Q5: A 8.5-hp, 200 V d.c. shunt motor running light takes a current of 4 A. The fieldresistance is 100 Q armature resistance is 0.6 Q. Calculate:(i) Speed regulation.(ii) The efficiency.(iii) The armature current at which efficiency is maximum

Answered: Q5: A 8.5-hp, 200 V d.c. shunt motor…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

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Q5: A 8.5-hp, 200 V d.c. shunt motor running light takes a current of 4 A. The field
resistance is 100 Q armature resistance is 0.6 Q. Calculate:
(i) Speed regulation.
(ii) The efficiency.
(iii) The armature current at which efficiency is maximum

Expert Solution

Step 1

Given that

$V_{t} = 200 V I_{L} = 4 A R_{f} = 100 Ω R_{a} = 0.6 Ω$

Figure : shunt DC Generator

shunt or field current

$I_{f} = \frac{V_{t}}{R_{f}} = \frac{200}{100} = 2 A$

Armature current

$I_{a} = I_{L} - I_{f} = 4 - 2 = 2 A$

then we will found back EMF

$E_{b} = V_{t} - I_{a} R_{a} = 200 - 2 \times 0.6 = 198.8 V$

(i) Speed regulation $= \frac{N_{n l} - N_{f l}}{N_{f l}}$

$= \frac{N_{n l}}{N_{f l}} - 1 . . . . . . . . . . . . . . . . . . . . . (1)$

voltage at no load $E_{b 1} = V_{t}$

voltage at full load $E_{b 2} = E_{b}$

As we know for dc shunt motor

$\frac{E_{b 1}}{E_{b 2}} = \frac{N_{1}}{N_{2}} \frac{V_{t}}{E_{b}} = \frac{N_{n L}}{N_{f l}}$

then from equation (1)

$Voltage regulat i o n = [\frac{V_{t}}{E_{b}} - 1] \times 100 = [\frac{200}{198.8} - 1] \times 100 V o l t a g e R e g u l a t i o n = 0.603 %$

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Q5: A 8.5-hp, 200 V d.c. shunt motor running light takes a current of 4 A. The field resistance is 100 Q armature resistance is 0.6 Q. Calculate: (i) Speed regulation. (ii) The efficiency. (iii) The armature current at which efficiency is maximum