Answer with Explanation:
We are given that
[tex]V_1=120 V[/tex]
[tex]V_2=13000 V[/tex]
[tex]I_{2}=8.5 mA=8.5\times 10^{-3} A[/tex]
[tex]1 mA=10^{-3} A[/tex]
a.We know that
[tex]\frac{N_2}{N_1}=\frac{V_2}{V_1}[/tex]
[tex]\frac{N_2}{N_1}=\frac{13000}{120}=108.3[/tex]
b.[tex]P_{avg}=I_2V_2[/tex]
[tex]P_{avg}=13000\times 8.5\times 10^{-3}=110.5 Watt[/tex]
c.[tex]I_1=\frac{P_{avg}}{V_1}[/tex]
[tex]I_1=\frac{110.5}{120}=0.92 A[/tex]
Answer:
Explanation:
Voltage in primary, Vp = 120 V
Voltage in secondary, Vs = 13000 V
Current in secondary, Is = 8.5 mA
(a)
[tex]\frac{N_{s}}{N_{p}}=\frac{V_{s}}{V_{p}}[/tex]
[tex]\frac{N_{s}}{N_{p}}=\frac{13000}{120}[/tex]
Ns : Np = 108.33
(b)
For an ideal transformer
Input power = Output power
Input Power = Vs x Is = 13000 x 0.0085
Input power = 110.5 Watt
(c)
[tex]\frac{I_{p}}{I_{s}}=\frac{V_{s}}{V_{p}}[/tex]
[tex]I_{p}=\frac{13000}{120}\times 8.5[/tex]
Ip = 920.8 mA
