In general,
**Fig.no.1: **Terminal characteristics of an ideal voltage source
**Fig.no.2: **Terminal characteristics of an ideal current source
As shown in Fig.no.1, an ideal voltage source provides a fixed voltage
to the network no matter the level of current drawn from the supply.
Take note in Fig.no.1 that at every level of current drawn from the
supply the terminal voltage of the battery is still E volts. The current
source of Fig.no.2 will establish a fixed level of current that will
define the resulting terminal voltage of the attached network as shown in
Fig.no.2. Note that the symbol for a current source includes an arrow
to show the direction in which it is supplying the current. Take special
note of the fact that the current supplied by the source is fixed no matter
what the resulting voltage is across the network. A voltage source and
current source are often said to have a dual relationship. The term dual
reveals that what was true for the voltage of one is true for the current of
the other and vice versa.
Furthermore,
A few examples will demonstrate the similarities between solving for
the source current of a voltage source and the terminal voltage of a current
source.
**Example 1: **Find the source voltage, the voltage $V_1$, and current $I_1$
for the circuit in Fig.no.3.
**Fig.no.3: **Circuit for Example 1.
**Solution: **Since the current source establishes the current in the branch
in which it is located, the current $I_1$ must equal I, and
$$I_1 = I = 10 mA$$
The voltage across R1 is then determined by Ohm's law:
$$V_1 = I_1 R_1 = (10 mA)(20 kΩ) = 200 V$$
Since resistor R1 and the current source are in parallel, the voltage across
each must be the same, and
$$Vs = V_1 = 200 V$$
with the polarity shown.
**Example 2: **Find the voltage Vs and currents $I_1$ and $I_2$ for the network
in Fig.no.4.
**Fig.no.4: **Circuit for Example 2.
**Solution: **This is an interesting problem because it has both a current
source and a voltage source. For each source, the dependent (a function
of something else) variable will be determined. That is, for the current
source, Vs must be determined, and for the voltage source, Is must be
determined.
Since the current source and voltage source are in parallel,
$$Vs = E = 12 V$$
Further, since the voltage source and resistor R are in parallel,
$$V_R = E = 12 V$$
and
$$I_2 = {V_R \over R} = {12 V \over 4Ω}= 3 A$$
The current $I_1$ of the voltage source can then be determined by applying
Kirchhoff's current law at the top of the network as follows:
$$\sum {I_i} = \sum {I_o}$$
$$I = I_1 + I_2$$
and
$$I_1 = I - I_2 = 7 A - 3 A = 4 A$$

Because the current source is not a typical piece of laboratory equipment
and has not been employed in the analysis thus far, it will take
some time before you are confident in understanding its characteristics
and the impact it will have on the network to which it is attached. For the
moment, simply keep in mind that a voltage source sets the voltage
between two points in a network and the other parameters have to
respond to the applied level. A current source sets the current in the
branch in which it is located and the other parameters, such as voltages
and currents in other branches, have to be in tune with this set level of
current.