Assuming that the armature resistance remains constant at 0.25 ohms, we can use the formula:

Load current = (terminal voltage - back emf) / armature resistance

At 1500 rpm, the back emf can be calculated using the formula:

Back emf = K * N * Ф

where K is a constant (dependent on the machine's design), N is the number of turns in the armature winding, and Ф is the magnetic flux in the air gap. Since these values are not given in the problem, we cannot calculate the back emf directly.

However, we can use the fact that at 1500 rpm, the generator supplies 200 A at 150 V to find the product of K * N * Ф:

K * N * Ф = (terminal voltage - armature voltage) * generator current

At 1500 rpm, the armature voltage is equal to the product of armature resistance and load current:

Armature voltage = armature resistance * load current

Substituting these values, we get:

K * N * Ф = (150 - 0.25 * load current) * 200

Simplifying, we get:

K * N * Ф = 30000 - 50 * load current

Now, at 1000 rpm, the back emf can be calculated using the formula:

Back emf = K * N * Ф'

where Ф' is the new magnetic flux in the air gap. Since the generator is separately excited, we can assume that the field current remains constant and therefore the magnetic flux is proportional to the speed. Thus:

Ф' / Ф = 1000 / 1500

Ф' = Ф * 2/3

Substituting this into the formula for back emf, we get:

Back emf = K * N * Ф * 2/3

Using the formula for load current, we get:

Load current = (terminal voltage - back emf) / armature resistance

Substituting the given values and simplifying, we get:

Load current = (150 - K * N * Ф * 2/3) / 0.25

We still cannot solve for load current directly, because we do not know the value of K * N * Ф. However, we can use the earlier equation relating K * N * Ф to load current, and solve for K * N * Ф:

K * N * Ф = 30000 - 50 * load current

Substituting this into the formula for load current, we get:

Load current = (150 - (30000 - 50 * load current) * 2/3) / 0.25

Simplifying and solving for load current, we get:

Load current = 105.88 A

Therefore, the load current decreases from 200 A to 105.88 A when the speed drops from 1500 rpm to 1000 rpm, assuming that the armature resistance remains constant.