Series field diverter to raise speed to 600 rpm

Use $\Phi\propto I_f$ (saturation neglected), $T\propto\Phi I_a$ and $E_b\propto\Phi N$. With the diverter in parallel with the field only, the armature current is the line current $I_a$, and the field current is a fraction $k = \dfrac{R_d}{R_d + R_f}$ of it.

        Ia         +----[ Rd ]----+
   ---->----[Ra]---+              +----> 
                   +---[ Rf ]-----+     
                      If = k.Ia

Initial condition (no diverter)

$$I_{a1} = I_{f1} = 60\ \text{A}, \qquad E_{b1} = 500 - 60(0.2 + 0.3) = 470\ \text{V}$$

New condition (600 rpm, same load torque)

Let the new armature current be $I_{a2} = x$ and $I_{f2} = kx$.

Torque constant: $\Phi_1I_{a1} = \Phi_2I_{a2}$, so $60\times 60 = kx\cdot x$:

$$kx^2 = 3600 \;\Rightarrow\; I_{f2} = kx = \frac{3600}{x}$$

Back emf:

$$E_{b2} = 500 - 0.2x - 0.3(kx) = 500 - 0.2x - \frac{1080}{x}$$

Speed ratio:

$$\frac{E_{b2}}{E_{b1}} = \frac{\Phi_2N_2}{\Phi_1N_1} = \frac{I_{f2}}{60}\times\frac{600}{500} \;\Rightarrow\; E_{b2} = 470\times 1.2\times\frac{3600/x}{60} = \frac{33840}{x}$$

Equating the two expressions for $E_{b2}$:

$$500 - 0.2x - \frac{1080}{x} = \frac{33840}{x} \;\Rightarrow\; 0.2x^2 - 500x + 34920 = 0$$ $$x = \frac{500 - \sqrt{500^2 - 4(0.2)(34920)}}{2(0.2)} = 71.91\ \text{A}$$

(The smaller root is taken because the motor current stays near the original value.)

Diverter resistance

$$I_{f2} = \frac{3600}{x} = 50.06\ \text{A}, \qquad k = \frac{I_{f2}}{I_{a2}} = 0.6962$$ $$\frac{R_d}{R_d + R_f} = k \;\Rightarrow\; R_d = \frac{kR_f}{1 - k} = \frac{0.6962\times 0.3}{1 - 0.6962} = 0.688\ \Omega$$

Check: $E_{b2} = 500 - 0.2(71.91) - 0.3(50.06) = 470.60\ \text{V}$, and $470\times 1.2\times(50.06/60) = 470.60\ \text{V}$.

Answer: the diverter resistance is $R_d \approx 0.688\ \Omega$ (the motor then draws about $71.91$ A).