3 phase V

Question 1

𝐸𝐴, 𝐸𝐵 and 𝐸𝐶 can be
mathematically written as

Answer 1

𝑒𝐴 = √2𝐸𝑝ℎ sin 𝜔𝑡
* 𝑒𝐵 = √2𝐸𝑝ℎ sin(𝜔𝑡 − 120°)
* 𝑒𝐶 = √2𝐸𝑝ℎ sin(𝜔𝑡 + 120°)

Question 2

how to minimise number of conductors

Answer 2

adding all conductors together gives sum of 0, if all connected from series a to b to c
voltage sum across give 0 and that point A is short circuited to 0

Question 3

delta connection

Answer 3

all connected from series a to b to c
voltage sum across give 0 and that point A is short circuited to 0, but now only 3 gives power to load not 6

Question 4

time varying currents for respective ea, eb and ec

Answer 4

• 𝑖𝐴 = √2𝐼𝑝ℎ sin(𝜔𝑡 ± 𝜙)
• 𝑖𝐵 = √2𝐼𝑝ℎ sin(𝜔𝑡 − 120° ± 𝜙)
• 𝑖𝐶 = √2𝐼𝑝ℎ sin(𝜔𝑡 + 120° ± 𝜙)
  ia +ib+ic = 0

Question 5

star connection

Answer 5

all 3 windings are connected from one point, no need to have another conductor connected to this node

Question 6

what are lines

Answer 6

end terminals of X, Y, Z

Question 7

what is a line voltage

Answer 7

potential difference between 2 lines giveing voltages eXY, eYZ and eZX

Question 8

line V for delta connection formula sheet

Answer 8

line V = phase V so EL=Eph

Question 9

line current for delta formula sheet

Answer 9

𝑖𝑋 = 𝑖𝐴 − 𝑖𝐶
* 𝑖𝑌 = 𝑖𝐵 − 𝑖𝐴
* 𝑖𝑍 = 𝑖𝐶 − 𝑖B
|𝐼𝑙|= √3|𝐼𝑝ℎ|.

Question 10

star connection line current formula sheet

Answer 10

𝑖𝑋 = 𝑖𝐴, 𝑖𝑌 = 𝑖𝐵 and 𝑖𝑍 = 𝑖𝐶 so 𝐼𝑝ℎ = 𝐼L

Question 11

star connection line voltage formula sheet

Answer 11

𝑒𝑋𝑌 = 𝑒𝐴 − 𝑒𝐵
* 𝑒𝑌𝑍 = 𝑒𝐵 − 𝑒𝐶
* 𝑒𝑍𝑋 = 𝑒𝐶 − 𝑒A
|𝐸𝑙| = √3|𝐸𝑝ℎ|