Chapter 28B - EMF and Terminal P.D. A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University 2007
Objectives: After completin this module, you should be able to: Solve problems involvin emf, terminal potential difference, internal resistance, and load resistance. Solve problems involvin power ains and losses in a simple circuit containin internal and load resistances. Work problems involvin the use of ammeters and voltmeters in dc circuits.
EMF and Terminal Potential Difference The emf E is the open-circuit potential difference. The terminal voltae V T for closed circuit is reduced due to internal resistance r inside source. Open Circuit E = 1.5 V Closed Circuit V T = 1.45 V r Applyin Ohm s s law to battery r, ives: V T = E - r r
Findin Current in Simple Circuit Ohm s s law: Current is the ratio of emf E to total resistance + r. r = E Cross multiplyin ives: + r = E; V T = r V T = E - r r r + V T = V T E r Battery -
Example 2. A 3-V battery has an internal resistance of 0.5 and is connected to a load resistance of 4.. What current is delivered and what is the terminal potential difference V T? E 3 V = r r 40.5 = 0.667 A V T = E r V T = 3 V (0.667 A)(0.5 ) V T T = 2.67 V = 4 E = 3 V + - r = 0.5
Power in Circuits ecall that the definition of of power is is work or or enery per unit of of time. The followin apply: V 2 2 PV; P ; P The first of of these is is normally associated with the power ains and losses throuh emf s; ; the latter two are more often associated with external loads.
Power, Potential, and EMF Consider simple circuit: Terminal Voltae V T = E - r r Multiply each term by : E V T r + Battery - V T = E E - 22 rr The power delivered to to the external circuit is is equal to to the power developed in in the emf less the power lost throuh internal resistance.
Example 3. The 3-V battery in Ex. 2 had an internal resistance of 0.5 and a load resistance of 4.. Discuss the power used in the circuit. From Ex. 2, we found: = 0.667 A V T T = 2.67 V Power developed in emf: E = (3.0 V)(0.667 A) = 2.0 W Power lost in internal r: 2 r = (0.667 A) 2 (0.5 ) ) = 0.222 W r = 4 E = 3 V + - r = 0.5
Example 3 (Cont.). Discuss the power used in the simple circuit below. Power in emf: Power loss: E E = 2.00 W 22 r r = 0.222 W r Power lost in external load : 2 = (0.667) 2 (4 ) ) = 1.78 W This power can also be found usin V T = 2.67 V V T = (2.67)(0.667 A) = 1.78 W = 4 E = 3 V + - r = 0.5 Actual power used externally.
Example 3 (Cont.). Discuss the power used in the simple circuit below. Power in emf: E E = 2.00 W r Power loss in internal r: 22 r r = 0.222 W Power lost in external load : 22 = V T T = 1.78 W = 4 E = 3 V + - r = 0.5 V T = E E - 22 rr 1.78 W = 2.00 W 0.222 W
A Discharin EMF When a battery is is discharin, there is is a GAN in in enery E as chemical enery is is converted to to electrical enery. At the same time, enery is is LOST throuh internal resistance r. r. A 12 V, 1 + - E r = 2 A Discharin B Discharin: V BA BA = E --r r 12 V - (2 A)(1 GAN LOSS (2 A)(1 ) ) = 12 V - 2 V = 10 V f V B = 20 V, then V A = 30 V; Net Gain = 10 V
Charin: eversin Flow Throuh EMF When a battery is is chared (current aainst normal output), enery is is lost throuh chemical chanes E and also throuh internal resistance r. r. A 12 V, 1 + - E r = 2 A Charin B Charin: V AB AB = E + r r -12 V - (2 A)(1 ) = LOSS LOSS ) = -12 V - 2 V = -14 V f V A = 20 V, then V B = 6.0 V; Net Loss = 14 V
Power Gain for Discharin EMF ecall that electric power is is either V Vor 22 When a battery is is discharin, there is is a GAN in in power E Eas chemical enery is is converted to to electrical enery. At the same time, power is is LOST throuh internal resistance 22 r. r. A 12 V, 1 + - E r = 2 A Discharin B Net Power Gain: V BA BA = E - - 22 rr (12 V)(2 A) - (2 A) 2 (1 ) ) = 24 W - 4 W = 20 W
Power Lost on Charin a Battery ecall that electric power is is either V Vor 22 When a battery is is chared (current aainst normal output), power is is lost throuh chemical chanes E E and throuh internal resistance r r 22.. A 12 V, 1 + - E r = 2 A Charin B Net Power Lost= E E + 22 rr (12 V)(2 A) + (2 A) 2 (1 ) ) = 24 W + 4 W = 24 W
Example 4: A 24-V enerator is used to chare a 12-V battery. For the enerator, r 1 = 0.4 and for the battery r 2 = 0.6. The load resistance is 5. First find current : E 24V 12V 50.40.6 12 V.6 + - E 2 r 2 5 Circuit current: = 2.00 A What is the terminal voltae V G across the enerator? 24 V.4 + - E 1 r 1 V T = E r = 24 V (2 A)(0.4 ) V G = 23.2 V
Example 4: Find the terminal voltae V B across the battery. Circuit current: = 2.00 A V B = E + r = 12 V + (2 A)(0.4 ) Terminal V B B = 13.6 V Note: The terminal voltae across a device in which the current is reversed is reater than its emf. 12 V + - E 2 r 2 24 V.6 5.4 + - E 1 r 1 For a discharin device, the terminal voltae is less than the emf because of internal resistance.
Ammeters and Voltmeters V Emf - + A heostat Voltmeter Source of EMF Ammeter heostat
The Ammeter An ammeter is is an instrument used to to measure currents. t t is is always connected in in series and its resistance must be small (neliible chane in in ). E - + A Diital read- out indicates current in A r Ammeter has nternal r The ammeter draws just enouh current to operate the meter; V = r
Galvanometer: A Simple Ammeter The alvanometer uses torque created by small currents as a means to indicate electric current. 20 10 0 10 20 A current causes the needle to deflect left or riht. ts resistance is. N S The sensitivity is determined by the current required for deflection. (Units are in Amps/div.) Examples: 5 A/div; 4 ma/div.
Example 5. f 0.05 A causes full-scale deflection for the alvanometer below, what is its sensitivity? Sensitivity 0.05A ma 2.50 20 div div 20 10 0 10 20 Assume = 0.6 and that a current causes the pointer to move to 10. What is the voltae drop across the alvanometer? 2.5mA 10div div 25mA N S V = (25 ma)(0.6 V = 15 mv
Operation of an Ammeter The alvanometer is often the workin element of both ammeters and voltmeters. A shunt resistance in parallel with the alvanometer allows most of the current to by- pass the meter. The whole device must be connected in series with the main circuit. s s = s s + The current is neliible and only enouh to operate the alvanometer. [ s >> ]
Shunt esistance Current causes full-scale deflection of ammeter of resistance. What s is needed to read current + from battery V B? V B- Ammeter A s s Junction rule at A: Or = + s s = - ( ) s = Voltae rule for Ammeter: 0 = s s ; s s = = 10 A s
Example 6. An ammeter has an internal resistance of 5 and ives full- scale defection for 1 ma. To read 10 A full scale, what shunt resistance s is needed? (see fiure) s V B- + 1 ma Ammeter A r r = 10 A 5 s (0.001A)(5 ) 10A (0.001 s = 5.0005 x 10-4 -4 The shunt draws 99.999% of the external current.
Operation of an Voltmeter The voltmeter must be connected in parallel and must have hih resistance so as not to disturb the main circuit. A multiplier resistance m is added in series with the alvanometer so that very little current is drawn from the main circuit. m V B The voltae rule ives: V B B = + m
Multiplier esistance Current causes full-scale deflection of meter whose resistance is. What m is needed to read voltae V B of the battery? V B B = + m Voltmeter V B m m = V B - m V B Which simplifies to: m V B
Example 7. A voltmeter has an internal resistance of 5 and ives full- scale deflection for 1 ma. To read 50 V full scale, what multiplier resistance m is needed? (see fiure) 1 ma Voltmeter m 5 m V B V B m 50 V 0.001A 5 m = 49995 The hih resistance draws neliible current in meter.
Summary of Formulas: Discharin: V T T = E --r r Power: V T = E E - 22 rr + - E r Discharin + - E r Charin Charin: V T T = E + r r Power: V T T = E E + 22 rr
Summary (Continued) Ammeter Voltmeter m V B- + A s V B s m V B
CONCLUSON: Chapter 28B EMF and Terminal P.D.