Resistive Network Analysis in Software

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Resistive Network Analysis
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FOCUS ON MEASUREMENTS
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Experimental Determination of Thevenin Equivalent Circuit
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Determine the Th venin equivalent of an unknown circuit from e measurements of open-circuit voltage and short-circuit current
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Solution: Known Quantities Measurement of short-circuit current and open-circuit
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voltage Internal resistance of measuring instrument Find Equivalent resistance, RT ; Th venin voltage, vT = vOC e Schematics, Diagrams, Circuits, and Given Data Measured vOC = 65 V; Measured iSC = 375 mA; rm = 15 Assumptions The unknown circuit is a linear circuit containing ideal sources and resistors only Analysis The unknown circuit, shown on the top left in Figure 372, is replaced by its Th venin equivalent, and is connected to an ammeter for a e measurement of the short-circuit current (Figure 372, top right), and then to a voltmeter for the measurement of the open-circuit voltage (Figure 372, bottom) The open-circuit voltage measurement yields the Th venin voltage: e vOC = vT = 65 V To determine the equivalent resistance, we observe in the gure depicting the voltage measurement that, according to the circuit diagram, vOC = RT + rm iSC Thus, vOC rm = 1,733 15 = 1,718 RT = iSC
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RT a A An unknown circuit Load terminals b b Network connected for measurement of short-circuit current (practical ammeter) RT a + vT + _ vT
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vOC
b Network connected for measurement of open-circuit voltage (ideal voltmeter)
Part I
Circuits
Comments Note how easy the experimental method is, provided
we are careful to account for the internal resistance of the measuring instruments
One last comment is in order concerning the practical measurement of the internal resistance of a network In most cases, it is not advisable to actually shortcircuit a network by inserting a series ammeter as shown in Figure 371; permanent damage to the circuit or to the ammeter may be a consequence For example, imagine that you wanted to estimate the internal resistance of an automotive battery; connecting a laboratory ammeter between the battery terminals would surely result in immediate loss of the instrument Most ammeters are not designed to withstand currents of such magnitude Thus, the experimenter should pay attention to the capabilities of the ammeters and voltmeters used in measurements of this type, as well as to the (approximate) power ratings of any sources present However, there are established techniques especially designed to measure large currents
MAXIMUM POWER TRANSFER
The reduction of any linear resistive circuit to its Th venin or Norton equivae lent form is a very convenient conceptualization, as far as the computation of load-related quantities is concerned One such computation is that of the power absorbed by the load The Th venin and Norton models imply that some of the e power generated by the source will necessarily be dissipated by the internal circuits within the source Given this unavoidable power loss, a logical question to ask is, how much power can be transferred to the load from the source under the most ideal conditions Or, alternatively, what is the value of the load resistance that will absorb maximum power from the source The answer to these questions is contained in the maximum power transfer theorem, which is the subject of the present section The model employed in the discussion of power transfer is illustrated in Figure 373, where a practical source is represented by means of its Th venin e equivalent circuit The maximum power transfer problem is easily formulated if we consider that the power absorbed by the load, PL , is given by the expression
2 PL = iL RL
Practical source
RL Load
(336)
vT + _
and that the load current is given by the familiar expression vT iL = RL + R T Combining the two expressions, we can compute the load power as PL =
2 vT RL (RL + RT )2
RL iL Source equivalent
(337)
(338)
Given vT and RT, what value of RL will allow for maximum power transfer
To nd the value of RL that maximizes the expression for PL (assuming that VT and RT are xed), the simple maximization problem dPL =0 dRL (339)
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