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x-rays generated depends almost entirely on the x-ray tube potential, and the number produced is proportional to Z of the target, the square of the kilovoltage ((kVp)2), and the product of x-ray tube current (mA) and the exposure time (in seconds), or mA s. Increasing the mA s if kVp is held constant, as shown in Figure 15-12, increases the number of x-rays emitted; however, the distribution of x-ray energies remains essentially the same.

/i sc , since the ratio is indetermin ate, We must look for another approach Notice th at if Voc = 0, then the equi va lent circuit is merely the un know n resistance R Th If we app ly an ex ternal source to circuit A- a test source v,-and determine the current, ill which flows into circ uit A fro m VII the n RTh can be determined fro m RTh := v,/it Although the numerical value of v, need not be specified, we could let VI = I V and then RTh = I / i( Altern at ively, we could use a current source as a test source and let it := I A; then vt = ( 1)RTh .

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6.3 COMPACT MICROSTRIP ANTENNAS WITH ACTIVE CIRCUITRY An ampli er-type active microstrip antenna integrates a two-port ampli er circuitry into a passive microstrip antenna and can work as a transmitting or receiving antenna [4, 5, 9]. In this section, the con guration of an ampli er-type active microstrip antenna for enhanced antenna gain and wider impedance bandwidth is presented. The active microstrip antenna studied here consists of a passive microstrip antenna and ampli er circuitry with a DC bias feedback resistor. With the selection of a possible maximum feedback resistance in the active circuitry for biasing the ampli er into

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For copper surfaces in the 400 MHz region, this expression gives Q of order lo4 for the geometry that we are using. The Q is one figure of merit for a cavity. Another has to do with seeking high energy gain per unit loss. The ratio of the square of the energy gain to the loss rate has the dimensions of resistance; this quantity is called the shunt impedance of the cavity, R,:

6

Before we begin our analysis of several examples that will demonstrate the uti lity of these theorems, remember th at these theorems, in addi tion to being another approach, often permi t us to solve several small problems rather than one large one They allow us to replace a network, no matter how large, at a pair of tel1l1ina/s with a Thevenin or NOIlon equi valent circui t In fact, we could represent the entire US power grid at a pair of terminals wi th one of the equivalent circui ts_ Once this is done, we can quickly analyze the effect of differe nt loads on a network Th us, these theorems provide us with additional insight into the operation of a specific network..

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