Example 1-21
(a) Analysis from the article, we can evaluate the specific information of
the current that in the very beginning is:
𝑖=
𝑉𝐵
100[𝑉]
=
= 400[𝐴]
𝑅
0.25[𝛺]
For the initial force, the value is:
𝐹 = 𝐵𝑙𝑖 = 0.5[𝑇] × 1.0[𝑚] × 400[𝐴] = 200[𝑁]
(b) The speed of no-load is means the very beginning of the moto, and:
𝑒𝑖𝑛𝑑 = 𝑉𝐵 = 𝜐 𝐵 𝑙 = 100[𝑉]
Meanwhile:
𝜐=
[𝑉 ]
𝑉𝐵
= 100
= 200[𝑚/𝑠]
𝐵𝑙
0.5 [𝑇] × 1.0[𝑚]
(c) If we give it a 25𝑁 force on that bar, there will be an anti-direction
current exist in the bar for:
𝑖𝑙𝑜𝑎𝑑 =
𝐹𝑎𝑝𝑝
25[𝑁]
=
= 50[𝐴]
𝐵𝑙
0.5[𝑇] × 1.0[𝑚]
Apply with @KVL, we have the formal of:
𝑒𝑖𝑛𝑑 + 𝑖𝑙𝑜𝑎𝑑 𝑅 = 𝑉𝐵
With:
𝑒𝑖𝑛𝑑−𝑛𝑒𝑤 = 100 [𝑉 ] − 50[𝐴] × 0.25[𝛺] = 87.5 [𝑉 ]
For the final speed of the bar:
𝜐=
𝑒𝑖𝑛𝑑−𝑛𝑒𝑤
87.5[𝑉 ]
=
= 175[𝑚/𝑠]
𝐵𝑙
0.5[𝑇] × 1.0[𝑚]
The efficacy of the machine will be calculated with the 𝑖𝑙𝑜𝑎𝑑 , and the
circuit is in series:
𝜂=
87.5[𝑤]
= 87.5%
100[𝑤]
Figure P1-14
(a) When switch is open, we can assume that we need to measure the
current between 𝑍1 and 𝑍2 , then sum them up:
𝐼1 =
𝐼2 =
120∠0°[𝑉]
5∠30°[𝛺]
120∠0°[𝑉]
5∠45°[𝛺]
= 24∠ − 30° [𝐴]
= 24∠ − 45° [𝐴]
And:
𝐼 = 𝐼1 + 𝐼2 = 24 × sin(−30°) + 24 × sin(−45°)
+ 𝑖(24 × cos(−30°) + 24 × cos(−45°))
And:
𝐼 = −28.97 + 37.75𝑖 = 47.59∠ − 37.5°[𝐴]
Then for the power factor:
𝑃𝐹 = cos(−37.5°) = 0.793 [𝑙𝑎𝑔𝑔𝑖𝑛𝑔]
Then for the (1) real, (2) reactive and (3) apparent power, we
assumed that:
(1) Real
𝑃 = 𝑉𝐼𝑐𝑜𝑠 (𝜃 ) = 120[𝑉 ] × 47.59[𝐴] × cos(−37.5°) = 4531[𝑊 ]
(2) With the analysis for the triangle shape displacement of:
The Reactive is
𝑄 = 120[𝑉 ] × 47.59[𝐴] × sin(−37.5°) = −3477 [𝑣𝑎𝑟]
(3) The apparent power is:
𝑆 = 120[𝑉 ] × 47.59[𝐴] = 5711 [𝑉𝐴]
(b) When the switch is closed, the current and the idea would followed the
same method:
𝐼3 = 24∠90° [𝐴]
𝐼𝑛𝑒𝑤 = 𝐼1 + 𝐼2 + 𝐼3 = 38.08∠ − 7.5°[𝐴]
And the PF is:
𝑃𝐹 = cos(∠ − 7.5°) = 0.991 [𝑙𝑎𝑔𝑔𝑖𝑛𝑔]
The ‘real’
𝑃 = 4531 [𝑊 ]
The ‘reactive’
𝑄 = −596 [𝑣𝑎𝑟]
The ‘apparent’
𝑆 = 4570 [𝑉𝐴]