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

A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007

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**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 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.

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**EMF and Terminal Potential Difference**

The emf E is the open-circuit potential difference. The terminal voltage VT for closed circuit is reduced due to internal resistance r inside source. Open Circuit E = 1.5 V Closed Circuit VT = 1.45 V r Applying Ohm’s law to battery r, gives: VT = E - Ir

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**Finding Current in Simple Circuit**

Ohm’s law: Current I is the ratio of emf E to total resistance R + r. VT VT = IR r R I + Battery r E - Cross multiplying gives: IR + Ir = E; VT = IR VT = E - Ir

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**Example 2. A 3-V battery has an internal resistance of 0**

Example 2. A 3-V battery has an internal resistance of 0.5 W and is connected to a load resistance of 4 W. What current is delivered and what is the terminal potential difference VT? r = 0.5 W R = 4 W I + - E = 3 V r R I = A VT = E – Ir VT = 3 V – (0.667 A)(0.5 W) VT = 2.67 V

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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 emf’s; the latter two are more often associated with external loads.

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**Power, Potential, and EMF**

Consider simple circuit: I + Battery r E - VT R VT = E - Ir Terminal Voltage Multiply each term by I: VTI = EI - I2r The power delivered to the external circuit is equal to the power developed in the emf less the power lost through internal resistance.

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**Example 3. The 3-V battery in Ex. 2 had an internal resistance of 0**

Example 3. The 3-V battery in Ex. 2 had an internal resistance of 0.5 W and a load resistance of 4 W. Discuss the power used in the circuit. r = 0.5 W R = 4 W I + - E = 3 V r R I = A VT = 2.67 V From Ex. 2, we found: Power developed in emf: EI = (3.0 V)(0.667 A) = 2.0 W Power lost in internal r: I2r = (0.667 A)2(0.5 W) = W

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

Power in emf: EI = 2.00 W Power loss: I2r = W r = 0.5 W R = 4 W I + - E = 3 V r R Power lost in external load R: I2R = (0.667)2(4 W) = 1.78 W This power can also be found using VT = 2.67 V Actual power used externally. VTI = (2.67)(0.667 A) = 1.78 W

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

r = 0.5 W R = 4 W I + - E = 3 V r R Power in emf: EI = 2.00 W Power loss in internal r: I2r = W Power lost in external load R: I2R = VTI = 1.78 W VTI = EI - I2r 1.78 W = 2.00 W – W

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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. r + - E I = 2 A Discharging 12 V, 1 W A B Discharging: VBA = E - Ir GAIN LOSS 12 V - (2 A)(1 W) = 12 V - 2 V = 10 V If VB= 20 V, then VA = 30 V; Net Gain = 10 V

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**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. r + - E I = 2 A Charging 12 V, 1 W A B Charging: VAB = E + Ir LOSS LOSS -12 V - (2 A)(1 W) = -12 V - 2 V = -14 V If VA= 20 V, then VB = 6.0 V; Net Loss = 14 V

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**Power Gain for Discharging EMF**

Recall that electric power is either VI or 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. r + - E I = 2 A Discharging 12 V, 1 W A B Net Power Gain: VBAI = E I- I2r (12 V)(2 A) - (2 A)2(1 W) = 24 W - 4 W = 20 W

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**Power Lost on Charging a Battery**

Recall that electric power is either VI or I2R When a battery is charged (current against normal output), power is lost through chemical changes EI and through internal resistance Ir2. r + - E I = 2 A Charging 12 V, 1 W A B Net Power Lost= EI + I2r (12 V)(2 A) + (2 A)2(1 W) = 24 W + 4 W = 24 W

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**What is the terminal voltage VG across the generator?**

Example 4: A 24-V generator is used to charge a 12-V battery. For the generator, r1 = 0.4 W and for the battery r2 = 0.6 W The load resistance is 5 W. r2 + - E2 I r1 E1 R 24 V 12 V .4 W .6 W 5 W First find current I: Circuit current: I = 2.00 A What is the terminal voltage VG across the generator? VT = E – Ir = 24 V – (2 A)(0.4 W) VG = 23.2 V

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**Example 4: Find the terminal voltage VB across the battery.**

+ - E2 I r1 E1 R 24 V 12 V .4 W .6 W 5 W Circuit current: I = 2.00 A VB = E + Ir = 12 V + (2 A)(0.4 W) Terminal VB = 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.

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**Ammeters and Voltmeters**

Rheostat Emf - + V Source of EMF Voltmeter Ammeter Rheostat

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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 - + rg Ammeter has Internal rg The ammeter draws just enough current Ig to operate the meter; Vg = Ig rg

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**Galvanometer: A Simple Ammeter**

The galvanometer uses torque created by small currents as a means to indicate electric current. 10 20 N S A current Ig causes the needle to deflect left or right. Its resistance is Rg. The sensitivity is determined by the current required for deflection. (Units are in Amps/div.) Examples: 5 A/div; 4 mA/div.

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Example 5. If 0.05 A causes full-scale deflection for the galvanometer below, what is its sensitivity? 10 20 N S Assume Rg = 0.6 W and that a current causes the pointer to move to “10.” What is the voltage drop across the galvanometer? Vg = (25 mA)(0.6 W) Vg = 15 mV

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**Operation of an Ammeter**

The galvanometer is often the working element of both ammeters and voltmeters. Rg I Rs Is Ig 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 = Is + Ig The current Ig is negligible and only enough to operate the galvanometer. [ Is >> Ig ]

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Shunt Resistance VB - + Ammeter R Ig I = 10 A Rs A Rg Is Current Ig causes full-scale deflection of ammeter of resistance Rg. What Rs is needed to read current I from battery VB? Junction rule at A: I = Ig + Is Or Is = I - Ig (I – Ig)Rs = IgRg Voltage rule for Ammeter: 0 = IgRg – IsRs; IsRs = IgRg

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

Example 6. An ammeter has an internal resistance of 5 W and gives full- scale defection for 1 mA. To read 10 A full scale, what shunt resistance Rs is needed? (see figure) VB - + rg Ammeter R 5 W Ig 1 mA I = 10 A A Rs = x 10-4 W The shunt draws % of the external current.

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**Operation of an Voltmeter**

The voltmeter must be connected in parallel and must have high resistance so as not to disturb the main circuit. Rg I VB Ig A multiplier resistance Rm is added in series with the galvanometer so that very little current is drawn from the main circuit. Rm VB = IgRg + IgRm The voltage rule gives:

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**Multiplier Resistance**

VB Voltmeter R I Rm Rg Current Ig causes full-scale deflection of meter whose resistance is Rg. What Rm is needed to read voltage VB of the battery? VB = IgRg + IgRm IgRm = VB - IgRg Which simplifies to:

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

Example 7. A voltmeter has an internal resistance of 5 W and gives full- scale deflection for 1 mA. To read 50 V full scale, what multiplier resistance Rm is needed? (see figure) VB Voltmeter R Ig 1 mA I Rm 5 W Rg Rm = W The high resistance draws negligible current in meter.

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**Summary of Formulas: Discharging: VT = E - Ir - I**

+ - E I Discharging Discharging: VT = E - Ir Power: VTI = EI - I2r r + - E I Charging Charging: VT = E + Ir Power: VTI = EI + I2r

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**Summary (Continued) VB VB - Ammeter Voltmeter Ig Rg Rg Rm A Rs + R R I**

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**CONCLUSION: Chapter 28B EMF and Terminal P.D.**

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