Hello!

(https://drive.google.com/file/d/15IfznjMnKxnxrZXoO6ujigj2Q2SvOjky/view?usp=sharing)

Please consider circuit diagram uploaded in the above google drive link.

i2 = 30 A (given)

i2*R = 220 V

Let E = required voltage across armature, and

Let i1 = current flowing through the armature

We have to find E and i1

i1 = i2 + i3 (Equation 1)

The contact drop per brush is 1 V (given)

Therefore, for two brushes, contact drop will be 2 * 1 V = 2 V

Applying Kirchhoff’s Voltage Law (KVL) in loop 1, we get

0.03*i2 + i2*R + 0.05*i1 + 2 = E

Substituting the value of i2 and i2*R, we get:

0.03*30 + 220 + 0.05*i1 + 2 = E

222.9 + 0.05*i1 = E

Applying Kirchhoff’s Voltage Law (KVL) in loop 2, we get:

0.05*i1 + 250*i3 = 0

0.05*i1 + 250*(i1-i2) = 0

0.05*i1 + 250*i1 – 250*i2 = 0

Substituting the value of i2, we get:

250.05*i1 = 7500

Therefore, i1 = 7500/250.05 = 29.99

Current flowing through the armature, i1 = 29.99 A (Answer)

Now, substituting the value of i1 in equation 2, we get the value of E as:

222.9 + 0.05*(7500/250.05) = E

E = 222.9 + 1.49

Voltage across the armature, E = 224.399 ~ 225 V (Answer)

Hope this helps.