Source: http://raise.spd.louisville.edu/ECE252/L1.htm
Timestamp: 2019-04-23 08:12:41+00:00

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The above equation is the definition of voltage. v is voltage, w is work (energy) in joules, and q is charge in coulombs. Voltage is work done per unit charge. Separating charges requires work. for simple cases (DC), we can use V = w/Q and w = VQ.
The above equation relates current to charge. i is current in amps, q is charge in coulombs, and t is time in seconds. Current is nothing more than the flow of charge. Note that current is defined as the flow of positive charges. For simple cases (DC), we can use I = Q/t and Q = It.
This is Ohm's Law. v is in volts, R is resistance in ohms, and i is current in amps. It's an important equation; commit it to memory.
This is the basic power equation. p is power in watts, w is work (energy) in joules, and t is time in seconds. Don't confuse the symbol w (work) with the power unit W (watts). Power can be positive or negative. When a device absorbs power, the power is positive. When it provides power (as with a battery), the power is negative. For simple cases (DC), we can use P = w/t and w = Pt.
This gives us the familiar power equation p = vi. Lower case is used to show that this equation is true for time-varying voltages and currents as well as DC voltages and currents.
This equation is useful when we know the current through the resistor.
This equation is especially useful when we know the voltage across a resistor. Take care when you use this equation. Make sure the voltage you use is the actual voltage across the resistor and not some voltage elsewhere in the circuit.
There are three ideal passive circuit elements: resistors, inductors, and capacitors. Resistors obey Ohm's Law, which is discussed above. We will leave the discussion of inductors and capacitors for another lesson.
An important point about passive circuit elements in general, and resistors in particular, is that current flows into the positive end, as shown below.
The Ohm's law equation, v = ir, assumes that current flows into the + end as shown.
There are two ideal active circuit elements: voltage sources and current sources.
The one on the left is the standard representation for a DC voltage source. The one on the right is, technically, the symbol for a battery. For our purposes, we can consider both symbols to represent general, ideal, DC voltage sources. An ideal voltage source produces a defined voltage, regardless of the current magnitude and direction. For example, an ideal 6 V battery has 6 V across it if the current is 0 A, 100 A, or -1,000 A. Ideal voltage sources do not obey Ohm's Law. The voltage of a real voltage source will vary slightly depending on the current, because the real device has internal resistance. Unless otherwise specified, we will always assume our circuit elements are ideal. Also note that current can flow either way through a voltage. It supplies energy when current flows out of the positive end; it absorbs energy ("charging" the battery) when current flows into the positive end.
An ideal current source is shown below.
This example is a 3 A current source. It will produce 3 A in the direction of the arrow, regardless of the magnitude and polarity of the voltage across it. For example, it could have 30 V across it or -700 V across it; that will depend on the circuit it is connected to. For example, if the 3 A current source is connected to a 50 Ω resistor, its voltage from Ohm's Law will be 3 × 50 = 150 V. Current sources do not obey Ohm's Law.
In addition to the independent sources shown above, there are also dependent sources, those that depend on what else is happening in the circuit. An ideal dependent voltage source is shown below.
The voltage vs may perhaps be 5V6 where V6 is a voltage elsewhere in the circuit; if V6 happens to be 3 V, then vs would be 5 × 3 = 15 V. vs doesn't have to be dependent on a voltage. For example, it could have a value of 2i7, where i7 is 2.5 A. In this instance, vs would be 2 × 2.5 = 5 V. Yes, although vs is dependent upon a current, it is still a voltage.
The device below is an ideal dependent current source.
The current is may perhaps be 3V2 where V2 is a voltage elsewhere in the circuit. If V2 is 7 V, is would be 3 × 7 = 21 A. The current does not have to be dependent on another current.
Be careful when dealing with dependent sources in circuit equations. If it's got a + sign inside, it's a voltage source, even if it's dependent on a current. If it has an arrow inside, it's a current source, even if it is dependent on a voltage.
Kirchhoff's Voltage Law and Kirchhoff's Current Law are powerful tools in circuit analysis. Before we state these laws, we must understand some terminology. Consider the circuit below.
Node: A node is a point in a circuit where two or more circuit elements join. Points A, B, C, and D are nodes.
Essential Node: An essential node is a point in a circuit where three or more circuit elements join. Points B and D are essential nodes. We usually care more about the essential nodes than about the simple nodes.
Branch: A branch is a portion of a circuit that connects two nodes. AB,AD, BD, BC, CD, DAB, and BCD are branches.
Essential Branch: An essential branch is a portion of a circuit that connects two essential nodes. BD, DAB, and BCD are essential branches.
Loop: A loop is a closed path. ABDA, BCDB, and ABCDA are loops.
Mesh: A mesh is a loop with no loops inside it. ABDA and BCDB are meshes. We usually choose to write Kirchhoff's Voltage Law equations for meshes rather than loops.
There is a subscripting convention for naming voltages. VAB is the voltage across the 100 Ω resistor. It is also the voltage of point A with respect to point B. It is positive if point A is more positive than point B. It is negative if point B is more positive than point A. For example if a 2 A current were flowing from A to B, VAB would be 200 V. If a 4 A current were flowing from B to A, VAB would be -400 V.
What we mean is that we'll count the currents going out of the node (i.e. leaving the node) as positive and those entering as negative. This is just a convention, but we will stick to it.
The current entering the node (7 A) was taken as negative. The others were leaving the node, so they were positive.
To determine the number of KCL equations you need to solve a circuit, count the essential nodes and subtract one. For example, if there are 5 essential nodes, you will need 4 KCL equations.
What we mean is that we'll count the voltage drops as positive and the voltage rises as negative. This is just a convention, but we will stick to it.
A portion of a circuit is shown below.
Going clockwise, starting from point A, the first circuit element encountered is the 40 V source. It is entered in the KVL as positive, because voltage drops (from + to -) are taken as positive. Some people remember this rule by noting that the + sign is the first thing you come to. Next comes the 20 Ω resistor; its voltage is 25 V, and it is also positive because it is a voltage drop. The 6 V battery is next, and its voltage is entered as negative because it is a voltage rise. The gap is next. There will be no current, but certainly there may be a voltage across the gap. There can be a voltage across anything: resistors, inductors, batteries, bananas, gaps ... anything. After the gap we come to the 2 A current source with a voltage rise (therefore negative) of 18 V. This completes the loop from point A to point A, and we set the KVL equal to zero. We then solve for the unknown Vgap.
Although Kirchhoff's Voltage Law will work for any direction chosen, for consistency, we will always go around loops in the clockwise direction, and we will count voltage drops as positive.
Here is a demo of how voltage, resistance, and current are related: http://micro.magnet.fsu.edu/electromag/java/ohmslaw/.
To determine the number of KVL equations you need to solve a circuit, count the meshes. For example, if there are 5 meshes, you will need 5 KVL equations.
KCL can be extended beyond just junction points in circuits. It will work for any portion of a circuit. Consider the circuit below.
We don't need to know the values of the resistors, the voltage of the battery, or the current of the other current source to solve for I.
Any part of a circuit can be circled as a supernode. The circuit elements inside the supernode will not contribute to the resulting KCL equation. The only currents to use in the KCL will be those that interface with the supernode.
I = V/R = (Va - Vb)/R = (500 - 600)/20 = -5 A.
At this time you should complete Tutorial 1 on basic circuit laws.

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