Answer:
(a) 1.984 V
(b) 0.064 A
(c) 7.94 Watt
Explanation:
number of turns in primary coil, Np = 500
number of turns in secondary coil, Ns = 8
Voltage in primary coil, Vp = 124 V
(a)
Let the output voltage is Vs.
By using the formula for the transformer
[tex]\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}[/tex]
[tex]\frac{V_{s}}{124}=\frac{8}{500}[/tex]
Vs = 1.984 V
(b)
Let the input current is Ip.
Output current, Is = 4 A
[tex]\frac{I_{p}}{I_{s}}=\frac{N_{s}}{N_{p}}[/tex]
[tex]\frac{I_{p}}{4}=\frac{8}{500}[/tex]
Ip = 0.064 A
(c)
Input power, P = Vp x Ip
P = 124 x 0.064
P = 7.94 Watt
A. The output voltage for the transformer is 1.984 V
B. The input current required to produce 4 A is 64 mA
C. The input power of the transformer is 7.936 W
Nₛ / Nₚ = Vₛ / Vₚ
8 / 500 = Vₛ / 124
Cross multiply
500 × Vₛ = 8 × 124
Divide both side by 500
Vₛ = (8 × 124) / 500
Vₛ = 1.984 V
Iₚ / Iₛ = Nₛ / Nₚ
Iₚ / 4 = 8 / 500
Cross multiply
Iₚ × 500 = 4 × 8
Divide both side by 500
Iₚ = (4 × 8) / 500
Iₚ = 0.064
Multiply by 1000 to express in mA
Iₚ = 0.064 × 1000
Iₚ = 64 mA
Pₚ = IₚVₚ
Pₚ = 0.064 × 124
Pₚ = 7.936 W
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