Direct-Current Circuits · We now deal with currents through the resistors ... we use the fact that both resistors have the ... After the switch is closed the net resistance is given

Fall 2008Physics 231 Lecture 6-1

Direct-Current Circuits


Resistors in Series and ParallelAs with capacitors, resistors are often in series and parallel configurations in circuits

Series

Parallel

The question then is what is the equivalent resistance


Resistors in Series

Since these resistors are in series, we have the same current inall three resistors

IIII === 321

We also have that the sum of the potential differences across the three resistors must be the same as the potential differencebetween points a and b

ybxyaxab VVVV ++=


Resistors in Series321 ;; RIVRIVRIV ybxyax ===Then using

( )321 RRRIVab ++=We have that

Now the equivalent resistor, R, will also have the same potential difference across it as Vab, and it will also have the same current I RIVab =

Equating these last two results, we then have that

∑=++=i

iRRRRR 321

The equivalent resistance for a sequence of resistors in series is just the sum of the individual resistances


Resistors in Parallel

Here we have that the voltage across each resistor has to be the same (work done in going from a to b is independent of the path, independent of which resistor you go through)

abVVVV === 321


Resistors in ParallelWe now deal with currents through the resistors

At point a the current splits up into three distinct currents

We have that the sum of theses three currents must add to the value coming into this point

321 IIII ++=

33

22

11 ;;

RVI

RVI

RVI ababab ===We also have that

The equivalent resistor, R, will have also have the current I going through it


Resistors in Parallel

RVI ab=Using

and combining with the previous equations, we then have

321 RV

RV

RV

RV abababab ++=

or

∑=++=i iRRRRR

11111

321

The inverse of the effective resistance is given by the sum of the inverses of the individual resistances


Solving Resistor NetworksMake a drawing of the resistor networkDetermine whether the resistors are in series or

parallel or some combinationDetermine what is being asked

Equivalent resistancePotential difference across a particular

resistance Current through a particular resistor


Solving Resistor NetworksSolve simplest parts of the network firstThen redraw network using the just calculated

effective resistanceRepeat calculating effective resistances until only

one effective resistance is left


Solving Resistor Networks

Given the following circuit

What is the equivalent resistance and what is the current through each resistor

We see that we have two resistors in parallel with each other and the effective resistance of these two is in series with the remaining resistor


Step 1: Combine the two resistors that are in parallel

yielding

Step 2: Combine the two resistors that are in series

ΩΩΩ 624 =+=effR yielding

Solving Resistor Networks

ΩΩΩΩ

2;21

31

611

==+= effeff

RR


Solving Resistor NetworksCurrent through this effective resistor is given by

AmpsRVIeff

36

18===

The current through the resistors in the intermediate circuit of Step 1 is also 3 Amps with the voltage drop across the individual resistors being given by

VoltsV

VoltsV

623

;1243

2

4

=⋅=

=⋅=

Ω

Ω


Solving Resistor NetworksTo find the current through the resistors of the parallel section of the initial circuit, we use the fact that both resistors have the same voltage drop – 6 Volts

AmpsVoltsI

AmpVoltsI

23

6

;16

6

3

6

=Ω

=

=Ω

=

Ω

Ω


Consistency CheckThere is a check that can be made to see if the answers for the currents make sense:

The power supplied by the battery should equal the totalpower being dissipated by the resistors

The power being supplied by the battery is given bywhere I is the total current

VIP =

WattsVIP 54183 =⋅==

The power being dissipated by each of the resistors is given by RIP 2=

WattsPWattsP

WattsPWattsP

Total 54;661

;1232;36432

6

23

24

==⋅=

=⋅==⋅=

Ω

ΩΩ


Two identical light bulbs are represented by the resistors R2 and R3 (R2 = R3 ). The switch S is initially open.

If switch S is closed, what happens to the brightness of the bulb R2?a) It increases b) It decreases c) It doesn’t change

Example 1

RVP

2=The power dissipated in R2 is given by

When the switch is closed neither V nor R changes

So the brightness does not change


Example 2Two identical light bulbs are represented by the resistors R2 and R3 (R2 = R3 ). The switch S is initially open.

What happens to the current I, after the switch is closed ?

a) Iafter = 1/2 Ibefore b) Iafter = Ibefore c) Iafter = 2 Ibefore

Initially the current is given by 2RIbefore ε=

After the switch is closed the net resistance is given by

32232

2111 RRRRRRnet

==+= since

beforenet

after IRRI 222

=⎟⎠⎞⎜

⎝⎛== εε

22RRnet =

The new current is then


Kirchoff’s RulesNot all circuits are reducible

There is no way to reduce the four resistors to one effective resistance or to combine the three voltage sources to one voltage source


Kirchoff’s RulesFirst some terminology

A junction, also called anode or branch point, is is a point where three or more conductors meet

A loop is any closed conducting path


Kirchoff’s RulesKirchoff’s Rules are basically two statements

1. The algebraic sum of the currents into any junction is zero ∑ = 0I

A sign convention:A current heading towards a junction, is considered to bepositive,

0321 =−+ III

A current heading away from a junction, is considered to be negative

Be aware that all the junction equations for a circuit may not be independent of each other


Kirchoff’s Rules2. The algebraic sum of the potential differences in any loop including those associated with emfs and those of resistive elements must equal zero

∑ = 0VProcedures to apply this rule:

Pick a direction for the current in each branchIf you picked the wrong direction, the current will come out negative


Kirchoff’s RulesPick a direction for traversing a loop – this direction must be the same for all loops

Note that there is a third loop along the outside branches

As with the junction equations not all the loop equations will be independent of each other.


Kirchoff’s RulesStarting at any point on the loop add the emfs and IR terms

An IR term is negative if we traverse it in the same sense as the current that is going through it, otherwise it is positive

An emf is considered to be positive if we go in the direction - to +, otherwise it is negative

Need to have as many independent equations as there are unknowns


Kirchoff’s Rules

For loop I we have 031432111 =−+−−− εεRIRIRI

For loop II we have 0324332 =+−+− εεRIRI

Junction equation at a gives us 0321 =−− III

We now have three equations for the three unknown currents


Kirchoff’s RulesAssume that the batteries are: ε1 = 19 V; ε2 = 6 V; ε3 = 2 V

and the resistors are: R1 = 6Ω; R2 = 4Ω; R3 = 4Ω; R4 = 1Ω

you should end up with: I1 = 1.5 A; I2 = -0.5 A; I3 = 2.0 A

The minus sign on I2 indicates that the current is in fact in the opposite direction to that shown on the diagram

Complete details can be found here


RC CircuitsUp until now we have assumed that the emfs and resistances are constant in time so that all potentials, currents, and powers are constant in timeHowever, whenever we have a capacitor that is being charged or discharged this is not the case

Now consider a circuit that consists of a source of emf, a resistor and a capacitor but with an open switch

With the switch open the current in the circuit is zero and zero charge accumulates on the capacitor


RC CircuitsInitially the full potential will be across the resistor as the potential across the capacitor is zero since q is zero

Now close the switch

Initially the full potential is across the resistorThe initial current in the circuit is then given by RI /0 ε=

As the current flows a charge will accumulate on the capacitorAt some time t, the current in the circuit will be I and the charge on the capacitor will be q


RC CircuitsAccording to Kirchoff’s 2nd rule we have

Using a counterclockwise loop

0

0

=−−

=−−

CqIR

VV capacitorresistor

ε

ε

RCq

RI −=

εSolving for the current

As time increases, the charge on the capacitor increases, therefore the current in the circuit decreases

Current will flow until the capacitor has a charge on it given by εCQ =


RC CircuitstdqdI =We remember that

( )εε CqCRCR

qRtd

qd−−=−=

1So we then have

RCtd

Cqqd

−=− ε

Rearranging we have

∫∫ −=−

tq

RCtd

Cqqd

00 εSetting up the integration we have

RCCCq 1ln −=⎟

⎠⎞

⎜⎝⎛

−−

εεThe resultant integration yields


RC CircuitsWe exponentiate both sides of this last equation and rearrange to obtain

( ) ( )RCtf

RCt eQeCq // 11 −− −=−= ε

where Qf is the final charge on the capacitor given by Cε

We see that the charge on the capacitor increases exponentially

The constant RC is known as the time constant of the circuit


Example 3

What will be the voltage across the capacitor a long time after the switch is closed?

(a) VC = 0 (b) VC = ε R2/(R1+ R2) (c) VC = ε

I1I3

I2

ε R2C

R1

At t = 0 the switch is closed in the circuit shown. The initially uncharged capacitor then begins to charge.

After a long time the capacitor is completely charged, so no current flows through itThe circuit is then equivalent to a battery with two resistors in seriesThe voltage across the capacitor equals the voltage across R2(since C and R2 are in parallel)


RC CircuitsThe current in the circuit is given by

RCtRCt eIeRtd

qdI /0

/ −− ===ε

and looks like

Note that is also how the voltage across the resistor behaves

IRVresistor =


RC Circuits – Charging SummaryFor the simple RC circuit we have the following for the voltage drops across the capacitor and the resistor


RC Circuits

We now start from a situation where we have a charged capacitor in series with a resistor and an open switch

The capacitor will now act as a source of emf, but one whose value is not constant with time


RC CircuitsWe now close the switch and a current will flow

Kirchoff’s 2nd rule gives us

0=−−CqIR

RCq

tdqd

−=Rearranging we have

To find q as a function of time we integrate the above equation


RC Circuits

RCt

Qqtd

RCqqd tq

Q

−=⎟⎟⎠

⎞⎜⎜⎝

⎛⇒−= ∫∫

00

ln'1''

0

Exponentiation of both sides of the equation on the right yieldsRCteQq /

0−=

We see that the charge on the capacitor decreases exponentially


RC CircuitsThe current in the circuit is obtained by taking the derivative of the charge equation

RCt

eRCQ

dtdqI

−−== 0

The quantity Q0 / C is just the initial voltage, Vo , across the capacitor

But then V0 / R is the initial current I0

RCt

eII−

= 0So we then have that

RCt

eVV−

= 0The voltage across the resistor is given by


Example 4The two circuits shown below contain identical fully charged capacitors at t = 0. Circuit 2 has twice as much resistance as circuit 1.

Compare the charge on the two capacitors a short time after t = 0

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

Initially, the charges on the two capacitors are the same. But the two circuits have different time constants: τ1 = RC and τ2 = 2RC

Since τ2 > τ1 it takes circuit 2 longer to discharge its capacitor

Therefore, at any given time, the charge on capacitor 2 is larger than that on capacitor 1


Example 5

C

a b

R 3RThe capacitor in the circuit shown is initially

charged to Q = Q0. At t = 0 the switch is connected to position a.

At t = t0 the switch is immediately flipped from position a to position b.

a) Which of the following graphs best represents the time dependence of the charge on C?

time

QQ

0

t0 time

QQ

0

t0 time

QQ

0

t0

a) b) c)

b) Which of the following correctly relates the value of t0 to the time constant τa while the switch is at a?

(a) t0 < τa (b) t0 = τa (c) t0 > τa

Example 5The capacitor in the circuit shown is initially



C

a b

R 3R

a) Which of the following graphs best represents the time dependence of the charge on C?

time

QQ

0

t0 time

QQ

0

t0 time

QQ

0

t0

a) b) c)

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

For t > t0, the capacitor is discharging with time constant τ = 3RC, i.e., much more slowly Therefore, the answer is a)



Example 5

C

a b

R 3RThe capacitor in the circuit shown is initially



time

QQ

0

t0 time

QQ

0

t0 time

QQ

0

t0

a) b) c)

b) Which of the following correctly relates the value of t0 to the time constant τa while the switch is at a?

We know that for t = τa, the value of the charge is e-1 = 0.37 of the value at t = 0

(a) t0 < τa (b) t0 = τa (c) t0 > τa

Since the curve shows Q(t0) ~ 0.6 Q0, t0 must be less than τa


Capacitors Circuits, QualitativeBasic principle:

Capacitor resists rapid change in Q resists rapid changes in V

ChargingIt takes time to put the final charge onInitially, the capacitor behaves like a wire (∆V = 0, since Q = 0).As current starts to flow, charge builds up on the capacitor

it then becomes more difficult to add more chargethe current decreases

After a long time, the capacitor behaves like an open switch.

DischargingInitially, the capacitor behaves like a battery.After a long time, the capacitor behaves like a wire.

Documents

Direct-Current Circuits · We now deal with currents through the resistors ... we use the fact that both resistors have the ... After the switch is closed the net resistance is given