Next: Source and Load Up: Chapter 1: Basic Quantities Previous: Kirchhoff's Laws

## Energy Sources

Ideal Energy Sources:

• An ideal voltage source provides constant voltage independent of the current through it .
• An ideal current source provides constant current independent of the voltage across it .

Ideal sources do not exist in reality, due to such dilemmas:

• When (short circuit) is the load of a voltage source , the load voltage is , where is the current through .
• When (open circuit) is the load of a current source , the load current is , where is the voltage across .

Voltage Source:

In reality, all voltage sources (e.g., a battery or a voltage amplifier circuit) have a nonzero internal resistance in series with the voltage source that causes the actual output voltage to be lower than , depending on the load current . The load voltage and the load current are constrained by the following two relationships imposed by the voltage source and the load , respectively:

Solving for and , we get:

The internal resistance of a voltage source should be as small as possible, ideally .

The slope of the first curve is the internal resistance and the slope of the second curve is . Solving these two equations we get load voltage and current .

Only in the case of an ideal voltage source with will . For , the heavier the load, i.e., the smaller , the larger the load current , and the lower the load voltage :

In particular, when the load is a short circuit (), the output voltage of a battery is zero (instead of the specified ), as the voltage drop across the nonzero internal resistance is the same as , i.e., the electric energy of the battery is consumed internally, with no energy delivered to external circuit.

Current Source:

In reality, all current sources (e.g., a solar cell or a current amplifier circuit) have an internal resistance in parallel to the current source that causes the actual output current to be lower than , depending on the load voltage . The load voltage and the load current are constrained by the following two relationships imposed by the current source and the load , respectively:

Solving for and , we get

The internal resistance of a current source should be as large as possible, ideally .

The slope of the first curve is the internal resistance and the slope of the second curve is . Solving these two equations we get load voltage and current .

Only in the case of an ideal current source will and . For , the larger the load resistance , the smaller the current .

Equivalent Circuits and Source Transformation:

Two circuits with the same voltage-current relation (V-I characteristics) at the output port are considered equivalent, as they have the same external behavior, although they may be different internally.

Comparing the two equations and describing the voltage and current sources respectively, we see that a voltage source can be equivalently converted into a current source and vise versa:

• The voltage-current relation of a voltage source () is

• The voltage-current relation of a current source () is

Comparing the two equations above, we see that if we let and or , then the two sources have the same V-I characteristics and are therefore equivalent, i.e., the voltage and current sources can be equivalently converted to each other by

Finding internal resistance:

The internal resistance of a voltage source is the slope of the straight line , while the internal resistance of a current source is the slope of the straight line . In both cases this slope can be obtained as the ratio of the open-circuit voltage (when ) and the short-circuit current (when ):

• For voltage source , the open-circuit voltage is , the short-circuit current is , and their ratio is .
• For current source with , the open-circuit voltage is , the short-circuit current is , and their ratio is .
While this method can be used to find theoretically, it is not practical experimentally, as the short circuit current is difficult to get (as the voltage source may be damaged). Instead, we can find some other two voltages and currents and () associated with each of two load resistors and . Then the internal resistance can also be found as the slope of the straight line determined by the two points and . We see that the previous method can be considered as a special case when (open circuit) and (short circuit).

Example 1:

A voltage source of and can be converted to a current source of with the same (and vice versa). For a load of , both energy sources will provide the load current and load voltage . As is low, this is a good voltage source but a poor current source.

Example 2:

A current source of and can be converted to a voltage source of with the same (and vice versa). For a load of , both energy sources will provide the load current and load voltage . As is high, this is a good current source but a poor voltage source.

Power Delivery/Absorption

• Voltage source
• If the direction of the current in a circuit is such that it goes internally through a voltage source from its low potential (-) to high (+), and externally through the rest of the circuit from high to low, then the polarities of the voltage and the current are consistent (both positive or negative depending on the assumed polarity), and the source is delivering power .
• If the direction of the current is reversed, then the power delivered will be negative, i.e., the voltage source is actually receiving power. A typical example is the rechargeable battery in your car that works in either of the two states.
• Current source
• If the polarity of the voltage across a current source is such that the head of the arrow of the current source is at high potential (+) and the tail of the arrow is at low potential (-), then the polarity of the voltage and the current is consistent, and the current source is delivering power .
• If the direction of the current is reversed, then the power delivered will be negative, i.e., the current source is actually receiving power.

Example 3:

The current in a circuit composed of an ideal voltage source and a resistor is . The power consumption of the resistor and voltage source are and , respectively. The negative value of indicates the power is actually not consumed but generated by the voltage source (converted from other forms of energy, e.g., chemical, mechanical, etc.)

Example 4:

In the circuit shown below, the ideal current source is , and the ideal voltage source is , the resistor is . Find the current through and voltage across each of the three components. Find the power delivered, absorbed, or dissipated by each of the three components.

The current source provides current through the left branch (upward), while the voltage source provides across all three components. The current through is (downward), the current through the voltage source is . Therefore

• Power dissipated by is ,
• Power received by voltage source is .
• Power delivered by current source is .

(Homework) Redo the above with the polarity of reversed. Find:

• Power dissipated by :
• Power received/delivered by voltage source:
• Power delivered/received by current source:
Verify your results so that total power delivered is equal to total power received and dissipated.

Comment: While various voltage sources such as batteries are very common in everyday life, current sources do not seem to be widely available. One type of current source is solar-cell, which generates current proportional to the intensity of the incoming light. Also, certain transistor circuits are designed to output constant current. Moreover, as discussed above, any voltage source can be converted into a current source. For example, a current source with mA and can be implemented by a voltage source of in series with .

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

while the internal resistance can be obtained as the ratio of the open-circuit voltage to the shirt-circuit current , which can be measured by an ammeter:

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.

Next: Source and Load Up: Chapter 1: Basic Quantities Previous: Kirchhoff's Laws
Ruye Wang 2014-03-04