A 220 V, DC shunt motor is operating at a speed of 1440

A 220 V, DC shunt motor is operating at a speed of 1440 rpm. The armature resistance is 1.0 Ω and armature current is 10 A. If the excitation of the machine is reduced by 10%, the extra resistance to be put in the armature circuit to maintain the same speed and torque will be

I_a1 = 10
Now flux is decreased by 10%, so φ₂ = 0.9 φ₁
Torque is constant, & T = k_T φ I_a
Then, I_a1 φ₁ = I _a2 φ₂

⇒ I_a2 =	10	= 11.11 A
	0.9

Now , N ∝	E_b
	φ

⇒	N₁	=	E_b1	×	φ₂
	N₂		E_b2		φ₁

=	220 - I_a1 r₁	×	0.9φ₁
	220 - I_a2 (r₁ + R)		φ₁

∴ 1 =	210 × 0.9
	220 - 11.11 (1 + R)

⇒ 1 + R = 2.79
⇒ R = 1.79 Ω