By John D. Kraus, Keith R. Carver
This fourth version of the vintage electromagnetics textual content has been largely revised to provide balanced, thorough assurance of either static and dynamic electromagnetics for modern classes within the topic. It explains either the elemental conception and its software to useful designs and includes many laptop courses for working on a private computing device. Many new issues are offered together with center dipole; superconductors; wave absorption, fiber optics and dielectric resonators.
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1 mA, but its direction is opposite to that assumed. The voltage V bd can be calculated using either of the closed paths abdea or bcdb. l mA in either equation yields V bd = 10 V. 3mW Now from the standpoint of determining the voltage V be, we can simply add the sources since they are in series, add the remaining resistors since they are in series, and reduce the network to that shown in Fig. 20b. 8 Find I and V hd in the circuit in Fig. 8. 9 40kD ANSWER In the network in Fig. 9, if V ad is 3 V, find Vs.
1 Sa is equivalent to Fig. 1 Sb. In other words, the sum of several voltage sources in series can be replaced by one source whose value is the algebraic sum of the individual sources. This analysis can, of course, be generalized to a circuit with N series sources. Now consider the circuit with N resistors in series, as shown in Fig. 19a. Applying Kirchhoff's voltage law to this circuit yields v(t) = vR, + vR, + ... 19 Equivalent circuits. 15 which is the voltage-division property for multiple resistors in series.
Since all of the circuit elements are in parallel, the voltage v( t) appears across each of them. Furthermore, an examination of the circuit indicates that the current i( t) is into the upper node of the circuit and the currents i 1(t) and i 2 ( t) are out of the node. Since KCL essentially states that what goes in must come out, the question we must answer is how i 1( t) and i2 ( t) divide the input current i ( t). 21 Simple parallel circuit. 4 SINGLE-NODE-PAIR CIRCUITS 31 and, employing Ohm's law, we have .