# Linear Circuit Theory: Matrices in Computer Applications by Jiri Vlach

By Jiri Vlach

This finished textbook covers all topics on linear circuit thought, with the emphasis on studying the topic with no quite a lot of details. This new angle stresses wisdom instead of laptop use to begin and differs from different books via introducing matrix algebra early within the booklet. The book’s 290 difficulties are supposed to be solved utilizing matrix algebra, which supplies the reader with a robust origin on which to construct.

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Extra resources for Linear Circuit Theory: Matrices in Computer Applications

Example text

A planar network is such that we can draw its diagram on paper without any connection crossing some other connection. The two networks we considered so far were clearly planar. The network in Fig. 3a seems not to be planar, but we can redraw it as shown in Fig. 3b and discover that it actually is planar. The network in Fig. 4 is not planar because some line will cross another line no matter how we redraw the network. Why did we use so much space to discuss planar networks? Because only planar networks can be analyzed by the method we will discuss here.

3. The bottom line is the reference node (ground), always assumed to have zero voltage. Assign voltages to the nodes which are not grounded. Normally we use subscripted variables, V1 and V2. The KCL states that the sum of currents leaving a node is zero. Consider the first node. We do not know the voltage V1, but if we assume that it is the largest voltage in the network, then the currents through all resistors connected to the node must flow away from the node. Since the first resistor is connected to ground, its current will be, as follows from the Ohm’s law, IG1 = G1V1.

Once we make this assumption, all currents through the resistors must flow away from node 2. The voltage across the second resistor is now V2 – V1 (because of our assumption we must put first the voltage of the second node and subtract the voltage of the first node). The current through this resistor will be IG2 = (V2 – V1)G2. The current through the third resistor must flow down and IG3 = G3V2. 2) Rearrange the terms in both equations and transfer the independent source to the right, with a change of sign.