An iron transformer assumed to be $100 %$ efficient. The ration of the secondary truns to the primary truns is $1 : 20$ . A $240 V$ ac supply is connected to the primary coil. What is the current in the primary coil?

0.1 A

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0.14 A

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2 A

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40 A

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Solution

The correct option is A $0.1 A$
As this is $100 %$ efficient this means the power in the secondary coil will be same as the power in the primary coil
i.e, $I_{p} V_{p} = I_{s} V_{s}$
$I_{p} = \frac{I_{s} V_{s}}{V_{p}}$
As we know that the ratio of emf is equal to the ratio of the number of turns.
So, $I_{p} = I_{s} . \frac{N_{s}}{N_{p}} = \frac{V_{s}}{R_{s}} . \frac{N_{s}}{N_{p}} = (\frac{N_{S}}{N_{P}}) . \frac{V_{p}}{R_{s}} . \frac{N_{s}}{N_{p}}$
$I_{p} = \frac{N_{s}^{2}}{N_{p}^{2}} \frac{V_{p}}{R_{s}}$
$I_{p} = {(\frac{1}{20})}^{2} . \frac{240}{6}$
$I_{p} = \frac{1}{10} = 0.1 A$

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An iron transformer assumed to be 100 % efficient. The ration of the secondary truns to the primary truns is 1:20. A 240 V ac supply is connected to the primary coil. What is the current in the primary coil?

An iron transformer assumed to be $100 %$ efficient. The ration of the secondary truns to the primary truns is $1 : 20$ . A $240 V$ ac supply is connected to the primary coil. What is the current in the primary coil?