Given
Number of turns in primary coil, n=7500
Number of turns in secondary coil, n'=125
The voltage across the primary coil, V=7200 V
To find
a. What is the voltage across the secondary? Answer
b. The current in the secondary is 36 A. What current flows in the primary?
Explanation
a.
Voltage across secondary coil is
[tex]\begin{gathered} V^{\prime}=\frac{n^{\prime}}{n^\text{ }}V \\ \text{ }\Rightarrow V^{\prime}=\frac{125}{7500}\times\text{ }7200 \\ \text{ }\Rightarrow V^{\prime}=120\text{ }V \end{gathered}[/tex]b. The current in the secondary coil is I'=36A
We know,
[tex]\begin{gathered} VI=V^{\prime}I^{\prime} \\ \Rightarrow7200I=120\times36 \\ \Rightarrow I=0.60\text{ A} \end{gathered}[/tex]Conclusion
a. The voltage across the secondary coil is 120 V
b. The current across the primary coil is 0.60 A
