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

Energy Sources

Ideal Energy Sources:

VIsources.gif

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

Voltage Source:

In reality, all voltage sources (e.g., a battery or a voltage amplifier circuit) have a nonzero internal resistance $R_0>0$ in series with the voltage source that causes the actual output voltage $V$ to be lower than $V_0$, depending on the load current $I$. The load voltage $V$ and the load current $I$ are constrained by the following two relationships imposed by the voltage source $(V_0, R_0)$ and the load $R_L$, respectively:

\begin{displaymath}
\left\{ \begin{array}{ll}
V=V_0-IR_0 & \mbox{(source)}  V=R_LI & \mbox{(load)} \end{array} \right.
\end{displaymath}

Solving for $I$ and $V$, we get:

\begin{displaymath}\left\{ \begin{array}{l}
I=V_0/(R_0+R_L)  V=V_0 R_L/(R_0+R_L) \end{array} \right.
\end{displaymath}

voltage_source.gif

The internal resistance $R_0$ of a voltage source should be as small as possible, ideally $R_0=0$.

voltagesource1.gif

The slope of the first curve is the internal resistance $R_0$ and the slope of the second curve is $R_L$. Solving these two equations we get load voltage $V$ and current $I$.

Only in the case of an ideal voltage source with $R_0=0$ will $V=V_0$. For $R_0>0$, the heavier the load, i.e., the smaller $R_L$, the larger the load current $I$, and the lower the load voltage $V$:

\begin{displaymath}R_L \downarrow \Longrightarrow I \uparrow \Longrightarrow R_0I \uparrow
\Longrightarrow V \downarrow \end{displaymath}

In particular, when the load is a short circuit ($R_L=0$), the output voltage of a battery is zero (instead of the specified $V_0=1.5V$), as the voltage drop across the nonzero internal resistance $R_0$ is the same as $V_0$, 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 $R_0<\infty$ in parallel to the current source that causes the actual output current $I$ to be lower than $I_0$, depending on the load voltage $V$. The load voltage $V$ and the load current $I$ are constrained by the following two relationships imposed by the current source $(I_0, R_0)$ and the load $R_L$, respectively:

\begin{displaymath}
\left\{ \begin{array}{ll} I=I_0-V/R_0 & \mbox{(source)} \\
...
...y}{l} V=I_0R_0-IR_0=(I_0-I)R_0  I=V/R_L \end{array} \right.
\end{displaymath}

Solving for $V$ and $I$, we get

\begin{displaymath}
\left\{ \begin{array}{l} V=I_0R_0R_L/(R_0+R_L)=I_0\;R_0\vert\vert R_L \\
I=I_0R_0/(R_0+R_L)\end{array} \right.
\end{displaymath}

current_source.gif

The internal resistance $R_0$ of a current source should be as large as possible, ideally $R_0=\infty$.

currentsource1.gif

The slope of the first curve is the internal resistance $R_0$ and the slope of the second curve is $R_L$. Solving these two equations we get load voltage $V$ and current $I$.

Only in the case of an ideal current source will $R_0=\infty$ and $I=I_0$. For $R_0<\infty$, the larger the load resistance $R_L$, the smaller the current $I_L$.

\begin{displaymath}R_L \uparrow \Longrightarrow V \uparrow \Longrightarrow V/R_0 \uparrow
\Longrightarrow I \downarrow \end{displaymath}

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.

sourcetransform.gif

Comparing the two equations $V=V_0-IR_0$ and $V=I_0R_0-IR_0$ describing the voltage and current sources respectively, we see that a voltage source can be equivalently converted into a current source and vise versa:

Comparing the two equations above, we see that if we let $R_0=R'_0$ and $V_0=R_0I_0$ or $I_0=V_0/R_0$, 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

\begin{displaymath}\left\{ \begin{array}{l} R_0=R'_0  I_0=V_0/R_0
\;\;\;\mbox{or}\;\;\; V_0=I_0 R_0 \end{array} \right. \end{displaymath}

Finding internal resistance:

The internal resistance $R_0$ of a voltage source is the slope of the straight line $V=V_0-IR_0$, while the internal resistance $R_0$ of a current source is the slope of the straight line $I=I_0-V/R_0$. In both cases this slope can be obtained as the ratio of the open-circuit voltage $V_{oc}$ (when $I=0$) and the short-circuit current $I_{sc}$ (when $V=0$):

\begin{displaymath}R_0=\frac{\mbox{open-circuit voltage}}{\mbox{short-circuit current}}
=\frac{V_{oc}}{I_{sc}} \end{displaymath}

While this method can be used to find $R_0$ 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 $V_i$ and $I_i$ ($i=1,2$) associated with each of two load resistors $R_1$ and $R_2$. Then the internal resistance $R_0$ can also be found as the slope of the straight line determined by the two points $(V_1, I_1)$ and $(V_2, I_2)$. We see that the previous method can be considered as a special case when $R_1=\infty$ (open circuit) and $R_2=0$ (short circuit).

InternalResistance.gif

Example 1:

A voltage source of $V_0=10V$ and $R_0=10\Omega$ can be converted to a current source of $I_0=V_0/R_0=1A$ with the same $R_0$ (and vice versa). For a load of $R_L=40\Omega$, both energy sources will provide the load current $I_L=0.2A$ and load voltage $V_L=8V$. As $R_0$ is low, this is a good voltage source but a poor current source.

sourceex1.gif PowerSources1.gif

Example 2:

A current source of $I_0=1 mA$ and $R_0=8 K\Omega$ can be converted to a voltage source of $V_0=I_0 R_0=8V$ with the same $R_0$ (and vice versa). For a load of $R_L=2 K\Omega$, both energy sources will provide the load current $I_L=0.8 mA$ and load voltage $V_L=1.6V$. As $R_0$ is high, this is a good current source but a poor voltage source.

sourceex2.gif PowerSources2.gif

Power Delivery/Absorption

Example 3:

The current in a circuit composed of an ideal voltage source $V_0=5 V$ and a resistor $R=5\Omega$ is $I=V/R=1 A$. The power consumption of the resistor and voltage source are $W_R=I^2R=V^2/R=IV=5$ and $W_V=-IV=-5$, respectively. The negative value of $W_V$ 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 $I_0=1A$, and the ideal voltage source is $V_0=2V$, the resistor is $R=3\Omega$. 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.

powerdeliveryexample.gif

The current source provides $I_0=1A$ current through the left branch (upward), while the voltage source provides $V_0=2V$ across all three components. The current through $R$ is $2V/3\Omega=2/3A$ (downward), the current through the voltage source is $I_v=1-2/3=1/3A$. Therefore

(Homework) Redo the above with the polarity of $V_0$ reversed. Find:

Verify your results so that total power delivered is equal to total power received and dissipated.

Answer

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 $I_0=1$ mA and $R_0=1\;M\Omega$ can be implemented by a voltage source of $V_0=1000\; V$ in series with $R_0=1\;M\Omega$.

Example 0 (Homework)

sourcemeter.gif

A realistic voltage source (e.g., a battery) can be modeled as an ideal voltage source $V_s$ in series with an internal resistance $R_s$. Ideally, the voltage $V_s$ can be obtained by measuring the open-circuit voltage $V_{oc}$ with a voltmeter

\begin{displaymath}V_s=V_{oc} \end{displaymath}

while the internal resistance $R_s$ can be obtained as the ratio of the open-circuit voltage $V_{oc}$ to the shirt-circuit current $I_{sc}$, which can be measured by an ammeter:

\begin{displaymath}R_s=\frac{V_{oc}}{I_{sc}} \end{displaymath}

However, in reality, any voltmeter has an internal resistance $r_v$ in parallel with the meter, any ammeter has an internal resistance $r_a$ in series with the meter. For better measurement accuracy, should $r_v$ be small or large, how about $r_a$? Why? Give the expression of the measured open-circuit voltage $V$ and short-circuit current $I$ in terms of the true $R_s$ and $V_s$, as well as $r_v$ and $r_a$.

Design a method to obtain the true source voltage $V_s$ and source resistance $R_s$ by a voltmeter and an ammeter with known $r_v$ and $r_a$. Give the expression of $V_s$ and $R_s$ in terms of the measured open-circuit voltage $V$, short-circuit $I$, and $r_v$ and $r_a$.

Now Assume $V_s=6V, R_s=100\Omega, r_v=10,000 \Omega, r_a=200 \Omega$ what are the measured open-circuit voltage $V$, and the short-circuit current $I$? Given $V$, $I$, and the known $r_v$ and $r_a$, how do your get the true $V_s$ and $R_s$ 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 $R_s$ and $V_s$ of a voltage source, we can measure the voltage $V_L$ across two different load resistors $R_L$ connected to the voltage source. If the values of $R_L$ are significantly smaller than that of the internal resistance $r_v$ of the voltmeter, the voltmeter can be considered to be ideal with $r_v\rightarrow \infty$.

Assume when the load resistor is $R_L=1 k\Omega$, the voltage across it is found to be $V_L=9.1V$, then a different load $R_L=2 k\Omega$ is used and the voltage across it is $V_L=9.52V$. Find $V_s$ and $R_s$ of the voltage source.

Answer


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