Question

Question:
For a DC motor, open circuit voltage at rated speed of 1000 rpm is 200 V. What is the armature resistive drop at
800
rpm
when terminal voltage is 180 V.​

Answer 1

Answer:

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Step-by-step explanation:

Let r(x) = ax+b, when p(x) is divided by (x-2)(x-3).

According to division lemma,

⇒ p(x) = (x-2)(x-3) q(x) + r(x)

⇒ p(x) = (x-2)(x-3) q(x) +ax+b

Now, p(2) = 1

⇒ (2-2)(2-3) q(3) + a×2 + b = 1

⇒ 2a+b = 1 ⇢ (i)

Also, p(3) = 3

⇒ (3-2)(3-3) q(3) + a×3 + b = 3

⇒ 3a+b = 3 ⇢ (ii)

Subtracting (i) and (ii), we get

a = 2, b = -3 ⇒ r(x) = 2x-3

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