SOLUTION

Concept:

Turns ratio for the transformer is given by

$\frac{{{V_p}}}{{{V_s}}} = \frac{{{I_s}}}{{{I_p}}} = \frac{{{N_p}}}{{{N_s}}}$

Where,

V_p is voltage on primary side

V_s is voltage on secondary side

I_p is current on primary side

I_s is current on secondary side

N_p is turns on primary side

N_s is turns on secondary side

In transformer if R₂ is the resistance connected on the secondary side.

Then the value of the secondary resistance referred to primary is $R_1’$

$\frac{{R_1′}}{{{R_2}}} = {\left( {\frac{{{N_p}}}{{{N_s}}}} \right)^2} = {\left( {\frac{{{V_p}}}{{{V_s}}}} \right)^2} = {\left( {\frac{{{I_s}}}{{{I_p}}}} \right)^2}$

Calculation:

Given that,

Voltage source is $V = 10\sqrt 2 \sin \left( {100\pi t} \right)\;V$

Referring 10Ω resistor to primary side of transformer with the transformation ratio of 10 : 1

$\frac{{R_1′}}{{10}} = {\left( {\frac{{10}}{1}} \right)^2}$

$R_1′ = 1000\;{\rm{\Omega }}$

For the transformer with 1 : 10 transformer ratio the total equivalent resistance at secondary is

$R_2′ = 1 + 1000 = 1001\;{\rm{\Omega }}$

Now referring AC voltage source $V = 10\sqrt 2 \sin \left( {100\pi t} \right)$ V to the secondary side of transformer with the transformation ratio of 1 : 10

$\frac{{10\sqrt 2 \sin \left( {100\pi t} \right)}}{{{V_s}}} = \frac{1}{{10}}$

${V_s} = 100\sqrt 2 \sin \left( {100\pi t} \right)$

RMS value of  ${V_s} = 100\sqrt 2 \sin \left( {100\pi t} \right)$

V_rms = 100 V

Then the RMS value of the current flowing in the 1 Ω resistor and source RMS voltage at the secondary side is 100 V.

Therefore, the RMS value of the current flowing in the above circuit is

$I = \frac{{100}}{{1001}} = 0.099\;A \approx 0.1\;A$