The current source appearing in the previous section is called an ideal source
due to the absence of any internal resistance. In reality, all sources—whether
they are voltage sources or current sources — have some internal resistance
in the relative positions shown in Fig.no.1. For the voltage source, if
Rs = 0Ω, or if it is so small compared to any series resistors that it can be
ignored, then we have an "ideal" voltage source for all practical purposes.
For the current source, since the resistor $R_P$ is in parallel, if $R_P = \infty$, or if it is large enough compared to any parallel resistive elements that it can be ignored, then we have an ideal current source.
**Fig.no.1: **Practical sources: (a) voltage; (b) current.
Unfortunately, however, ideal sources cannot be converted from one
type to another. That is, a voltage source cannot be converted to a current
source, and vice versa—the internal resistance must be present. If
the voltage source in Fig.no.1(a) is to be equivalent to the source in Fig.no.1(b), any load connected to the sources such as $R_L$ should receive the
same current, voltage, and power from each configuration. In other
words, if the source were enclosed in a container, the load $R_L$ would not
know which source it was connected to.
**Fig.no.2: **Source conversion.
This type of equivalence is established using the equations appearing in
Fig.no.2. First note that the resistance is the same in each configuration—a
nice advantage. For the voltage source equivalent, the voltage is determined
by a simple application of Ohm's law to the current source: $E = I R_p$. For
the current source equivalent, the current is again determined by applying Ohm's law to the voltage source: I = E/Rs. At first glance, it all seems too
simple, but Example 1 verifies the results.
The internal characteristics of each are quite different.
**Example 1: **For the circuit in Fig.no3:

a. Determine the current $I_L$.

b. Convert the voltage source to a current source.

c. Using the resulting current source of part (b), calculate the current through the load resistor, and compare your answer to the result of part (a). **Fig.no.3: **Practical voltage source and load for Example 1.
**Solution: **

a. Applying Ohm's law gives $$I_L = {E \over Rs + R_L}$$ $$ = {6 V \over 2 Ω + 4 Ω} = {6 V \over 6 Ω} = 1 A$$ b. Using Ohm's law again gives $$I = {E \over Rs} = {6 V \over 2 Ω} = 3 A$$ and the equivalent source appears in Fig.no.4 with the load reapplied. **Fig.no.4: **Equivalent current source and load for the voltage source in Fig.no.3.
c. Using the current divider rule gives
$$I_L = {Rp\over R_p + R_L}(I)$$
$$ ={(2 Ω)(3 A) \over 2 Ω + 4 Ω} = {6A \over 6} = 1 A$$
We find that the current $I_L$ is the same for the voltage source as it was
for the equivalent current source—the sources are therefore equivalent.

a. Determine the current $I_L$.

b. Convert the voltage source to a current source.

c. Using the resulting current source of part (b), calculate the current through the load resistor, and compare your answer to the result of part (a).

a. Applying Ohm's law gives $$I_L = {E \over Rs + R_L}$$ $$ = {6 V \over 2 Ω + 4 Ω} = {6 V \over 6 Ω} = 1 A$$ b. Using Ohm's law again gives $$I = {E \over Rs} = {6 V \over 2 Ω} = 3 A$$ and the equivalent source appears in Fig.no.4 with the load reapplied.