Chapter 28B - EMF and Terminal P.D. A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University A PowerPoint.

Chapter 28B - EMF and Terminal Chapter 28B - EMF and Terminal P.D.P.D.

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics

Southern Polytechnic State Southern Polytechnic State UniversityUniversity

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics

Southern Polytechnic State Southern Polytechnic State UniversityUniversity© 2007

Objectives: Objectives: After completing After completing this module, you should be this module, you should be

able to:able to:• Solve problems involving Solve problems involving emfemf, ,

terminal potential differenceterminal potential difference, , internal internal resistanceresistance, and , and load resistanceload resistance..

• Solve problems involving Solve problems involving power gainspower gains and lossesand losses in a simple circuit containing in a simple circuit containing internal and load resistances.internal and load resistances.

• Work problems involving the use of Work problems involving the use of ammetersammeters and and voltmetersvoltmeters in dc circuits. in dc circuits.

EMF and Terminal EMF and Terminal Potential DifferencePotential Difference

Open Circuit E = 1.5

V

The The emf emf EE is the is the open-circuitopen-circuit potential potential difference. The difference. The terminal voltageterminal voltage VVTT for for closed circuit is reduced due to closed circuit is reduced due to internal internal resistanceresistance r r inside source.inside source.Close

d Circui

t

VT = 1.45 V

r

Applying Ohm’s Applying Ohm’s law to battery law to battery r, r,

gives:gives:

VT = E - IrVT = E - Ir

Finding Current in Simple Finding Current in Simple CircuitCircuitOhm’s law:Ohm’s law: Current Current II is the ratio of emf is the ratio of emf E E to total resistance to total resistance R R + r+ r..

IR r

E=IR r

E=

Cross multiplying Cross multiplying gives:gives:

IR + IrIR + Ir = = EE; ; VVTT = IR = IR

VT = E - IrVT = E - Ir

r

RI

+Battery

rE

-

VT

VT = IR

Example 2. Example 2. A A 3-V3-V battery has an battery has an internal resistance of internal resistance of 0.5 0.5 and is and is connected to a load resistance of connected to a load resistance of 4 4 . What current is delivered and . What current is delivered and what is the terminal potential what is the terminal potential difference difference VVTT??

II = 0.667 A = 0.667 AII = 0.667 A = 0.667 A

3 V

4 0.5I

R r

E

=

VVTT = = E E – Ir – Ir

VVTT = 3 = 3 V – (0.667 A)(0.5 V – (0.667 A)(0.5 ))

VVTT = 2.67 = 2.67 VV

VVTT = 2.67 = 2.67 VV

r = 0.5

R = 4 I

+ -

E = 3 V

rR

Power in CircuitsPower in Circuits

Recall that the definition of power is work Recall that the definition of power is work or energy per unit of time. The following or energy per unit of time. The following apply:apply:

Recall that the definition of power is work Recall that the definition of power is work or energy per unit of time. The following or energy per unit of time. The following apply:apply:

22; ;

VP VI P I R P

R

22; ;

VP VI P I R P

R

The first of these is normally associated with The first of these is normally associated with the the powerpower gains and losses through gains and losses through emf’semf’s; ; the latter two are more often associated with the latter two are more often associated with external loadsexternal loads..

The first of these is normally associated with The first of these is normally associated with the the powerpower gains and losses through gains and losses through emf’semf’s; ; the latter two are more often associated with the latter two are more often associated with external loadsexternal loads..

Power, Potential, and EMFPower, Potential, and EMF

I

+Battery

rE

-

VT

R

Consider simple Consider simple circuit:circuit:

VT = E - IrVT = E - IrTerminal Terminal VoltageVoltage

Multiply each term Multiply each term by by I:I:

VTI = EI - I2rVTI = EI - I2r

The The power deliveredpower delivered to the external to the external circuit is equal to the circuit is equal to the power developedpower developed in the emf less the in the emf less the power lostpower lost through through internal resistance.internal resistance.

The The power deliveredpower delivered to the external to the external circuit is equal to the circuit is equal to the power developedpower developed in the emf less the in the emf less the power lostpower lost through through internal resistance.internal resistance.

Example 3. Example 3. The The 3-V3-V battery in battery in Ex. 2Ex. 2 had an internal resistance of had an internal resistance of 0.5 0.5 and a load resistance of and a load resistance of 4 4 . . Discuss the power used in the Discuss the power used in the circuit.circuit.

r = 0.5

R = 4 I

+ -

E = 3 V

r R

II = 0.667 A = 0.667 AII = 0.667 A = 0.667 A VVTT = 2.67 = 2.67 VV

VVTT = 2.67 = 2.67 VV

From Ex. 2, we found:From Ex. 2, we found:

Power developed in emf:Power developed in emf:

EEI = I = (3.0(3.0 V)(0.667 A) = 2.0 V)(0.667 A) = 2.0 WW

Power lost in internal Power lost in internal rr::

II22r = r = (0.667 A)(0.667 A)22(0.5 (0.5 ) = 0.222 ) = 0.222 WW

Example 3 (Cont.). Example 3 (Cont.). Discuss the Discuss the power used in the simple circuit power used in the simple circuit below.below.

r = 0.5

R = 4 I

+ -

E = 3 V

r RPower in Power in emf:emf:

EI = 2.00 WEI = 2.00 W

Power loss:Power loss: I2r = 0.222 WI2r = 0.222 W

Power lost in external Power lost in external load R:load R:II22R = R = (0.667)(0.667)22(4 (4 ) = ) = 1.78 1.78 WW

VVTTI = I = (2.67)(0.667 A) = (2.67)(0.667 A) = 1.78 1.78 WW

This power can also be This power can also be found using found using VVT T == 2.672.67 VV

Actual power Actual power used used

externally.externally.

Example 3 (Cont.). Example 3 (Cont.). Discuss the Discuss the power used in the simple circuit power used in the simple circuit below.below.

r = 0.5

R = 4 I

+ -

E = 3 V

r RPower in Power in emf:emf:

EI = 2.00 WEI = 2.00 W

Power loss Power loss in internal in internal

r:r:

I2r = 0.222 WI2r = 0.222 W

VTI = EI - I2rVTI = EI - I2r 1.78 W = 2.00 W – 0.222 W

1.78 W = 2.00 W – 0.222 W

Power lost in external load Power lost in external load R:R:

I2R = VTI = 1.78 W

I2R = VTI = 1.78 W

A A DischargingDischarging EMF EMF

When a battery is discharging, there is a GAIN in energy E as chemical energy is converted to electrical energy. At the same time, energy is LOST through internal resistance Ir.

When a battery is discharging, there is a GAIN in energy E as chemical energy is converted to electrical energy. At the same time, energy is LOST through internal resistance Ir. Discharging: VBA = E - IrDischarging: VBA = E - Ir

12 V - (2 A)(1 12 V - (2 A)(1 ) = 12 V - 2 V = 10 ) = 12 V - 2 V = 10 V V

r+ -

E

I = I = 2 A2 ADischarginDischargin

gg

12 V, 1 12 V, 1 AA BB

If VIf VBB= 20 V, then V= 20 V, then VAA = 30 V; = 30 V; Net GainNet Gain = = 10 V10 V

GAINGAIN LOSLOSSS

Charging:Charging: Reversing Flow Through Reversing Flow Through EMFEMF

When a battery is charged (current against normal output), energy is lost through chemical changes E and also through internal resistance Ir.

When a battery is charged (current against normal output), energy is lost through chemical changes E and also through internal resistance Ir.

Charging: VAB = E + IrCharging: VAB = E + Ir

-12 V - (2 A)(1 -12 V - (2 A)(1 ) = -12 V - 2 V = -) = -12 V - 2 V = -14 V 14 V

r+ -

E

I = I = 2 A2 AChargingCharging

12 V, 1 12 V, 1 AA BB

If VIf VAA= 20 V, then V= 20 V, then VBB = 6.0 V; = 6.0 V; Net LossNet Loss = = 14 V14 V

LOSLOSSS

LOSLOSSS

Power GainPower Gain for for DischargingDischarging EMFEMF

When a battery is discharging, there is a GAIN in power EI as chemical energy is converted to electrical energy. At the same time, power is LOST through internal resistance I2r.

When a battery is discharging, there is a GAIN in power EI as chemical energy is converted to electrical energy. At the same time, power is LOST through internal resistance I2r. Net Power Gain: VBAI = E I-

I2r

Net Power Gain: VBAI = E I- I2r

(12 V)(2 A) - (2 A)(12 V)(2 A) - (2 A)22(1 (1 ) = 24 W - 4 W = 20 ) = 24 W - 4 W = 20 W W

r+ -

E

I = I = 2 A2 ADischarginDischargin

gg

12 V, 1 12 V, 1 AA BB

Recall that electric power is either VI or I2R

Recall that electric power is either VI or I2R

PowerPower LostLost on on ChargingCharging a a BatteryBattery

When a battery is charged (current against normal output), power is lost through chemical changes EI and through internal resistance Ir2.

When a battery is charged (current against normal output), power is lost through chemical changes EI and through internal resistance Ir2.

Net Power Lost= EI + I2rNet Power Lost= EI + I2r

(12 V)(2 A) + (2 A)(12 V)(2 A) + (2 A)22(1 (1 ) = 24 W + 4 W = ) = 24 W + 4 W = 24 24 WW

r+ -

E

I = I = 2 A2 AChargingCharging

12 V, 1 12 V, 1 AA BB

Recall that electric power is either VI or I2R

Recall that electric power is either VI or I2R

Example 4:Example 4: A A 24-V24-V generator is used to generator is used to charge a charge a 12-V12-V battery. For the generator, battery. For the generator, rr11 = 0.4 = 0.4 and for the battery and for the battery rr22 = = 0.6 0.6 . . The load resistance is The load resistance is 5 5 ..

rr22

+ -E2

II

rr11

+ -E1 II

RR

24 V24 V

12 V12 V

.4 .4

.6 .6

5 5

First find current First find current I:I:

24V 12V

5 0.4 0.6I

R

E

Circuit current: I = 2.00 A

Circuit current: I = 2.00 A

What is the terminal What is the terminal voltage voltage VVGG across the across the

generator?generator?VVTT = = E E – Ir = – Ir = 24 V – (2 A)(0.4 24 V – (2 A)(0.4 )) VG = 23.2

V

VG = 23.2 V

Example 4:Example 4: Find the Find the terminal voltage terminal voltage VVBB across across the battery.the battery.

rr22

+ -E2

II

rr11

+ -E1 II

RR

24 V24 V

12 V12 V

.4 .4

.6 .6

5 5

Circuit current: I = 2.00 A

Circuit current: I = 2.00 A

VVB B = = E E ++ Ir = Ir = 12 V + (2 A)(0.4 12 V + (2 A)(0.4 ))

Terminal VB = 13.6 VTerminal VB = 13.6 V

Note:Note: The terminal voltage The terminal voltage across a device in which the across a device in which the current is reversed is current is reversed is greatergreater than its emf.than its emf.

For a discharging device, the terminal For a discharging device, the terminal voltage is voltage is lessless than the emf because of than the emf because of internal resistance.internal resistance.

AmmeterAmmeterVoltmeterVoltmeter RheostatRheostatSource Source of EMFof EMF

Rheostat

A

Ammeters and VoltmetersAmmeters and Voltmeters

V Emf-

+

The AmmeterThe AmmeterAn An ammeterammeter is an instrument used to is an instrument used to measure currents. It is always connected measure currents. It is always connected in in series series and its resistance must be and its resistance must be small small (negligible change in (negligible change in II).).

An An ammeterammeter is an instrument used to is an instrument used to measure currents. It is always connected measure currents. It is always connected in in series series and its resistance must be and its resistance must be small small (negligible change in (negligible change in II).).

Digital Digital read- out read- out indicates indicates

current in Acurrent in A

A

E-

+rrgg

Ammeter Ammeter has Internal has Internal

rrgg

The ammeter draws just enough current The ammeter draws just enough current IIgg to operate the meter; to operate the meter; VVgg = I = Igg r rgg

Galvanometer: A Simple Galvanometer: A Simple AmmeterAmmeter

00 10102020

10102020

N S

The galvanometer The galvanometer uses torque created uses torque created by small currents as a by small currents as a means to indicate means to indicate electric current.electric current.A current A current IIgg causes the causes the needle to deflect left needle to deflect left or right. Its resistance or right. Its resistance is is RRgg..

The sensitivity is determined by the The sensitivity is determined by the current required for deflection. (Units are current required for deflection. (Units are in in Amps/div.Amps/div.)) Examples:Examples: 5 A/div; 4 5 A/div; 4 mA/div.mA/div.

Example 5.Example 5. If 0.05 A causes full-scale If 0.05 A causes full-scale deflection for the galvanometer below, deflection for the galvanometer below, what is its sensitivity?what is its sensitivity?

00 10102020

10102020

N S

0.05A mA2.50

20 div divSensitivity

0.05A mA2.50

20 div divSensitivity

Assume Assume RRgg = 0.6 = 0.6 and and that a current causes that a current causes the pointer to move to the pointer to move to ““1010.” .” What is the What is the voltage drop across the voltage drop across the galvanometergalvanometer??

2.5mA10div 25mA

divI

VVgg = (25 mA)(0.6 = (25 mA)(0.6

Vg = 15 mV

Vg = 15 mV

Operation of an AmmeterOperation of an AmmeterThe The galvanometergalvanometer is often the working is often the working element of both ammeters and element of both ammeters and voltmeters.voltmeters.A A shunt resistanceshunt resistance in in parallel with the parallel with the galvanometer allows most galvanometer allows most of the current of the current II to by- pass to by- pass the meter. The whole the meter. The whole device must be connected device must be connected in series with the main in series with the main circuit. circuit.

I = Is + IgI = Is + Ig

RRgg

II RRss

IIss

IIgg

The current The current IIgg is negligible and only is negligible and only enough to operate the galvanometer. enough to operate the galvanometer. [ [ IIss >> I>> Ig g ]]

Junction rule at A:Junction rule at A:

I = II = Ig g + I+ Iss

Voltage rule for Voltage rule for Ammeter:Ammeter:0 = I0 = IggRRgg – I – IssRRss; I; IssRRss = I = IggRRgg

Or Or IIss = I - I = I - Igg (I – I(I – Igg)R)Rs s = I= IggRRgg

g gs

g

I RR

I I

g g

sg

I RR

I I

Shunt ResistanceShunt Resistance

Current Current IIg g causes full-causes full-scale deflection of scale deflection of ammeter of resistance ammeter of resistance RRgg. . What What RRss is needed to is needed to read current read current I I from from battery battery VVBB??

VB-

+

AmmetAmmeterer

RR

IIgg

I = 10 A

RRssAA

RRgg IIss

Example 6.Example 6. An An ammeter has an ammeter has an internal resistance of internal resistance of 5 5 and gives full- scale and gives full- scale defection for defection for 1 mA1 mA. To . To read read 10 A10 A full scale, full scale, what shunt resistance what shunt resistance RRss is needed? (see is needed? (see figure)figure) VB

-

+

rrgg

AmmetAmmeterer

RR

5 5 IIgg

1 mA1 mA

I = 10 A

rrggAA

g gs

g

I RR

I I

g g

sg

I RR

I I

(0.001A)(5 )

10A (0.001sR

Rs = 5.0005 x 10-4 Rs = 5.0005 x 10-4

The shunt draws The shunt draws 99.999%99.999% of the external of the external current.current.

Operation of an VoltmeterOperation of an VoltmeterThe The voltmetervoltmeter must be connected in must be connected in parallel parallel and must have and must have high resistancehigh resistance so so as not to disturb the main circuit.as not to disturb the main circuit.

A A multiplier resistancemultiplier resistance RRmm

is added in series with the is added in series with the galvanometer so that very galvanometer so that very little current is drawn from little current is drawn from the main circuit. the main circuit.

VB = IgRg + IgRmVB = IgRg + IgRm

RRgg

II

VVBB

IIgg

The voltage rule The voltage rule gives:gives:

RRmm

IIggRRm m = V= VBB - I - IggRRgg

B g gm

g

V I RR

I

Multiplier ResistanceMultiplier Resistance

Current Current IIg g causes full-causes full-scale deflection of scale deflection of meter whose resistance meter whose resistance is is RRgg. . What What RRmm is needed is needed to read voltage to read voltage VVBB of the of the battery?battery?

VB = IgRg + IgRmVB = IgRg + IgRm

Bm g

g

VR R

I B

m gg

VR R

I

Which simplifies Which simplifies to:to:

VB

VoltmeterVoltmeter

RRI

RRmmRRgg

Example 7.Example 7. A A voltmeter has an voltmeter has an internal resistance of internal resistance of 5 5 and gives full- scale and gives full- scale deflection for deflection for 1 mA1 mA. To . To read read 50 V50 V full scale, full scale, what multiplier what multiplier resistance resistance RRmm is is needed? (see figure)needed? (see figure)

Rm = 49995 Rm = 49995

The high resistance draws negligible current in The high resistance draws negligible current in meter.meter.

Bm g

g

VR R

I B

m gg

VR R

I

50V5

0.001AmR

VB

VoltmeterVoltmeter

RR

IIgg

1 mA1 mA

I

RRmm

5 5

RRgg

Summary of Formulas:Summary of Formulas:

Power: VTI = EI - I2rPower: VTI = EI - I2r

Charging: VT = E + IrCharging: VT = E + Ir

Discharging: VT = E - Ir

Discharging: VT = E - Ir

r+ -

E

IIChargingCharging

r+ -

E

IIDischarginDischargin

gg

Power: VTI = EI + I2rPower: VTI = EI + I2r

Summary (Continued)Summary (Continued)

g gs

g

I RR

I I

g g

sg

I RR

I I

B

m gg

VR R

I B

m gg

VR R

I

VB

VoltmeterVoltmeter

RRI

RRmmRRgg

VB-

+

AmmetAmmeterer

RR

IIgg

I

RRssAA

RRgg

CONCLUSION: Chapter 28BCONCLUSION: Chapter 28BEMF and Terminal P.D.EMF and Terminal P.D.