1-21was a variation in plate current of 7.5 milliamperes. Instead of amplification, De Forest had obtained"conversion," or in other words, converted a signal voltage to a current variation. This wasn’t exactly whathe had in mind. As it stood, the circuit wasn’t very useful. Obviously, something was needed. Afterexamining the circuit, De Forest discovered the answer—Ohm's law. Remember E = I × R? De Forestwanted a voltage change, not a current change. The answer was simple:In other words, run the plate current variation (caused by the voltage on the grid) through a resistor,and cause a varying voltage drop across the resistor. This is shown in figure 1-16.Figure 1-16.—Operation of the plate load resistor.The circuit is identical to the one in figure 1-15 except that now a resistor (called a plate-loadresistor, R_{L}) has been added to the plate circuit, and a voltmeter has been added to measure the voltagedrop across R_{L}.In view (A) of figure 1-16, the control grid is at 0 volts. Once again 5 milliamperes flow in the platecircuit. Now, the 5 milliamperes must flow through R_{L}. The voltage drop is equal to:E = I × RE = (5 × 10^{-3} amperes) × (10 × 10^{3} ohms)E = (5 × 10^{-3}) × (10 × 10^{3})E = 5 × 10E = 50 voltsThus the voltage drop across the plate-load resistor, R_{L}, is 50 volts when no voltage is applied to thegrid. In view (B) of the figure, +3 volts is applied to the control grid. Once again plate current increases to10 milliamperes. The voltage drop across R_{L} is