(a) Star connection
Inductive reactance
X L = 2 π f L X L = 2 π × 50 × 42 × 1 0 − 3 = 13.19 Ω X_L = 2πfL \\
X_L = 2π \times 50 \times 42 \times 10^{-3} = 13.19 \;Ω X L = 2 π f L X L = 2 π × 50 × 42 × 1 0 − 3 = 13.19 Ω
Phase impedance
Z p h = R 2 + X L 2 Z p h = 1 5 2 + 13.1 9 2 = 19.97 Z_{ph} = \sqrt{R^2 + X_L^2} \\
Z_{ph} = \sqrt{15^2 + 13.19^2} = 19.97 Z p h = R 2 + X L 2 Z p h = 1 5 2 + 13.1 9 2 = 19.97
Line voltage
V l i n e = 3 V p h V p h = V l i n e 3 = 415 3 = 240 v o l t s V_{line} = \sqrt{3}V_{ph} \\
V_{ph} = \frac{V_{line}}{\sqrt{3}} = \frac{415}{\sqrt{3}} = 240 \;volts V l in e = 3 V p h V p h = 3 V l in e = 3 415 = 240 v o lt s
Phase current
I p h = V p h Z p h I p h = 240 19.97 = 12.02 A I_{ph } = \frac{V_{ph}}{Z_{ph}} \\
I_{ph} = \frac{240}{19.97} = 12.02 \;A I p h = Z p h V p h I p h = 19.97 240 = 12.02 A
Line current
I l i n e = I p h = 12.02 A I_{line} = I_{ph} = 12.02 \;A I l in e = I p h = 12.02 A
Dissipated power
P = 3 I l i n e 2 R P = 3 × ( 12.02 ) 2 × 15 = 6.6 k W P = 3I_{line}^2R \\
P = 3 \times (12.02)^2 \times 15 = 6.6 \;kW P = 3 I l in e 2 R P = 3 × ( 12.02 ) 2 × 15 = 6.6 kW
(b) Delta connection
V l i n e = V p h = 415 V Z p h = 19.97 I p h = V p h Z p h I p h = 415 19.97 = 20.78 A V_{line} = V_{ph} = 415 \;V \\
Z_{ph} = 19.97 \\
I_{ph} = \frac{V_{ph}}{Z_{ph}} \\
I_{ph} = \frac{415}{19.97} = 20.78 \;A V l in e = V p h = 415 V Z p h = 19.97 I p h = Z p h V p h I p h = 19.97 415 = 20.78 A
Dissipated power
P = 3 I p h 2 R P = 3 × 20.7 8 2 × 15 = 11.22 k W P = \sqrt{3}I_{ph}^2R \\
P = \sqrt{3} \times 20.78^2 \times 15 = 11.22 \;kW P = 3 I p h 2 R P = 3 × 20.7 8 2 × 15 = 11.22 kW
Comments
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