In a step up transformer, the ratio of turns of primary to secondary is $5 : 6$ and the primary voltage is $410 V$ . If the current in the primary is $4 A$ , then what is the load current?

$1.15 A$

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$2.15 A$

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$3.33 A$

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$4.33 A$

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Solution

The correct option is C
$3.33 A$

Consider,
$V_{1}$ : Voltage across the primary coil
$V_{2}$ : voltage across the secondary coil
$N_{1}$ : Number of turns in primary coil
$N_{2}$ : Number of turns in secondary coil

So, the relation between these terms:-

$\frac{V_{1}}{N_{1}} = \frac{V_{2}}{N_{2}}$

The turn ratio of the transformer is :-

$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}}$

$\frac{N_{1}}{N_{2}} = \frac{5}{6}; \frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}}$

$\frac{410}{V_{2}} = \frac{5}{6}$

$V_{2} = 492 V$

We have known that,
$V_{1} I_{1} = V_{2} I_{2}$
$I_{2} = \frac{V_{1} I_{1}}{V_{2}} = \frac{410 \times 4}{492} = 3.33 A$

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In a step up transformer, the ratio of turns of primary to secondary is 5:6 and the primary voltage is 410 V. If the current in the primary is 4 A, then what is the load current?

In a step up transformer, the ratio of turns of primary to secondary is $5 : 6$ and the primary voltage is $410 V$ . If the current in the primary is $4 A$ , then what is the load current?