule (the junction rule).

"At any junction point in a circuit where the current

can divide (also called a node), the sum of the currentsinto the node must equal the sum of the currents out ofthe node."

This is just a statement of the conservation of charge at anygiven node.

I Iin out

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Resistors

in Series

1ba IRVV - 2cb IRVV -

)RR(IVV 21ca +-

The Voltage drops:

Whenever devices are in SERIES, thecurrent is the same through both !

This reduces the circuit to:

a

c

Reffective

)RR(R 21effective +Hence:

a

b

c

R1

R2

I

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Resistors in Parallel What to do?

But current through R1 is not I !

Call it I1. Similarly, R

2I

2.

How is I related to I1 & I2 ??Current is conserved!

a

d

a

d

I

I

I

I

R1 R2

I1 I2

R

V

V

IRV

KVL I R V1 1 0- I R V2 2 0-

I I I +1 2

V

R

V

R

V

R

+

1 2

1 1 1

1 2R R R

+

Very generally, devices in parallel havethe same voltage drop

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Loop Demo

a

d

b

ec

f

R1

I

R2 R3

R4

I

IR IR IR IR1 2 2 3 4 1 0+ + + + - e e

IR R R R

-

+ + +

e e1 2

1 2 3 4

Vnloop

0KVL:

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Lecture 11, CQ 2 Consider the circuit shown.

The switch is initially open and the currentflowing through the bottom resistor is I

0

.

Just after the switch is closed, the currentflowing through the bottom resistor is I1.

What is the relation between I0 and I1?

(a)I1 < I0 (b)I1 = I0 (c)I1 > I0

R

12 V

12 V

R

12 V

I

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Lecture 11, CQ 2 Consider the circuit shown.


0

.

After the switch is closed, the currentflowing through the bottom resistor is I1.


(a) I1 < I0 (b) I1 = I0 (c) I1 > I0

Write a loop law for original loop:

R

12 V

12 V

R

12 V

Ia

b

-12V + I1R = 0

I1 = 12V/R

Write a loop law for the new loop:

-12V -12V + I0R + I0R = 0

I0 = 12V/R

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or, Lecture 11, CQ 2 Consider the circuit shown.


0

.

After the switch is closed, the currentflowing through the bottom resistor is I1.


(a) I1 < I0 (b) I1 = I0 (c) I1 > I0

The key here is to determine the potential (Va-Vb) before the switch isclosed.

From symmetry, (Va-Vb) = +12V.

Therefore, when the switch is closed, NO additional current will flow!

Therefore, the current after the switch is closed is equal to the currentafter the switch is closed.

R

12 V

12 V

R

12 V

Ia

b

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Junction Demo

Outside loop:

Top loop:

Junction:

I1

R

R

R

I2

I3

I R I R1 3 3 1 0+ + - e e

I R I R1 2 2 1 0+ + - e e

I I I1 2 3 +

IR

11 2 32

3

- -e e e

I

R2

1 3 22

3

+ -e e e

IR

31 2 32

3

+ -e e e

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a

bR

C

II

t

q

RC 2RC

0

C

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Overview of Lecture

RC Circuit: Charging of capacitor through aResistor

RC Circuit: Discharging of capacitor through aResistor

Text Reference: Chapter 27.4, 28.2, 28.6

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RC Circuits

R

C

II

Add a Capacitor to a simplecircuit with a resistor

Recall voltage drop on C?

Upon closing circuit Loop rule gives:

Recall that Substituting:

Differential Equation for q!

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Compare with simple resistance circuit Simple resistance circuit:

Main Feature: Currents are attained instantaneously and

do not vary with time!!

Circuit with a capacitor:

KVL yields a differential equation with a term

proportional to q and a term proportional toI = dq/dt.

Physically, whats happening is that the final charge cannot be

placedon a capacitor instantly.

Initially, the voltage drop across an uncharged capacitor = 0

because the charge on it is zero !

As current starts to flow, charge builds up on the capacitor,

the voltage drop is proportional to this charge and increases;

it then becomes more difficult to add morecharge so the

current slows

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The differential equation is easy to solve if we re-write in the form:

dq C q

dt RC

e -

dq dt

C q RC e

-or equivalently,

Integrating both sides we obtain,

Exponentiating both sides we obtain,

If there is no initial charge on C then:

Thus,

Integration

constant

We have to find q such that is satisfied.

Determinesintegration constant

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Charging the Capacitor

Charge on C

Max = C

63% Max at t=RCt

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

0

c

Idq

dt R et RC

-e /

RC 2RC

I

0 t

RCurrent

Max = e/R

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Lecture 12, CQ 1

At t=0 the switch is thrown from position b toposition a in the circuit shown: The capacitor

is initially uncharged.

What is the value of the current I0+ justafter the switch is thrown?

(a) I0+ = 0 (b) I0+ = /2R (c) I0+ = 2 /R

(a) I

= 0 (b) I

= /2R (c) I

> 2 /R

1B What is the value of the current I

after a very long time?

1A

a

b

R

C

II

R

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Lecture 12, CQ 1

At t=0 the switch is thrown from position b toposition a in the circuit shown: The capacitor is

initially uncharged.


(a) I0+ = 0 (b) I0+ = /2R (c) I0+ = 2 /R

1A

a

b

R

C

II

R

Just after the switch is thrown, the capacitor still has no charge,therefore the voltage drop across the capacitor = 0! Applying KVL to the loop at t=0+, IR + 0 + IR - = 0 I = /2R

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Lecture 12, CQ 1

At t=0 the switch is thrown from position b toposition a in the circuit shown: The capacitor is

initially uncharged.


(a) I0+ = 0 (b) I0+ = /2R (c) I0+ = 2 /R

1A

1B

(a) I

= 0 (b) I

= /2R (c) I

> 2 /R

What is the value of the current I

after a very long time?

a

b

R

C

II

R

The key here is to realize that as the current continues to flow, thecharge on the capacitor continues to grow. As the charge on the capacitor continues to grow, the voltage across

the capacitor will increase. The voltage across the capacitor is limited to ; the current goes to 0.

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Lecture 12, CQ 2 At t=0 the switch is thrown from position b to

position a in the circuit shown: The capacitoris initially uncharged.

At timet=t1=t, the chargeQ1on the capacitoris (1-1/e) of its asymptotic charge Qf=Ce.

What is the relation betweenQ1andQ2, thecharge on the capacitor at timet=t2=2t?

(a) Q2 < 2 Q1 (b) Q2 = 2 Q1 (c) Q2 > 2 Q1

a

b

R

C

II

R

Hint: think graphically!

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(a) Q2 < 2 Q1 (b) Q2 = 2 Q1 (c) Q2 > 2 Q1

a

b

R

C

II

R

From the graph at the right, it is clear thatthe charge increase is not as fast as linear. In fact the rate of increase is justproportional to the current (dq/dt) whichdecreases with time. Therefore, Q2< 2Q1.

The charge q on the capacitor increases with time as:

So the question is: how does this charge increase differ from a linearincrease?

At t=0 the switch is thrown from position b to position ain the circuit shown: The capacitor is initially uncharged.

At timet=t1=t, the charge Q1 on the capacitor

is (1-1/e) of its asymptotic charge Qf=Ce.

What is the relation between Q1 and Q2 , the

charge on the capacitor at time t=t2=2t?

q 1Q

2Q

12Q

t 2t

RC Ci it

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RC Circuits(Time-varying currents)

Discharge capacitor:

C initially charged withQ=Ce

Connect switch to b at t=0.

Calculate current and chargeas function of time.

Convert to differential equation for q:

C

a

b+ +

- -

R

I I

Loop theorem

RC Ci i

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Solution:

Check that it is a solution:

!

Note that this guessincorporates theboundary conditions:

C

a

b+ +

- -

R

I I Discharge capacitor:

RC Ci it

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Current is found fromdifferentiation:

Conclusion:

Capacitor discharges exponentiallywith time constant t = RC

Current decays from initial max value(= -e/R) with same time constant

Discharge capacitor:

C

a

b+

- -

R+

I I

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Discharging Capacitor

Charge on C

Max = C

37% Max at t=RC

q = C ee -t/RC

t

q

0

C

I

dq

dt R e

t RC

-

-e /

RC 2RC

0

- R

I

t

Current

Max = -e/R

37% Max at t=RC

L 12 CQ 3

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Lecture 12, CQ 3 At t=0 the switch is connected to position a

in the circuit shown: The capacitor isinitially uncharged.

At t = t0, the switch is thrownfrom position a to position b.

Which of the following graphsbest represents the timedependence of the charge on C?

C

a b

R 2R

(a) (b) (c)

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At t=0 the switch is connected to position ain the circuit shown: The capacitor isinitially uncharged.

At t = t0, the switch is thrownfrom position a to position b.

Which of the following graphsbest represents the timedependence of the charge on C?

(a) (b) (c)

C

a b

R 2R

t

q

0

C

t0 t

q

0

C

t0

q

0

C

tt0

For 0 < t < t0, the capacitor is charging with time constant = RC For t > t0, the capacitor is discharging with time constant = 2RC

(a) has equal charging and discharging time constants (b) has a larger discharging than a charging

(c) has a smaller discharging than a charging

Ch i i h i

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Charging DischargingRC

t

2RC

0

R

I

t

q

0

CRC

t

q

2RC

0

C

I

0

t

R

q = C ee -t/RC

Idq

dt Re

t RC - -e /

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

Idq

dt Re t RC -

e /

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