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Question: A transformer has 500 primary turns and 10 secondary turns. If the secondary has resistive load of \...

A transformer has 500 primary turns and 10 secondary turns. If the secondary has resistive load of 15Ω15\Omega, the currents in the primary and secondary respectively, are:
(A) 0.16  A,  3.2×103A0.16\;A,\;3.2 \times {10^{ - 3}}A
(B) 3.2×103A,  0.16A3.2 \times {10^{ - 3}}A,\;0.16A
(C) 0.16A,  0.16A0.16A,\;0.16A
(D) 3.2×103A,  3.2×103A3.2 \times {10^{ - 3}}A,\;3.2 \times {10^{ - 3}}A

Explanation

Solution

Hint The transformer is defined as a static electrical machine that transfers AC electrical power from one circuit to another circuit at a constant frequency, but the voltage level can be altered which means that the voltage can be either increased or decreased according to the requirement of the user. Based on this concept we have to answer this question.

Complete step by step answer:
Here it is given that, VSVP=NSNP\dfrac{{{V_S}}}{{{V_P}}} = \dfrac{{{N_S}}}{{{N_P}}}
Or
VS=NSNP×VP=10500×120=2.4V{V_S} = \dfrac{{{N_S}}}{{{N_P}}} \times {V_P} = \dfrac{{10}}{{500}} \times 120 = 2.4V
Secondary current is given as, IS=VSRS=2.415=0.16A{I_S} = \dfrac{{{V_S}}}{{{R_S}}} = \dfrac{{2.4}}{{15}} = 0.16A
Now we have to write that, ISIP=NPNS\dfrac{{{I_S}}}{{{I_P}}} = \dfrac{{{N_P}}}{{{N_S}}}
Or
IP=NSNP×IS=10500×0.16=0.0032A{I_P} = \dfrac{{{N_S}}}{{{N_P}}} \times {I_S} = \dfrac{{10}}{{500}} \times 0.16 = 0.0032A

Hence, the correct answer is Option B.

Note The primary winding of a transformer is connected to the input voltage supply and converts or transforms the electrical power into a magnetic field. On the other hand, the function of the secondary winding is to convert this alternating magnetic field into electrical power producing the required output voltage.