PHY 202 (Blum)1 More basic electricity Non-Ideal meters, Power, Power supplies

PHY 202 (Blum)2 What makes for ideal voltmeters and ammeters?

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.

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.

PHY 202 (Blum)5 Theoretical calculation 5 V = (1 k  k  ) I 5 V = (4.3 k  ) I I = mA V 3.3 = (3.3 k  ) ( mA) V 3.3 = V Slight discrepancy? Without the voltmeter, the two resistors are in series.

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.

PHY 202 (Blum)7 R V should be large If R v  , then Voltmeters should have large resistances. 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.

Electronics Workbench default resistance of voltmeter is 10 Mega-ohms PHY 202 (Blum)8

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.

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.

Electronics Workbench default resistance of ammeter is 1 nano-ohm PHY 202 (Blum)11

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.

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 = Watt = 2.2 mW Energy = Power * Time ( Joule/sec)*(3 minutes)*(60 seconds/minute) Joules PHY 202 (Blum)13

A kilowatt-hour is a measure of energy PHY 202 (Blum)14

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

References Physics, Paul Tipler CompTIA A+ Certification, Mike Meyers PHY 202 (Blum)16