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The FullWave Rectifier With Inductive LoadAuthor: E.E. Kimberly If the load were purely inductive on a fullwave rectifier, the current in the inductance would rise on the first conducting halfcycle on one tube to point m in Fig. 277 but would not fall again because it would be augmented by the current from the second tube on the succeeding halfcycle.
As a result the load current would continue to rise until some element of the circuit was destroyed or the rise was limited by some element other than the inductance. If, however, the load contains resistance in series with the inductance, as in Fig. 278, the current will rise only until the voltage drop across the resistance is equal to the dc component of the load voltage. The dc or average value of a halfwave of a sinusoid starting at zero is 0.636 times the maximum value, as derived on page 311. If the inductance of the load is large, the current through the load resistor and hence the voltage available between A and B will be nearly constant. A condenser connected across R will reduce still further whatever ripple remains. The foregoing statements are made on the assumption that the voltage drop through the tube is negligible.


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