Download Lecture 4: Meters and Power

More basic electricity
Non-Ideal meters, Power, Power
supplies
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What makes for ideal
voltmeters and ammeters?
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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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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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Theoretical calculation
5 V = (1 k + 3.3 k ) I
Without the
voltmeter, the two
 5 V = (4.3 k ) I
resistors are in
series.
 I = 1.16279 mA
 V3.3 = (3.3 k ) (1.16279 mA)
 V3.3 = 3.837 V
 Slight discrepancy?

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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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RV should be large
1
Req

=
1
R3.3
+
If Rv  , then
1
1

Req
R3.3

1
Rv
The voltmeter is in parallel with
the 3.3-k resistor and has an
equivalent resistance Req.
We want the circuit with and without the
voltmeter to be as close as possible.
Thus we want Req to be close to 3.3 k.
This is accomplished in Rv is very large.
Voltmeters should have large resistances.
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Electronics Workbench default resistance
of voltmeter is 10 Mega-ohms
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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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RA should be small
ammeter is in series with
Req = (RA + R1 + R3.3 ) The
the 1- and 3.3-k resistors.
 If RA  0
For the ammeter to have a minimal effect on
the equivalent resistance, its resistance
 Req  (R1 + R3.3 ) should be small.
 Ammeters should have small
resistances

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Electronics Workbench default resistance
of ammeter is 1 nano-ohm
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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 * I2
 P = V * ( V / R) = V2 / 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
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A kilowatt-hour is a measure of energy
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Power differences for elements in
“Equivalent” circuits
Same for circuit but
different for individual
resistors
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Resistor
dissipates
25 mW
Resistor
dissipates
100 mW
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References
Physics, Paul Tipler
 http://www.pcguide.com
 CompTIA A+ Certification, Mike Meyers

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