If the resistances equal each other, then each will share half the available voltage. If one resistance is twice the value of the other, then it will grab 2/3 of the available voltage.
    Graphically plotting the voltage ratio is easy enough. We start with the resistor that connects to ground, i.e. 0 volts. Fix the first point at 0 volts and 0 current and then place the second point at the intersection of the maximum voltage and the maximum current that this resistor would see at that voltage based on the formula:
       I = V/R.
     Now to plot the resistor that connects to the B+ voltage, we start at the other end of the graph at the maximum B+ voltage and zero current. (This makes sense, because if the value of the resistance were 0 ohms, then the current would be 0 mA.) Moving to the extreme left, we find the value of current this resistor would draw, if it experienced the full B+ voltage; once again, this equals I = V/R. The result of our efforts is two lines crossing each other. The intersection of the two lines defines the voltage at the connection of the two resistances and it defines the maximum current the circuit will see with the given B+ voltage.

Plate Curves
      Finally, we get to the triode's curving grid lines. Each line represents the triode's current conduction plotted across an increasing voltage, while a constant grid voltage is applied. The reason this set of curves goes by named "plate curves" is that each point on a curve reveals the triode's plate current at that point.  For example, in the graph below, we see that with a grid voltage of -6 volts and a cathode-to-plate voltage of 100 volts, the triode conducts 50 mA of current. So does the rp equal 2k, as 100 / 0.05 = 2000? The answer is no, as once the current climbs over 15 mA, the -6-volt grid line reflects a slope closer to 714 ohms. Yet a 714-ohm resistor would draw 140 mA at 100 volts, not just 50 mA.

Goldpoint

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