Presentation on theme: "Chapter 28B - EMF and Terminal P.D. A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University A PowerPoint."— Presentation transcript:
Objectives: After completing this module, you should be able to: Solve problems involving emf, terminal potential difference, internal resistance, and load resistance.Solve problems involving emf, terminal potential difference, internal resistance, and load resistance. Solve problems involving power gains and losses in a simple circuit containing internal and load resistances.Solve problems involving power gains and losses in a simple circuit containing internal and load resistances. Work problems involving the use of ammeters and voltmeters in dc circuits.Work problems involving the use of ammeters and voltmeters in dc circuits.
EMF and Terminal Potential Difference Open Circuit E = 1.5 V The emf E is the open-circuit potential difference. The terminal voltage V T for closed circuit is reduced due to internal resistance r inside source. Closed Circuit V T = 1.45 V r Applying Ohms law to battery r, gives: V T = E - Ir
Finding Current in Simple Circuit Ohms law: Current I is the ratio of emf E to total resistance R + r. Cross multiplying gives: IR + Ir = E ; V T = IR V T = E - Ir r R I + Battery r E - VTVT V T = IR
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 ? I = A V T = E – Ir V T = 3 V – (0.667 A)(0.5 ) V T = 2.67 V r = 0.5 R = 4 I + - E = 3 V r R
Power in Circuits Recall that the definition of power is work or energy per unit of time. The following apply: The first of these is normally associated with the power gains and losses through emfs; the latter two are more often associated with external loads.
Power, Potential, and EMF I + Battery r E - VTVT R Consider simple circuit: V T = E - Ir Terminal Voltage Multiply each term by I: V T I = E I - I 2 r The power delivered to the external circuit is equal to the power developed in the emf less the power lost through 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. r = 0.5 R = 4 I + - E = 3 V r R I = A V T = 2.67 V From Ex. 2, we found: Power developed in emf: E I = (3.0 V)(0.667 A) = 2.0 W Power lost in internal r: I 2 r = (0.667 A) 2 (0.5 ) = W
Example 3 (Cont.). Discuss the power used in the simple circuit below. r = 0.5 R = 4 I + - E = 3 V r R Power in emf: E I = 2.00 W Power loss: I 2 r = W Power lost in external load R: I 2 R = (0.667) 2 (4 ) = 1.78 W V T I = (2.67)(0.667 A) = 1.78 W This power can also be found using V T = 2.67 V Actual power used externally.
Example 3 (Cont.). Discuss the power used in the simple circuit below. r = 0.5 R = 4 I + - E = 3 V r R Power in emf: E I = 2.00 W Power loss in internal r: I 2 r = W V T I = E I - I 2 r 1.78 W = 2.00 W – W Power lost in external load R: I 2 R = V T I = 1.78 W
A Discharging 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. Discharging: V BA = E - Ir 12 V - (2 A)(1 ) = 12 V - 2 V = 10 V r + - E I = 2 A Discharging 12 V, 1 12 V, 1 AB If V B = 20 V, then V A = 30 V; Net Gain = 10 V GAINLOSS
Charging: Reversing Flow Through EMF When a battery is charged (current against normal output), energy is lost through chemical changes E and also through internal resistance Ir. Charging: V AB = E + Ir -12 V - (2 A)(1 ) = -12 V - 2 V = -14 V r + - E I = 2 A Charging 12 V, 1 12 V, 1 AB If V A = 20 V, then V B = 6.0 V; Net Loss = 14 V LOSSLOSS
Power Gain for Discharging EMF When a battery is discharging, there is a GAIN in power E I as chemical energy is converted to electrical energy. At the same time, power is LOST through internal resistance I 2 r. Net Power Gain: V BA I = E I- I 2 r (12 V)(2 A) - (2 A) 2 (1 ) = 24 W - 4 W = 20 W r + - E I = 2 A Discharging 12 V, 1 12 V, 1 AB Recall that electric power is either VI or I 2 R
Power Lost on Charging a Battery When a battery is charged (current against normal output), power is lost through chemical changes E I and through internal resistance Ir 2. Net Power Lost= E I + I 2 r (12 V)(2 A) + (2 A) 2 (1 ) = 24 W + 4 W = 24 W r + - E I = 2 A Charging 12 V, 1 12 V, 1 AB Recall that electric power is either VI or I 2 R
Example 4: A 24-V generator is used to charge a 12-V battery. For the generator, r 1 = 0.4 and for the battery r 2 = 0.6. The load resistance is 5. r2r2r2r2 + - E2E2 I r1r1r1r1 + - E1E1 I R 24 V 12 V First find current I: Circuit current: I = 2.00 A What is the terminal voltage V G across the generator? V T = E – Ir = 24 V – (2 A)(0.4 ) V G = 23.2 V
Example 4: Find the terminal voltage V B across the battery. r2r2r2r2 + - E2E2 I r1r1r1r1 + - E1E1 I R 24 V 12 V Circuit current: I = 2.00 A V B = E + Ir = 12 V + (2 A)(0.4 ) Terminal V B = 13.6 V Note: The terminal voltage across a device in which the current is reversed is greater than its emf. For a discharging device, the terminal voltage is less than the emf because of internal resistance.
Ammeter Voltmeter Rheostat Source of EMF Rheostat A Ammeters and Voltmeters V Emf - +
The Ammeter An ammeter is an instrument used to measure currents. It is always connected in series and its resistance must be small (negligible change in I). Digital read- out indicates current in A A E - + rgrgrgrg Ammeter has Internal r g The ammeter draws just enough current I g to operate the meter; V g = I g r g
Galvanometer: A Simple Ammeter NS The galvanometer uses torque created by small currents as a means to indicate electric current. A current I g causes the needle to deflect left or right. Its resistance is R g. The sensitivity is determined by the current required for deflection. (Units are in Amps/div.) Examples: 5 A/div; 4 mA/div.
Example 5. If 0.05 A causes full-scale deflection for the galvanometer below, what is its sensitivity? NS Assume R g = 0.6 and that a current causes the pointer to move to 10. What is the voltage drop across the galvanometer? V g = (25 mA)(0.6 V g = (25 mA)(0.6 V g = 15 mV
Operation of an Ammeter The galvanometer is often the working element of both ammeters and voltmeters. A shunt resistance in parallel with the galvanometer allows most of the current I to by- pass the meter. The whole device must be connected in series with the main circuit. I = I s + I g RgRgRgRgI RsRsRsRs IsIsIsIs IgIgIgIg The current I g is negligible and only enough to operate the galvanometer. [ I s >> I g ]
Junction rule at A: I = I g + I s Voltage rule for Ammeter: 0 = I g R g – I s R s ; I s R s = I g R g Or I s = I - I g (I – I g )R s = I g R g Shunt Resistance Current I g causes full-scale deflection of ammeter of resistance R g. What R s is needed to read current I from battery V B ? VBVB - + Ammeter R IgIgIgIg I = 10 A RsRsRsRs A RgRgRgRg IsIsIsIs
Example 6. An ammeter has an internal resistance of 5 and gives full- scale defection for 1 mA. To read 10 A full scale, what shunt resistance R s is needed? (see figure) VBVB - + rgrgrgrg Ammeter R 5 I g 1 mA I = 10 A rgrgrgrg A R s = x The shunt draws % of the external current.
Operation of an Voltmeter The voltmeter must be connected in parallel and must have high resistance so as not to disturb the main circuit. A multiplier resistance R m is added in series with the galvanometer so that very little current is drawn from the main circuit. V B = I g R g + I g R m RgRgRgRgI VBVBVBVB IgIgIgIg The voltage rule gives: RmRmRmRm
I g R m = V B - I g R g Multiplier Resistance Current I g causes full-scale deflection of meter whose resistance is R g. What R m is needed to read voltage V B of the battery? V B = I g R g + I g R m Which simplifies to: VBVB Voltmeter R I RmRmRmRm RgRgRgRg
Example 7. A voltmeter has an internal resistance of 5 and gives full- scale deflection for 1 mA. To read 50 V full scale, what multiplier resistance R m is needed? (see figure) R m = The high resistance draws negligible current in meter. VBVB Voltmeter R I g 1 mA I RmRmRmRm 5 RgRgRgRg
Summary of Formulas: Power: V T I = E I - I 2 r Charging: V T = E + Ir Discharging: V T = E - Ir r + - E I Charging r + - E I Discharging Power: V T I = E I + I 2 r
Summary (Continued) VBVB Voltmeter R I RmRmRmRm RgRgRgRg VBVB - + Ammeter R IgIgIgIg I RsRsRsRs A RgRgRgRg