Answer:
(a) 108
(b) 110.500 kW
(c) 920.84 A
Solution:
As per the question:
Voltage at primary, [tex]V_{p} = 120\ V[/tex] (rms voltage)
Voltage at secondary, [tex]V_{s} = 13000\ V[/tex] (rms voltage)
Current in the secondary, [tex]I_{s} = 8.50\ mA[/tex]
Now,
(a) The ratio of secondary to primary turns is given by the relation:
[tex]\frac{N_{s}}{N_{p}} = \frac{V_{s}}{V_{s}}[/tex]
where
[tex]N_{p}[/tex] = No. of turns in primary
[tex]N_{s}[/tex] = No. of turns in secondary
[tex]\frac{N_{s}}{N_{p}} = \frac{13000}{120}[/tex] ≈ 108
(b) The power supplied to the line is given by:
Power, P = [tex]V_{s}I_{s} = 13000\times 8.50 = 110.500\ kW[/tex]
(c) The current rating that the fuse should have is given by:
[tex]\frac{V_{s}}{V_{p}} = \frac{I_{p}}{I_{s}}[/tex]
[tex]\frac{13000}{120} = \frac{I_{p}}{8.50}[/tex]
[tex]I_{p} = \frac{13000}{120}\times 8.50 = 920.84\ A[/tex]
