Numerical Problem No. 15.3 (Electromagnetism) Physics SSC

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}

$\N_s$ :

N_{p}

$\N_p$ = 1 : 100

\frac{N_{s}}{N_{p}}

$\frac{N_s}{N_p}$ =

\frac{1}{100}

$\frac{1}{100}$

Supplied Voltage to the primary coil of transformer = $V_{p}$ $\V_p$ = 170 V

Current in the primary coil = $I_{p}$ $\I_p$ = 1 mA = 1 ｘ 10⁻³ A

To Find:

Current in the secondary coil = $I_{s}$ $\I_s$ = ?

Solution:

To find

I_{s}

$\I_s$ (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}

$\P_p$ =

P_{s}

$\P_s$

V_{p}

$\V_p$

I_{p}

$\I_p$ =

V_{s}

$\V_s$

I_{s}

$\I_s$ (∴ P = VI )

I_{s}

$\I_s$ =

\frac{V p}{V s}

$\frac{Vp}{Vs}$ ｘ

I_{p}

$\I_p$ --------equation (1)

To find

I_{s}

$\I_s$ 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}}

$\frac{N_s}{N_p}$ =

\frac{V_{s}}{V_{p}}

$\frac{V_s}{V_p}$

V_{s}

$\V_s$ =

\frac{N_{s}}{N_{p}}

$\frac{N_s}{N_p}$ ｘ

V_{p}

$\V_p$

by putting values

V_{s}

$\V_s$ =

\frac{1}{100}

$\frac{1}{100}$ ｘ 170 V

V_{s}

$\V_s$ = 1.7 Volts

Now using equation (1)

I_{s}

$\I_s$ =

\frac{V_{p}}{V_{s}}

$\frac{V_p}{V_s}$ ｘ

I_{p}

$\I_p$

By putting values

I_{s}

$\I_s$ =

\frac{170 V}{1.7 V}

$\frac{170V}{1.7V}$ ｘ 1 ｘ 10⁻³ A

by simplifying we get

I_{s}

$\I_s$ = 0.1 A------ Ans.

Thus the current in the secondary coil is 0.1 A

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