p212c26: 1

Ch26: Direct Current Circuits

Two Basic Principles:Conservation of ChargeConservation of Energy

Resistance Networks

V IR

R VI

ab eq

eqab

=

≡Req

Ia

b

p212c26: 2

Resistors in seriesConservation of Charge

I = I1 = I2 = I3

Conservation of Energy Vab = V1 + V2 + V3

RVI

V V VI

VI

VI

VI

VI

VI

VI

R R R R

eqab

eq

≡ =+ +

= + + = + +

= + +

1 2 3

1 2 3 1

1

2

2

3

3

1 2 3

a

b

I

R1

V1=I1R1

R2

V2=I2R2

R3

V3=I3R3

Voltage Divider: VV

RRab eq

1 1=

p212c26: 3

Resistors in parallelConservation of Charge

I = I1 + I2 + I3

Conservation of Energy Vab = V1 = V2 = V3

1

1 1 1 1

1 2 3

1 2 3 1

1

2

2

3

3

1 2 3

RI

VI I I

VI

VI

VI

VIV

IV

IV

R R R R

eq ab ab

ab ab ab

eq

≡ =+ +

= + + = + +

= + +

aI

R1

V1=I1R1

R2

V2=I2R2

R3

V3=I3R3

b

Current Divider: II

RR

eq1

1

=

p212c26: 4

R1=4Ω

R2=3Ω R3=6Ω

E=18V

Example: Determine the equivalent resistance of the circuit as shown.Determine the voltage across and current through each resistor.Determine the power dissipated in each resistorDetermine the power delivered by the battery

p212c26: 5

Kirchoff’s Rules

Some circuits cannot be represented in terms of series and parallel combinations

E1

E2

R1

R2

R3

R1

R3

R2

R4

R5E

Kirchoff’s rules are based upon conservation of energy and conservation of charge.

p212c26: 6

I1 I2

I3

Conservation of Charge: Junction ruleThe algebraic sum of the currents into any

junction is zero.

Σ I = 0 = + I1 + I2 − I3

Σ I = 0 = − I1 − I2 + I3

R1

R2

R3

Alternatively: The sum of the currents into a junction is equal to the sum of the currents out of that junction.

I1 + I2 = I3

p212c26: 7

Conservation of Energy: Loop ruleThe algebraic sum of the potential

differences around any closed loop is zero.

I1

I2

I3

E1

E2

R1

R2

R3

p212c26: 8

Potential Differences in the direction of travel

E

+ −

direction of travel

V = − E

E

+−

direction of travel

V = + E

I

+ −

direction of travel

V =− IR

I

+−

direction of travel V = + IR

p212c26: 9

I1

I2

I3

E1

E2 R2

R1

R3

+ I1 + I2 − I3 =0−E1 + I1 R1 − I2 R2 + E2 =0− I3 R3 − I2 R2 + E2 =0− I3 R3 − I1 R1 + E1 =0

p212c26: 10

I1

I2

I3

E1 =12V

E2 =5V R2 =1Ω

R1 =2Ω

R3 =3Ω

+ I1 + I2 − I3 =0− 12 + I1 2 − I2 1 + 5 =0− I3 3 −I2 1 + 5 =0

p212c26: 11

Standard linear form:+ 1 I1 + 1 I2 − 1 I3 = 0+ 2 I1 −1 I2 + 0 I3 = 7+ 0 I1 + 1 I2 + 3 I3 = 5

+ I1 + I2 − I3 =0− 12 + I1 2 − I2 1 + 5 =0− I3 3 −I2 1 + 5 =0

TI-89: [2nd] [MATH] [4](selects matrix) [5] (selects simult)

Entry line: simult([1,1,-1;2,-1,0;0,1,3],[0;7;5])

I1= 3

I2= −1

I3= 2

or (using [F2])

solve(i1+i2-i3=0 and -12+2*i1-i2+5=0 and -3*i3-i2+5=0,i1,i2,i3)

yields i1=3 and i2=-1 and i3 =2

−

−

−

21

3

570

,310012111

simult

p212c26: 12

R1 R2 R3

I 3A (-)1A 2AVP

P1

P2

I1

I2

I3

E1 =12V

E2 =5V R2 =1Ω

R1 =2Ω

R3 =3Ω

p212c26: 13

Electrical InstrumentsGalvanometer:

Torque proportional to current IRestoring torque proportional to angle=> (Equilibrium) angle proportional to angle

θ

Maximum Deflection =“Full Scale Deflection”Ifs = Current to produce Full Scale DeflectionRc = Resistance (of coil)Vfs = Ifs Rc

example: Ifs = 1.0 mA, Rc = 20.0 Ω Vfs = 20 mV = .020 V

G

p212c26: 14

Wheatstone Bridge

Circuit Balanced: Vab = 0

R2

bR1

R3 R4

E a G

4

2

3

1

44

22

33

11

4

2

3

1

4321

4231

0

RR

RR

RIRI

RIRI

VV

VV

VVVVIIII

IG

=

⇒=⇒=

====

=

p212c26: 15

AmmeterMeasures current through device

Measure currents larger than Ifs by bypassing the galvanometer with another resistor (shunt Resistor Rs).

Ia = ammeter full scale

When I = Ia , we need Ic = Ifs VG = Vsh => Ifs Rc = (Ia -Ifs)Rsh

G

Rsh

I

Ic

I-Ic

afs II=

θθ

Design a 50 mA ammeter from example galvanometerIfs = 1.0 mA, Rc = 20.0 Ω, Vfs = 20 mV = .020 V

I

θ

p212c26: 16

VoltmeterMeasures potential difference across device

Measure V larger than Vfs by adding a resistor (Rs) in series to the galvanometer.

Vm = Voltmeter full scaleWhen V = Vm , we need Ic = Ifs

Ic = Is , Vm = Vs + VG => Vm = Ifs Rs + Ifs Rc

Rs = (Vm / Ifs )- Rc

GRs

I

V

θθ fs fs m

II

VV

= =

V

Design a 10 V voltmeter from example galvanometerIfs = 1.0 mA, Rc = 20.0 Ω, Vfs = 20 mV = .020 V

p212c26: 17

Ohmmeter

Measures resistance across terminalsMeter supplies an EMF, resistance is determined

by response.

When R = 0 , we need Ic = Ifs E = (Ifs Rs + Ifs Rc)=> Vm = Ifs Rs + Ifs Rc

Rs = (E / Ifs )- Rc

GRs

I

R

θθ fs fs

s c

s c

II

R RR R R

= = ++ + R

p212c26: 18

p212c26: 19

p212c26: 20

Resistance Capacitance Circuits

Charging Capacitor

i E

RC

switch

+q -q

vc = q/C vR = iR

Capacitor is initially uncharged.Switch is closed at t=0 Apply Loop rule:

E − q/C− iR=0

p212c26: 21

tRCC

Ctq

uduu

CqudtRCCq

dq

dtRCCq

dq

CqRCdt

dqdtdq

RCq

Ri

iRCq

ttq

1))((ln

ln1,1)(

1)(

)(1

=-

0=--

0

)(

0

−=−

−

=−=−=−

−=−

−−=

=

∫∫∫

EE

EE

E

E

E

E

p212c26: 22

i E

RC

switch

+q -q

vc = q/C vR = iR

τ

τ

τ

//

/

/

/

)1(

"ntTimeConsta")1(

)1()(

1))((ln

to

RCt

RCt

tfinal

RCt

eIeR

eRC

Cdtdqi

RCeQ

eCtq

tRCC

Ctq

−−

−

−

−

==

==

==

−=−=

−=−

−

E

E

EEE

0 0.2 0.4 0.6 0.8 1

t

q

0 0.2 0.4 0.6 0.8 1

t

i

p212c26: 23

Resistance Capacitance Circuits

Discharging Capacitor

i

RC

switch

+q -q

vc = q/C vR = iR

Capacitor has an initial charge Qo.Switch is closed at t=0 Apply Loop rule:

+ q/C-iR=0

p212c26: 24

tRCQ

tq

dtRCq

dq

dtRCq

dq

qRCdt

dqdtdq

RCqi

iRCq

o

ttq

Qo

1)(ln

1

1

1

=

0=

0

)(

−=

−=

−=

−=

−=

−+

∫∫

p212c26: 25

τ−

−

τ−

−

=

=−=

==τ=

=

−=

/

/

/

/

Constant" Time" same

)(

1)(ln

to

RCto

to

RCto

o

eI

eRCQ

dtdqi

RCeQ

eQtq

tRCQ

tq

0 0.2 0.4 0.6 0.8 1

t

q

i

RC

switch

+q -q

vc = q/C vR = iR

0 0.2 0.4 0.6 0.8 1

t

i