A multipurpose transformer has a secondary coil with several points at which a voltage can be extracted, giving outputs of 5.60, 12.0, and 480 V. (a) The input voltage is 240 V to a primary coil of 280 turns. What are the numbers of turns in the parts of the secondary used to produce the output voltages? (b) If the maximum input current is 5.00 A, what are the maximum output currents (each used alone)?

• The number of turns in the primary coil:\({N_p} = 280\)
• Input voltage:\({V_p} = 240\;{\rm{V}}\)
• Maximum input current:\({I_p} = 5\;{\rm{A}}\)
• Output voltage: \({V_{s1}} = 5.6\;{\rm{V}}, {V_{s2}} = 12\;{\rm{V}}, {V_{s3}} = 480\;{\rm{V}}\)

An electrical device that works on the phenomenon of electromagnetic induction and can transfer electrical energy from one alternating circuit to another.

If\({V_p}\),\({N_p}\),\({I_p}\)are the input potential difference, the number of turns and the input current respectively, in the primary circuit and\({V_s}\),\({N_s}\),\({I_s}\)are the output potential difference, the number of coils and the output current respectively, in the secondary circuit. The relation between these variables is given as-

\(\frac{{{V_s}}}{{{V_p}}} = \frac{{{N_s}}}{{{N_p}}} = \frac{{{I_p}}}{{{I_s}}} \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (1)\)

(a.)

For\(\left( {{V_{s1}}} \right) = 5.60\;{\rm{V}}\), the number of turns in the secondary coil\(\left( {{N_{s1}}} \right)\)can be calculated from equation (1) as-

\(\begin{aligned} \frac{{{V_{s1}}}}{{{V_p}}} &= \frac{{{N_{s1}}}}{{{N_p}}}\\\frac{{{\rm{5}}{\rm{.6}}\;{\rm{V}}}}{{{\rm{240}}\;{\rm{V}}}} &= \frac{{{N_{s1}}}}{{280}}\\{N_{s1}} &= \frac{{{\rm{5}}{\rm{.6}}\;{\rm{V}}}}{{{\rm{240}}\;{\rm{V}}}} \times 280\\{N_{s1}} &= 6.53\end{aligned}\)

For\({V_{s2}} = 12\;{\rm{V}}\), the number of turns in the secondary coil\(\left( {{N_{s2}}} \right)\)can be calculated from equation (1) as-

\(\begin{aligned} \frac{{{V_{s2}}}}{{{V_p}}} &= \frac{{{N_{s2}}}}{{{N_p}}}\\\frac{{{\rm{12}}\;{\rm{V}}}}{{{\rm{240}}\;{\rm{V}}}} &= \frac{{{N_{s2}}}}{{280}}\\{N_{s2}} &= \frac{{{\rm{12}}\;{\rm{V}}}}{{{\rm{240}}\;{\rm{V}}}} \times 280\\{N_{s2}} &= 14\end{aligned}\)

For\({V_{s3}} = 480\;{\rm{V}}\), the number of turns in the secondary coil\(\left( {{N_{s3}}} \right)\)can be calculated from equation (1) as-

\(\begin{aligned} \frac{{{V_{s3}}}}{{{V_p}}} &= \frac{{{N_{s3}}}}{{{N_p}}}\\\frac{{480\;{\rm{V}}}}{{{\rm{240}}\;{\rm{V}}}} &= \frac{{{N_{s3}}}}{{280}}\\{N_{s3}} &= \frac{{480\;{\rm{V}}}}{{{\rm{240}}\;{\rm{V}}}} \times 280\\{N_{s3}} &= 560\end{aligned}\)

the number of turns in the secondary coil for input voltages \({V_{s1}} = 5.6\;{\rm{V}}, {V_{s2}} = 12\;{\rm{V}}, \)and \({V_{s3}} = 480\;{\rm{V}}\) are \({N_{s1}} = 6.53\), \({N_{s2}} = 14\) and \({N_{s3}} = 560\) respectively

For\(\left( {{V_{s1}}} \right) = 5.60\;{\rm{V}}\), the maximum output current\(\left( {{I_{s1}}} \right)\)can be calculated from equation (1) as-

\(\begin{aligned} \frac{{{V_{s1}}}}{{{V_p}}} &= \frac{{{I_p}}}{{{I_{s1}}}}\\\frac{{{\rm{5}}{\rm{.6}}\;{\rm{V}}}}{{{\rm{240}}\;{\rm{V}}}} &= \frac{{5\;{\rm{A}}}}{{{I_{s1}}}}\\\frac{{5\;{\rm{A}}}}{{{\rm{5}}{\rm{.6}}\;{\rm{V}}}} \times {\rm{240}}\;{\rm{V}} &= {I_{s1}}\\{I_{s1}} &= 214.28\;{\rm{A}}\end{aligned}\)

For\(\left( {{V_{s2}}} \right) = 12\;{\rm{V}}\), the maximum output current\(\left( {{I_{s2}}} \right)\)can be calculated from equation (1) as-

\(\begin{aligned} \frac{{{V_{s2}}}}{{{V_p}}} &= \frac{{{I_p}}}{{{I_{s2}}}}\\\frac{{{\rm{12}}\;{\rm{V}}}}{{{\rm{240}}\;{\rm{V}}}} &= \frac{{5\;{\rm{A}}}}{{{I_{s2}}}}\\{I_{s2}} &= \frac{{5\;{\rm{A}}}}{{{\rm{12}}\;{\rm{V}}}} \times {\rm{240}}\;{\rm{V}}\\{I_{s2}} &= 100\;{\rm{A}}\end{aligned}\)

For\(\left( {{V_{s3}}} \right) = 480\;{\rm{V}}\), the maximum output current\(\left( {{I_{s3}}} \right)\)can be calculated from equation (1) as-

\(\begin{aligned} \frac{{{V_{s3}}}}{{{V_p}}} &= \frac{{{I_p}}}{{{I_{s3}}}}\\\frac{{{\rm{480}}\;{\rm{V}}}}{{{\rm{240}}\;{\rm{V}}}} &= \frac{{5\;{\rm{A}}}}{{{I_{s3}}}}\\{I_{s3}} &= \frac{{5\;{\rm{A}}}}{{{\rm{480}}\;{\rm{V}}}} \times {\rm{240}}\;{\rm{V}}\\{I_{s3}} &= 2.5\;{\rm{A}}\end{aligned}\)

the maximum output current in the secondary coil for input voltages\({V_{s1}} = 5.6\;{\rm{V}},\)\({V_{s2}} = 12\;{\rm{V}}, \)and\({V_{s3}} = 480\;{\rm{V}}\)are\({I_{s1}} = 214.28\;{\rm{A}}\),\({I_{s2}} = 100\;{\rm{A}}\)and\({I_{s3}} = 2.5\;{\rm{A}}\)respectively.