Question

The circuit given in the figure is driven by a voltage source $V_s = 25\sqrt{2}\angle 30^\circ V$. The system is operating at a frequency of 50 Hz. The transformers are assumed to be ideal. The average power dissipated, in W, in the $50 k\Omega$ resistance is ________ (rounded off to two decimal places).

 

Hint:

For ideal transformers, impedances are scaled by the square of the turns ratio when reflected from one side to the other. This simplifies the analysis of circuits with multiple transformers.

Answer 1

Correct Answer: 31

Step 1: Reflect the $50 k\Omega$ resistor through the second transformer.
The impedance seen at the primary of the second transformer is $Z_{p2} = \frac{50 \times 10^3}{10^2} = 500 \Omega$. 
Step 2: Calculate the total impedance on the secondary side of the first transformer.
$Z_{s1} = 400 \Omega + Z_{p2} = 400 \Omega + 500 \Omega = 900 \Omega$. 
Step 3: Reflect this impedance through the first transformer to the primary side.
The impedance seen by the source (excluding the $1 \Omega$ resistor) is $Z_{p1} = \frac{Z_{s1}}{10^2} = \frac{900}{100} = 9 \Omega$. 
Step 4: Calculate the total impedance seen by the voltage source.
$Z_{total} = 1 \Omega + Z_{p1} = 1 \Omega + 9 \Omega = 10 \Omega$. 
Step 5: Calculate the current drawn from the voltage source.
$I_1 = \frac{V_s}{Z_{total}} = \frac{25\sqrt{2}\angle 30^\circ}{10} = 2.5\sqrt{2}\angle 30^\circ A$. 
Step 6: Calculate the current in the secondary of the first transformer.
$I_2 = \frac{I_1}{10} = \frac{2.5\sqrt{2}\angle 30^\circ}{10} = 0.25\sqrt{2}\angle 30^\circ A$. 
Step 7: Calculate the current in the secondary of the second transformer (through the $50 k\Omega$ resistor).
$I_4 = \frac{I_2}{10} = \frac{0.25\sqrt{2}\angle 30^\circ}{10} = 0.025\sqrt{2}\angle 30^\circ A$. 
Step 8: Calculate the average power dissipated in the $50 k\Omega$ resistor. $P = |I_{4,rms}|^2 R = \left(\frac{|I_4|}{\sqrt{2}}\right)^2 R = \left(\frac{0.025\sqrt{2}}{\sqrt{2}}\right)^2 \times 50 \times 10^3 = (0.025)^2 \times 50000 = 0.000625 \times 50000 = 31.25 W$.