Gm´=

       0.49 =  1650 / 3350
In all the previous examples the effective transconductance of each triode was the same, as the load impedance was set to zero:
      Gm´ = mu / (Rak + rp).
But when the load impedance represents some positive value, then the top triode's effective Gm must differ from the bottom triode, as the load resistance unequally subtracts from the Gm of both tubes. So, given a greater than zero load impedance, will the calculated value of Rak be equal to the inverse of the top tube's Gm? As a design example let us take a 6DJ8 with a Vb of 200 volts and a load of 1550 ohms. The first step is to find Rak:
       Rak = Rk =  rp + 2Rl
                              mu -1.
Substituting variables,
       Rak = Rk =  3300 + 2x1550
                                    33-1
       Rak = Rk = 200.
Now we find the effective Gm for the top tube, invert this value and compare to the value of Rak.

                           mu

Ra + rp + [(mu + 1)Rak + rp][(Rak+rp)||Rl]
                                      rp + Rak

    Gm´ = 0.005 = 5 mA/v,
and
    200 = 1 / 0.005.
And as Rak = 200, we know the formula worked in this case.
   The test I like to use is to try the formula with a 6AS7, as its puny mu of 2 will quickly reveal a mathematical absurdity that a higher mu tube may hide. Let's even give this test some practical results: Sennheiser makes some wonderful dynamic headphones with an impedance of 300 ohm, the HD-580 and HD-600 for example. These headphones require a just few volts to sing. An OTL headphone amplifier that used a 6AS7 might be just what someone needs. So now that we have the load impedance and the tube, what we need is the power supply voltage. Let's keep the 200 volt power supply from the previous example as it is a sane power supply voltage for a headphone amplifier. Now let's find Rak's value.