An ideal step-down transformer is connected to the main supply of 240 V. It is desired to operate a 12 V, 30 W lamp. Find the current in the primary and the transformation ratio?  (Ans: 0.125 A, 1/20). 



Data Given:


Voltage of the Primary = `\V_P` = 240 V

Voltage of the secondary = `\V_S` = 12 V

Power of secondary = `\P_S` = 30 W


To Find:


Current in primary= `\I_P` = 30 W


Solution:


For an ideal transformer, we know that 

Power Input = Power Out

As P = V`\I` So,

`\V_P` `\I_P` = `\V_s` `\I_s` 

or

`\I_P` = `\frac {V_s I_s}{V_P}`  ------(1)

but the value of `\I_s` is also unknown so, we have for the secondary coil

`\P_S` = `\V_S` `\I_S` 

Or

`\I_s` = `\frac {P_S}{V_S}`

Thus equation (1) after putting the relation of  `\I_s`

`\I_P` = `\frac {\cancel {V_s} (P_S/\cancel {V_s})}{V_P}`


`\I_P` = `\frac {P_S}{V_P}`


Now putting all the corresponding values in equation (1)

`\I_P` = `\frac {30 W}{240 V}`

`\I_P` = 0.125 A ----------------Ans. 1

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Now for the transformation ratio, we know that

`\frac {N_S}{N_P}` = `\frac {V_S}{V_P}`

By putting values 

`\frac {N_S}{N_P}` = `\frac {12 V}{240 V}`

`\frac {N_S}{N_P}` = `\frac {1}{20}` ------------Ans. 2



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