**Ideal Independent Current Source: **In addition to ideal independent voltage sources, we will often come across ideal independent current sources in the circuits we analyze. While many students find current sources a bit less intuitive than voltage sources, as you will see, the key point to remember is that an ideal independent current source will guarantee a specified *current * regardless of the voltage drop that may develop across the current source. A frequent error beginning students make is to assume there is zero voltage drop across a current source. While possible, a potential difference of zero volts across the element would indicate that the current source is neither providing nor absorbing power. The circuit symbol for a generic ideal independent current source is shown below in Figure 1.

An ideal independent current source provides a specified current regardless of the circuit to which it is connected. That is, the current of the current source is independent of the potential difference across the source. Consider Figure 2 which shows an ideal 0.5 A ideal independent current source connected to a hypothetical circuit.

The current source of Figure 2 is assumed to be ideal. Since it is an *independent* current source, we are guaranteed that the current in the source is 0.5A in the direction suggested by the arrow within the source's circuit symbol. What we don't know about the current source is the voltage across it. That is, we do not know whether node A or B is higher in potential and by how much. Why do we care about this potential difference? It may turn out that we need to know this voltage to determine other voltages and currents within the circuit. Also, to determine the value of the power associated with the current source we need to know the potential difference across it as recall that P=VI. In the next tab, we will examine the current-voltage characteristic of an ideal independent current source to better understand the role of the potential difference that exists across an ideal independent current source. Before moving on however, click on the circuit of Figure 2 to toggle between the possible cases of the voltage drop across the current source. In all cases shown in Figure 2, the voltage is assumed positive as drawn.

Just as in the case of an ideal independent voltage source, it may be helpful to think about the currrent-voltage (I-V) characteristic of an ideal independent current source. Recall that an I-V characteristic is a plot of the current associated with an element as a function of the voltage drop across the element. An ideal independent current source provides a specified *current* regardless of the potential difference across the terminals of the current source. Thus, as shown in Figure 3, the I-V characteristic of an independent current source is a horizontal line that cuts through the current axis at a value equal to I_{s} -- the value of the current source. The I-V characteristic tells us that when the source is placed in a circuit, the actual potential difference across its terminals will depend on the circuit as a whole and that we are only guaranteed that the current in the current source is I_{s}.

An important point to be made with regard to Figure 3 is that the voltage polarity is assumed to be in accord with the passive convention. When we sketch an I-V characteristic we must be clear about the assumed polarity of voltage and direction of current. Why? Consider the case of Figure 4 that displays the I-V characteritic of a 1.5A ideal independent current source.

On the left of the figure we see the circuit symbol of a 1.5A ideal independent current source in which the voltage polarity is drawn in accord with the passive convention. The I-V diagram on the right of the figure is the expected horizonatal line cutting through the current axis at 1.5A. If the potential difference across the current source is positive as drawn, that is current is directed from higher to lower potential through the current source, the source is absorbing energy. If on the other hand, the current is directed from lower to higher potential through the source, the voltage is negative as drawn and we are considering the *second* quadrant of the I-V characteritic. In this case, the source is supplying energy to the circuit. To visualize this more clearly, click on various sections of the line in Figure 4 to see whether the current source provides or absorbs energy. If you click on the point V=0, I=1.5A, you should see that under this condition, the source neither provides nor supplies power.

The important point to be made is that an ideal independent current source will provide a specified current, in a specified direction as given by its arrow, regardless of the potential difference across the current source. This point will be made clear when we consider a couple of examples.

Shown in Figure 5 is a 0.2A current source, the same current source connnected in a simple series circuit with a 100 ohm resistor, and the current-voltage characteritic of the current source. As expected, the I-V characteristic of the current source is a horizontal line with a value of 0.2A. When the current source is placed in a circuit, the current source has a specfic *operating point* on its current-voltage characteristic. The operating point is defined as a specific value of voltage and current. Since the current source is an ideal independent current source, we know its current is 0.2A, but what is its voltage?

The simple circuit in which the current source has been placed is a single loop with the current source and a 100 ohm resistor. Since the current is the same everywhere in this single-loop circuit, the resistor must be carrying the 0.2A of current. Looking at the direction of the arrow within the current source, we know that the current is directed in a clockwise fashion in the circuit. From our study of the passive convention, we recall that resistors are passive elements and current is **always** directed through a resistor from high to low potential. Finally, Ohm's law tells us that if there is 0.2A of current in a 100 ohm resistor, the voltage drop across the resistor must be 20V. In Figure 6 we combine these observations to redraw the circuit labeling voltages and currents.

In comparing the circuit diagrams in Figure 6 we see that the actual voltage drop across the current source is opposite that assumed in creating the current-voltage charactersitic of Figure 5. For this reason, in identifying the operating point of the current source we take the point V=-20V. Therefore, the current source is operating in the second quadrant -- a click on Figure 5 will show this. Looking back to the previous tab, this implies that the current source is providing power in this particular circuit. This should be unsurprising as the only other element in the circuit is a resistor and resistors only dissipate power. In the next example, we take up a more complicated case that will require us to use our understanding of the models of both an ideal independent voltage source and ideal independent current source as well as Ohm's law. You may find the example conceptually challenging; a solid understanding of the example will position you well for mastering the content to come.

Consider now the circuit of Figure 7 that consists of a parallel combination of an ideal independent voltage source, an ideal independent current source and a resistor. We want to determine the operating point of each of the three elements. To do so will require a proper understanding of the models of the three elements, to recall what elements in parallel share in common, and to correctly apply KCL. Perhaps you are ready to solve the problem on your own. Give it a try -- if you need help there will be plenty of hints to lead you to the correct understanding.

*What is the potential difference across each element in the circuit of Figure 7?*