Understanding circuit solving with Kirchhoff's Laws is crucial to everything that follows in this course.
To completely "solve" a circuit, we must know the voltage across and the current through each element. The technique is as follows:
We will employ the process above to completely solve the circuit shown in Figure 1 below.
Figure 1. Circuit to be solved.
Step 1a is to assign currents to every element. This is shown in Figure 2, below.
Figure 2. Circuit with currents defined.
The current in the 6 V battery and the 9 Ω resistor is already known to be 3 A, so it is not necessary to define a current in that branch. Note the source in the upper branch. This is a dependent voltage source. Its voltage is dependent on the current in the center branch. Although it is dependent on a current, it is still a voltage source, not a current source. It is therefore necessary to define a current in the branch with the dependent source. The direction of current I8 is arbitrary. The current may be defined in either direction.
Step 1b is to assign voltages, as shown below in Figure 3.
Figure 3. Circuit with currents and voltages defined.
For each of the resistors, a voltage is assigned. The polarity of each voltage is chosen with the + sign where the current comes in. A voltage must also be assigned to the 3 A current source. The polarity of this voltage is arbitrary.
Step1c requires that essential nodes be identified circled and labled. This is shown below in Figure 4. Step 1d is to identify and label meshes. This is also done in Figure 4.
Figure 4. Circuit with all necessary labels.
Now that the circuit is completely labeled, it's time to write the equations. Step 2 calls for writing the Ohm's Law equations. We write one for each resistor. These are shown below.
V4 = 4I4
V8 = 8I8
V9 = 3×9
In step 3, we write the Kirchhoff's Current Law Equations. There are two essential nodes. We write one fewer equation than there are nodes, so we need only one equation. We can write it at either node A or node B. Let's use node B:
3 + I4 - I8 = 0
In step 4, we write Kirchhoff's Voltage Law Equations. There are two meshes. We write one equation for each mesh:
Mesh 1: V8 - 5I4 + V4 = 0
Mesh 2: V9 - V4 + 6 - V3 = 0
In step 5, we count the equations and unknowns. There are 6 equations and 6 unknowns (V3, V4, V8 , V9, I4, I8), therefore we can be confident that we can solve the circuit.
Solving the circuit (step 6) yields the following results:
V3 = 46.714 V
V4 = -13.714 V
V8 = -3.429 V
V9 = 27 V
I4 = -3.429 A
I8 = -0.429 V
Most of the results were negative. Choosing the other direction for I8 would have resulted in additional positive values, but that doesn't matter.
Step 6, above, "Solve the set of simultaneous equations," can be greatly simplified by use of modern tools. High-end calculators can solve sets of simultaneous equations. If you have such a calculator, you should learn to use this feature. Computer algebra systems such as Maple can also be used to solve simultaneous equations.
Below are five links to simulations of Kirchhoff's Laws. They were
created by Sergey Kiselev and Tanya Yanovsky-Kiselev.
Double loop with three sources
Double loop with three resistors
Double loop with three resistors and two sources
Double loop with three resistors and three sources
Try this link: http://www.article19.com/shockwave/oz.htm for a site that allows you to design and build a circuit with resistors, light bulbs, ammeters, voltmeters, etc. It was created by the Article 19 Group. It requires the Shockwave plugin.
At this time you should complete Tutorial 3 on circuit solving with Kirchhoff's Laws.