The primary to secondary turns ratio of a transformer is given as 2 : 3. If the current through the primary coil is 3 A, then the current through load resistance is

1 A

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4.5 A

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2 A

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1.5 A

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Solution

The correct option is C 2 A
$\frac{N_{p}}{N_{S}} = \frac{2}{3} = \frac{I_{s}}{I_{p}}$
So, $I_{s} = \frac{2}{3} \times I_{p} = \frac{2 \times 3}{3} = 2 A$

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The primary to secondary turns ratio of a transformer is given as 2 : 3. If the current through the primary coil is 3 A, then the current through load resistance is

The correct option is C 2 ANpNS=23=IsIpSo, Is=23×Ip=2×33=2A

The correct option is C 2 A
$\frac{N_{p}}{N_{S}} = \frac{2}{3} = \frac{I_{s}}{I_{p}}$
So, $I_{s} = \frac{2}{3} \times I_{p} = \frac{2 \times 3}{3} = 2 A$