# conductance Electronics Help

The value of can be found in Figure 9-6 by recognizing that it equals the output conductance of the network, that is, the reciprocal of the resistance looking into the output terminals, Since one side of the 60-0 resistor is open-circuited, the output resistance is clearly 40 0, and Find the h parameters of the network shown in Figure 9-7. Note that the capacitor impedance 0 n, or 500/ -9000, at the frequency for which the analysis is to be performed. The val lies found for the II parameters will therefore be valid  only a hat one frequency of operation. Solon. When the output of Figure 9-7 is shorted, the capacities reactant is in parallel with the 1-kO resistance. See Figure 9-8. Therefore,T be value of Jill can be found by assuming that a voltage Vz is applied across the output terminals of Figure 9-7. Since the input terminals are open-circuited whee compelling, the voltage VI will be the voltage developed across the capacitor, who can he found using the voltage-divider rule Finally is the output admittance of the network with the input open-circuited. The output impedance under that condition is clearly the sum of 1000 ohms of resistance and 500 ohms of capacities reactant, so We emphasize again that the h-parameter values computed in this example are only valid at the frequency for which Xc = 500 n. The h-parameter values for any network having reactive components will always depend on frequency.

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Posted on November 19, 2015 in h AND y PARAMETERS