Example I, Maximize Voltage Delivery
Sometimes there is the need to concatenate two circuits in series. As some simple example, the source can be a battery and the load can be just a resistor, or the source is a signal (waveform) generator that can produce various waveforms (sinusoids, saw tooth, square wave, etc.) and the load can be an oscilloscope to display such waveforms. Alternatively, as a more sophisticated example, two voltage amplifiers can be cascaded together so that the output of the first one, the source, is treated as the input of the second one, the load, for the purpose of amplifying some very weak signals. Note that the amplifier circuit is active in the sense it contains a voltage source that depends on the input voltage (across the input impedance), e.g., , with being the voltage gain.
To maximize the output voltage, it is important to consider both the input and output impedances (resistances for now) of the circuits. The output impedance of the first circuit and the input impedance of the second circuit form a voltage divider. For the second circuit to receive maximum voltage from the first one, we want
Example II, Maximize power delivery
Sometimes we want to maximize the power delivered from the voltage source
to the load:
The efficiency of the circuit is defined as the ratio of the power
delivered to the load and the power generated by the source :
In power network, it is more desirable to have high efficiency to avoid wasting energy than delivering maximal power. But in communication network and other applications with small power, efficiency can be sacrificed to maximize the load power. For example, in an audio system, it is important for the speaker's impedance to match the output impedance of the power amplification circuit, so that the speaker can receive maximum power.
Example III, Minimize loss in power transmission line:
The power loss along the power transmission line between the power plant and the consumers should be minimized. Assume the resistance of the power line is and the consumer load is . Also assume the voltage on the consumer's side of the power line is .
Summary: The circuits in the three examples above are essentially the same, they all have a voltage source with an internal resistance (or ), and a load resistance . However, the circuit will be optimized differently according to different requirements:
Example 0 (Homework)
A realistic voltage source (e.g., a battery) can be modeled as an ideal
voltage source in series with an internal resistance . Ideally,
the voltage can be obtained by measuring the open-circuit voltage
with a voltmeter
However, in reality, any voltmeter has an internal resistance in parallel with the meter, any ammeter has an internal resistance in series with the meter. For better measurement accuracy, should be small or large, how about ? Why? Give the expression of the measured open-circuit voltage and short-circuit current in terms of the true and , as well as and .
Design a method to obtain the true source voltage and source resistance by a voltmeter and an ammeter with known and . Give the expression of and in terms of the measured open-circuit voltage , short-circuit , and and .
Now Assume what are the measured open-circuit voltage , and the short-circuit current ? Given , , and the known and , how do your get the true and using your method above? Show your numerical computations.
Example 1 (Homework)
Usually the internal resistances of the voltmeter and ammeter are not readily known (and the values may change depending on the scale used). As another method to find and of a voltage source, we can measure the voltage across two different load resistors connected to the voltage source. If the values of are significantly smaller than that of the internal resistance of the voltmeter, the voltmeter can be considered to be ideal with .
Assume when the load resistor is , the voltage across it is found to be , then a different load is used and the voltage across it is . Find and of the voltage source.