2Objectives: 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.
3EMF 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 CircuitE = 1.5 VClosed CircuitVT = 1.45 VrApplying Ohm’s law to battery r, gives:VT = E - Ir
4Finding Current in Simple Circuit Ohm’s law: Current I is the ratio of emf E to total resistance R + r.VTVT = IRrRI+BatteryrE-Cross multiplying gives:IR + Ir = E; VT = IRVT = E - Ir
5Example 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 WR = 4 WI+-E = 3 VrRI = AVT = E – IrVT = 3 V – (0.667 A)(0.5 W)VT = 2.67 V
6Power in CircuitsRecall 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.
7Power, Potential, and EMF Consider simple circuit:I+BatteryrE-VTRVT = E - IrTerminal VoltageMultiply each term by I:VTI = EI - I2rThe power delivered to the external circuit is equal to the power developed in the emf less the power lost through internal resistance.
8Example 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 WR = 4 WI+-E = 3 VrRI = AVT = 2.67 VFrom Ex. 2, we found:Power developed in emf:EI = (3.0 V)(0.667 A) = 2.0 WPower lost in internal r:I2r = (0.667 A)2(0.5 W) = W
9Example 3 (Cont.). Discuss the power used in the simple circuit below. Power in emf:EI = 2.00 WPower loss:I2r = Wr = 0.5 WR = 4 WI+-E = 3 VrRPower lost in external load R:I2R = (0.667)2(4 W) = 1.78 WThis power can also be found using VT = 2.67 VActual power used externally.VTI = (2.67)(0.667 A) = 1.78 W
10Example 3 (Cont.). Discuss the power used in the simple circuit below. r = 0.5 WR = 4 WI+-E = 3 VrRPower in emf:EI = 2.00 WPower loss in internal r:I2r = WPower lost in external load R:I2R = VTI = 1.78 WVTI = EI - I2r1.78 W = 2.00 W – W
11A Discharging EMFWhen 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+-EI = 2 ADischarging12 V, 1 WABDischarging: VBA = E - IrGAINLOSS12 V - (2 A)(1 W) = 12 V - 2 V = 10 VIf VB= 20 V, then VA = 30 V; Net Gain = 10 V
12Charging: 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+-EI = 2 ACharging12 V, 1 WABCharging: VAB = E + IrLOSSLOSS-12 V - (2 A)(1 W) = -12 V - 2 V = -14 VIf VA= 20 V, then VB = 6.0 V; Net Loss = 14 V
13Power Gain for Discharging EMF Recall that electric power is either VI or I2RWhen 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+-EI = 2 ADischarging12 V, 1 WABNet Power Gain: VBAI = E I- I2r(12 V)(2 A) - (2 A)2(1 W) = 24 W - 4 W = 20 W
14Power Lost on Charging a Battery Recall that electric power is either VI or I2RWhen a battery is charged (current against normal output), power is lost through chemical changes EI and through internal resistance Ir2.r+-EI = 2 ACharging12 V, 1 WABNet Power Lost= EI + I2r(12 V)(2 A) + (2 A)2(1 W) = 24 W + 4 W = 24 W
15What 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+-E2Ir1E1R24 V12 V.4 W.6 W5 WFirst find current I:Circuit current: I = 2.00 AWhat is the terminal voltage VG across the generator?VT = E – Ir = 24 V – (2 A)(0.4 W)VG = 23.2 V
16Example 4: Find the terminal voltage VB across the battery. +-E2Ir1E1R24 V12 V.4 W.6 W5 WCircuit current: I = 2.00 AVB = E + Ir = 12 V + (2 A)(0.4 W)Terminal VB = 13.6 VNote: 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.
17Ammeters and Voltmeters RheostatEmf-+VSource of EMFVoltmeterAmmeterRheostat
18The AmmeterAn 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 AAE-+rgAmmeter has Internal rgThe ammeter draws just enough current Ig to operate the meter; Vg = Ig rg
19Galvanometer: A Simple Ammeter The galvanometer uses torque created by small currents as a means to indicate electric current.1020NSA 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.
20Example 5. If 0.05 A causes full-scale deflection for the galvanometer below, what is its sensitivity?1020NSAssume 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
21Operation of an Ammeter The galvanometer is often the working element of both ammeters and voltmeters.RgIRsIsIgA 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 + IgThe current Ig is negligible and only enough to operate the galvanometer. [ Is >> Ig ]
22Shunt ResistanceVB-+AmmeterRIgI = 10 ARsARgIsCurrent 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 + IsOr Is = I - Ig(I – Ig)Rs = IgRgVoltage rule for Ammeter:0 = IgRg – IsRs; IsRs = IgRg
23The 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-+rgAmmeterR5 WIg1 mAI = 10 AARs = x 10-4 WThe shunt draws % of the external current.
24Operation of an Voltmeter The voltmeter must be connected in parallel and must have high resistance so as not to disturb the main circuit.RgIVBIgA multiplier resistance Rm is added in series with the galvanometer so that very little current is drawn from the main circuit.RmVB = IgRg + IgRmThe voltage rule gives:
25Multiplier Resistance VBVoltmeterRIRmRgCurrent Ig causes full-scale deflection of meter whose resistance is Rg. What Rm is needed to read voltage VB of the battery?VB = IgRg + IgRmIgRm = VB - IgRgWhich simplifies to:
26The 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)VBVoltmeterRIg1 mAIRm5 WRgRm = WThe high resistance draws negligible current in meter.
27Summary of Formulas: Discharging: VT = E - Ir - I +-EIDischargingDischarging: VT = E - IrPower: VTI = EI - I2rr+-EIChargingCharging: VT = E + IrPower: VTI = EI + I2r
28Summary (Continued) VB VB - Ammeter Voltmeter Ig Rg Rg Rm A Rs + R R I