Answer:
581.82 mA
145.75 V
Explanation:
N = Number of turns
I = Current
V = Voltage
p denotes primary (input)
s denotes secondary (output)
In a transformer the following relation is applied
[tex]\dfrac{I_p}{I_s}=\dfrac{N_s}{N_p}\\\Rightarrow I_p=I_s\dfrac{N_s}{N_p}=16\dfrac{800}{22}\\\Rightarrow I_p=581.82\ mA[/tex]
The input current is 581.82 mA
[tex]\dfrac{V_p}{V_s}=\dfrac{N_p}{N_s}\\\Rightarrow V_p=V_s\dfrac{N_p}{N_s}=5300\dfrac{22}{800}\\\Rightarrow V_p=145.75\ V[/tex]
The input voltage is 145.75 V
Answer:
[tex]I_p=0.582\ A[/tex]
[tex]V_p=145.75\ V[/tex]
Explanation:
Given:
No. of turns in the primary coil, [tex]N_p=22[/tex]
No. of turns in the secondary coil, [tex]N_s=800[/tex]
output voltage of the transformer, [tex]V_s=5300\ V[/tex]
output current of the transformer, [tex]I_s=0.016\ A[/tex]
from the equation of transformer we have:
[tex]I_s.N_s=I_p.N_p[/tex]
where:
[tex]I_p[/tex] = primary current on the input side
[tex]0.016\times 800=I_p\times 22[/tex]
[tex]I_p=0.582\ A[/tex]
Also
[tex]\frac{V_s}{N_s} =\frac{V_p}{N_p}[/tex]
where:
[tex]V_p=[/tex] input voltage on the primary coil
[tex]\frac{5300}{800} =\frac{V_p}{22}[/tex]
[tex]V_p=145.75\ V[/tex]
