Series motor: resistance to limit speed to 2400 rpm

Importance of back emf

• The back emf $E_b = \dfrac{\Phi ZNP}{60A}$ opposes the supply voltage, so $I_a = (V - E_b)/R_a$. It limits the armature current to a safe value once the motor runs.
• It is self-regulating: when the load increases the speed and $E_b$ fall, more current flows and the torque rises to meet the load; the reverse happens for a lighter load.
• It is the means of energy conversion: the electrical power converted to mechanical power is $E_bI_a$.
• At starting $E_b = 0$, so a starter is needed to limit the current.
• Its value shows the speed of the motor, since $N\propto E_b/\Phi$.

Numerical

For a series motor (unsaturated) $\Phi\propto I$, so $E_b\propto N I$.

Condition 1: 1800 rpm, 40 A

$$E_{b1} = V - IR = 240 - 40(0.2) = 232.0\ \text{V}$$

Condition 2: 2400 rpm, 10 A, with an extra series resistance $R_x$

$$\frac{E_{b2}}{E_{b1}} = \frac{N_2I_2}{N_1I_1} \;\Rightarrow\; E_{b2} = 232.0\times\frac{2400}{1800}\times\frac{10}{40} = 77.33\ \text{V}$$

The total resistance in the circuit is found from the voltage equation:

$$V = E_{b2} + I_2(R + R_x) \;\Rightarrow\; 240 = 77.33 + 10(0.2 + R_x)$$ $$0.2 + R_x = \frac{240 - 77.33}{10} = 16.267 \;\Rightarrow\; R_x = 16.07\ \Omega$$

Answer: an external resistance of $16.07\ \Omega$ must be connected in series with the armature.