volts. This amplifier would put out  only 1 watt of RMS power into an 8 ohm load. Still, if you have a pair of horn speakers, 1 watt might be all you need. 

Free Transformers (Sort of)
    In a previous issue we saw how twenty 6.3 volt heaters could be strung together like Christmas lights and plugged directly into the wall. The disadvantage is, of course like the Christmas lights, that if one heater opens, all the heaters stop working. This trick has some merit, although 20 tubes is a bit much. Still, twenty 6BQ5s in a push-pull amplifier would yield about 100 watts of Class A power and would probably be the best sounding 100 watt amplifier in existence. However, since only amplifiers with one 2A3 or one 300B are of any merit these days, we will drop this subject.
    The problem remains: how do we use fewer than twenty tubes and still power the heater string from the wall socket directly? The easy answer is to place a high wattage resistor in series with the heater string. This solution has promise, if you live in Canada and it is a particularly cold winter; otherwise, the heat wasted by the resistor will be excessive. No, the better solution is to use a reactive element in series with the heater string.

     Capacitors and inductors are reactive elements in that they oppose a flow of AC current. Both will limit the voltage across a heater string if placed in series with it. Since our aim was to eliminate iron, the capacitor is the better choice. A capacitor's reactance is equal to:
    Xc = 1 / (2piFC)
or
    Xc = 159155 / (FC).
    Our path seems clear enough: find the equivalent resistor value by using Ohm's law and place that value in the previous equation and we will arrive at our capacitor value except that it leads to the wrong answer. Remember, reactance is a complex or phasor quantity that can be fully defined by specifying a magnitude and a "direction" or phase angle. This something like the difference between speed, a scalar quantity, and velocity, a vector quantity. If you have ever built a simple high-pass filter for a tweeter, you have had some experience with (the capacitor is a reactive device) a reactance's  voltage division with a resistance (the speaker). A capacitor is chosen whose reactance equals the nominal resistance of the tweeter at the crossover frequency. Yet the tweeter is only -3 dB down at the crossover frequency not the -6 dB as expected. Where did we go wrong? We ignored the phase angle in the equation. 

     Rather than lose any more readers to either boredom from remembrances of high school trigonometry or to fear from remembrances of high school trigonometry, we will get to the punch line:
    C =                             1                           ,
             2 x ¶ x Fwall x \/¯(Vwall² - Vh²)
                                                Ih   
where
     C is in Farads,
     Fwall = the wall frequency
     Vwall = the wall voltage