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Physics Question on Alternating current

A transformer has 500500 primary turns and 1010 secondary turns. If the secondary has resistive load of 15 Ω15 \,\Omega, the currents in the primary and secondary respectively, are

A

0.16A,3.2×10−3A{0.16 A, 3.2 \times 10^{-3} A}

B

3.2×10−3A,0.16A{3.2 \times 10^{-3} A , 0.16 A}

C

0.16A,0.16A{0.16 A, 0.16 A}

D

3.2×10−3A,3.2×10−3A{3.2 \times 10^{-3} A , 3.2 \times 10^{-3} A}

Answer

3.2×10−3A,0.16A{3.2 \times 10^{-3} A , 0.16 A}

Explanation

Solution

As NsNp=IpIs⇒10500=IpIs\frac{N_s}{N_p} = \frac{I_p}{I_s} \Rightarrow \frac{10}{500} = \frac{I_p}{I_s}
or IpIs=150\frac{I_p}{I_s} = \frac{1}{50} or Is=50IpI_s = 50 I_p
This condition is satisfied only when current in primary is 3.2×10−33.2 \times 10^{-3} A and in secondary 0.16 A.