# Example Problem CourseWork

```Example 1-21
(a) Analysis from the article, we can evaluate the specific information of
the current that in the very beginning is:
π=
ππ΅
100[π]
=
= 400[π΄]
π
0.25[πΊ]
For the initial force, the value is:
πΉ = π΅ππ = 0.5[π] &times; 1.0[π] &times; 400[π΄] = 200[π]
(b) The speed of no-load is means the very beginning of the moto, and:
ππππ = ππ΅ = π π΅ π = 100[π]
Meanwhile:
π=
[π ]
ππ΅
= 100
= 200[π/π ]
π΅π
0.5 [π] &times; 1.0[π]
(c) If we give it a 25π force on that bar, there will be an anti-direction
current exist in the bar for:
πππππ =
πΉπππ
25[π]
=
= 50[π΄]
π΅π
0.5[π] &times; 1.0[π]
Apply with @KVL, we have the formal of:
ππππ + πππππ π = ππ΅
With:
ππππ−πππ€ = 100 [π ] − 50[π΄] &times; 0.25[πΊ] = 87.5 [π ]
For the final speed of the bar:
π=
ππππ−πππ€
87.5[π ]
=
= 175[π/π ]
π΅π
0.5[π] &times; 1.0[π]
The efficacy of the machine will be calculated with the πππππ , and the
circuit is in series:
π=
87.5[π€]
= 87.5%
100[π€]
Figure P1-14
(a) When switch is open, we can assume that we need to measure the
current between π1 and π2 , then sum them up:
πΌ1 =
πΌ2 =
120∠0&deg;[π]
5∠30&deg;[πΊ]
120∠0&deg;[π]
5∠45&deg;[πΊ]
= 24∠ − 30&deg; [π΄]
= 24∠ − 45&deg; [π΄]
And:
πΌ = πΌ1 + πΌ2 = 24 &times; sin(−30&deg;) + 24 &times; sin(−45&deg;)
+ π(24 &times; cos(−30&deg;) + 24 &times; cos(−45&deg;))
And:
πΌ = −28.97 + 37.75π = 47.59∠ − 37.5&deg;[π΄]
Then for the power factor:
ππΉ = cos(−37.5&deg;) = 0.793 [πππππππ]
Then for the (1) real, (2) reactive and (3) apparent power, we
assumed that:
(1) Real
π = ππΌπππ  (π ) = 120[π ] &times; 47.59[π΄] &times; cos(−37.5&deg;) = 4531[π ]
(2) With the analysis for the triangle shape displacement of:
The Reactive is
π = 120[π ] &times; 47.59[π΄] &times; sin(−37.5&deg;) = −3477 [π£ππ]
(3) The apparent power is:
π = 120[π ] &times; 47.59[π΄] = 5711 [ππ΄]
(b) When the switch is closed, the current and the idea would followed the
same method:
πΌ3 = 24∠90&deg; [π΄]
πΌπππ€ = πΌ1 + πΌ2 + πΌ3 = 38.08∠ − 7.5&deg;[π΄]
And the PF is:
ππΉ = cos(∠ − 7.5&deg;) = 0.991 [πππππππ]
The ‘real’
π = 4531 [π ]
The ‘reactive’
π = −596 [π£ππ]
The ‘apparent’
π = 4570 [ππ΄]
```