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PHY 202 (Blum)1 More basic electricity Non-Ideal meters, Power, Power supplies
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PHY 202 (Blum)2 What makes for ideal voltmeters and ammeters?
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PHY 202 (Blum)3 Ideal Meters Ideally when a voltmeter is added to a circuit, it should not alter the voltage or current of any of the circuit elements. These circuits should be the same.
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PHY 202 (Blum)4 Voltmeter Devices in parallel have the same voltage. Voltmeters are placed in parallel with a circuit element, so they will experience the same voltage as the element.
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PHY 202 (Blum)5 Theoretical calculation 5 V = (1 k + 3.3 k ) I 5 V = (4.3 k ) I I = 1.16279 mA V 3.3 = (3.3 k ) (1.16279 mA) V 3.3 = 3.837 V Slight discrepancy? Without the voltmeter, the two resistors are in series.
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PHY 202 (Blum)6 Non-Ideal Voltmeter Ideally the voltmeter should not affect current in resistor. Let us focus on the resistance of the voltmeter.
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PHY 202 (Blum)7 R V should be large If R v , then Voltmeters should have large resistances. 1 = 1 + 1 R eq R 3.3 RvRv 1 1 R eq R 3.3 The voltmeter is in parallel with the 3.3-k resistor and has an equivalent resistance R eq. We want the circuit with and without the voltmeter to be as close as possible. Thus we want R eq to be close to 3.3 k . This is accomplished in R v is very large.
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Electronics Workbench default resistance of voltmeter is 10 Mega-ohms PHY 202 (Blum)8
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9 Ammeter Devices in series have the same current. Ammeters are placed in series with a circuit element, so they will experience the same current as it.
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PHY 202 (Blum)10 R A should be small R eq = (R A + R 1 + R 3.3 ) If R A 0 R eq (R 1 + R 3.3 ) Ammeters should have small resistances The ammeter is in series with the 1- and 3.3-k resistors. For the ammeter to have a minimal effect on the equivalent resistance, its resistance should be small.
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Electronics Workbench default resistance of ammeter is 1 nano-ohm PHY 202 (Blum)11
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PHY 202 (Blum)12 Power Recall Voltage = Energy/Charge Current = Charge/Time Voltage Current = Energy/Time The rate of energy per time is known as power. It comes in units called watts.
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Power Formulas P = V * I P = (I * R) * I = R * I 2 P = V * ( V / R) = V 2 / R Example, a 5.2-kΩ resistor has a 0.65 mA current for 3 minutes? What is the corresponding power? The corresponding energy? Power = (5200)*(.00065) 2 = 0.002197 Watt = 2.2 mW Energy = Power * Time (.002197 Joule/sec)*(3 minutes)*(60 seconds/minute) 0.39546 Joules PHY 202 (Blum)13
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A kilowatt-hour is a measure of energy PHY 202 (Blum)14
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PHY 202 (Blum)15 Power differences for elements in “Equivalent” circuits Resistor dissipates 100 mW Resistor dissipates 25 mW Same for circuit but different for individual resistors
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References Physics, Paul Tipler http://www.pcguide.com CompTIA A+ Certification, Mike Meyers PHY 202 (Blum)16
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