A step-down transformer has a turns ratio of 100 : 1. An ac voltage of amplitude 170 V is applied to the primary. If the current in the primary is 1.0 mA, what is the current in the secondary? Ans. (0.1A)



Data Given:


A step-down transformer has a turn ratios = `\N_s`  :  `\N_p` = 1 : 100

`\frac{N_s}{N_p}` = `\frac{1}{100}`


Supplied Voltage to the primary coil of transformer  = `\V_p` = 170 V

Current in the primary coil = `\I_p` = 1 mA = 1 ｘ 10⁻³ A


To Find:


Current in the secondary coil = `\I_s` = ?


Solution:

  

To find `\I_s`  (Current in the secondary coil) by using the following formula for an ideal transformer that is 
Power of primary coils = power of the secondary coil

`\P_p` = `\P_s`   

or

`\V_p` `\I_p` = `\V_s` `\I_s`    (∴ P = VI )

or 

`\I_s` = `\frac{Vp}{Vs}` ｘ `\I_p`  --------equation (1)

To find `\I_s` we have to find first the Vs also So,
by using the formula between voltages and number of turns of Primary and secondary coils are as follows

`\frac{N_s}{N_p}` = `\frac{V_s}{V_p}`

or 

`\V_s` = `\frac{N_s}{N_p}` ｘ `\V_p`

by putting values

`\V_s` = `\frac{1}{100}` ｘ 170 V

`\V_s` = 1.7 Volts

Now using  equation (1)

`\I_s` = `\frac{V_p}{V_s}` ｘ `\I_p` 

By putting values

`\I_s` = `\frac{170V}{1.7V}` ｘ 1 ｘ 10⁻³ A

by simplifying we get

`\I_s` = 0.1 A------ Ans.

Thus the current in the secondary coil is 0.1 A

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