This problem requires calculating the required firing angle ( α α ) for a single-phase full-controlled converter operating in inverter mode (regenerative braking) to maintain a specified armature current.

1. Calculate the Rated Back EMF ($E_{b, rated}$)

Using the rated parameters of the DC motor:

$$V_{a, rated} = E_{b, rated} + I_{a, rated} R_a$$ $$E_{b, rated} = V_{a, rated} - I_{a, rated} R_a$$ $$E_{b, rated} = 210 \, \text{V} - (10 \, \text{A} \times 1 \, \Omega) = 210 - 10 = 200 \, \text{V}$$

2. Calculate the Back EMF at the Operating Speed ($E_{b, op}$)

Since the field current is constant, the back EMF is directly proportional to the speed ($E_b \propto N$).

$$E_{b, op} = E_{b, rated} \times \frac{N_{op}}{N_{rated}}$$ $$E_{b, op} = 200 \, \text{V} \times \frac{600 \, \text{rpm}}{1200 \, \text{rpm}} = 200 \, \text{V} \times 0.5 = 100 \, \text{V}$$

3. Determine the Required Converter DC Voltage ($V_{dc}$)

During regenerative braking, the motor acts as a generator, and the armature current ($I_a = 10 \, \text{A}$) flows out of the motor and back into the converter. The converter operates in the inversion mode, requiring a negative average DC voltage ($V_{dc}$).

The KVL equation for the armature circuit during braking (assuming the current $I_a$ is maintained flowing out of $E_b$ into $V_{dc}$):

$$E_{b, op} = |V_{dc}| + I_a R_a \quad \text{OR}$$ $$V_{dc} = -(E_{b, op} - I_a R_a)$$

We need the magnitude of $V_{dc}$ to limit the current to $10 \, \text{A}$:

$$|V_{dc}| = E_{b, op} - I_a R_a$$ $$|V_{dc}| = 100 \, \text{V} - (10 \, \text{A} \times 1 \, \Omega) = 100 \, \text{V} - 10 \, \text{V} = 90 \, \text{V}$$

Since the armature terminals are suitably reversed for regeneration, the converter must act as an inverter, meaning the required average DC voltage is negative:

$$V_{dc} = -90 \, \text{V}$$

4. Calculate the Peak AC Input Voltage ($V_m$)

The input AC voltage is $240 \, \text{V}$ RMS.

$$V_m = V_{rms} \sqrt{2} = 240 \sqrt{2} \, \text{V}$$ $$V_m \approx 339.411 \, \text{V}$$

5. Determine the Triggering Angle ($\alpha$)

The average DC output voltage for a single-phase full-controlled bridge converter is given by:

$$V_{dc} = \frac{2 V_m}{\pi} \cos(\alpha)$$

Substituting $V_{dc} = -90 \, \text{V}$ and $V_m = 240 \sqrt{2} \, \text{V}$:

$$-90 \, \text{V} = \frac{2 \cdot (240 \sqrt{2})}{\pi} \cos(\alpha)$$ $$-90 \, \text{V} = \frac{678.823}{\pi} \cos(\alpha)$$ $$-90 \, \text{V} \approx 215.916 \cdot \cos(\alpha)$$ $$\cos(\alpha) = \frac{-90}{215.916} \approx -0.416823$$ $$\alpha = \arccos(-0.416823)$$ $$\alpha \approx 114.6465^\circ$$

Rounding off to two decimal places, the triggering angle is $114.65^\circ$.