Question

A 440V shunt motor takes a total current of 95A and runs at 1200 rev/min. If the iron, friction and windage losses amount to 1.8kW, the shunt field resistance is 50Ω and the armature resistance is 0.3Ω, determine the overall efficiency of the motor.​

Answer 1

$\large\boxed{\fcolorbox{blue}{brown}{Answer:-}}$

$\rule{200}{2}$

Given :

For DC shunt motor

Line current = armature current + shunt current

= 20 A = ...........1

∵ Shunt current = = \dfrac{supply voltage}{shunt resistance}

shuntresistance

supplyvoltage

i.e Shunt current = = 1 Amp( a )

= 20 A - 1 A

= 19 A

So, Armature current = I_aI a = 19 amp

( b )For Motor

∵ EMF = V - I_aI a × R_aR a

= 250 - 19 × 0.3

= 250 - 5.7

= 244.3 volt

i.e EMF = 244.3 volt

Hence,

( a ) Armature current is 19 amp

( b ) Back EMF is 244.3 volt

$\rule{200}{2}$

$<marquee> Thank You </marquee>$

$\rule{200}{2}$

Answer 2

Answer:

$\huge \underline \bold \pink {♡IT'S Shefali ♡:-}$

$\huge{\mathbb{\red{Hello Mate!!}}}$

$\huge\boxed{\fcolorbox{red}{yellow}{answer}}$

$\huge\boxed{\fcolorbox{white}{pink}{thanks:-))}}$

$<marquee> please mark it as brainliest answer ✌️✌️</marquee>$

$<p style="color:cyan;font-family:cursive;background:black;font size 25px;">answer is in attachment ✌️✌️</p><p>$

$<marquee>Shefali here</marquee>$

$<marquee>(◍•ᴗ•◍)❤(◍•ᴗ•◍)❤(◍•ᴗ•◍)❤</marquee>$

$<marquee>♥╣[-_-]╠♥♥╣[-_-]╠♥♥╣[-_-]╠♥</marquee>$

$<marquee>^_________^^_________^^_________^(◠‿◕)</marquee>$