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A transformer of primary voltage 220 V and turns ratio of 11:1. The secondary of the transformer is connected to a 30 Ω inductive reluctance in series with a load resistance (R). If the current flowing in (R) must not be below 0.4 A. Then the value of (R) is:

The answer on my book is at most 40 Ω , but I found the answer to be at most 20 Ω

What is did is that , secondary voltage , which can be easily found by the turn ratio, is 20 V

Then, to find the current at the secondary side we divide the secondary voltage = 20 V over the total resistance = 30 + R Ω

after doing the math R should be at most 20Ω for the current at the secondary side not to drop below .4 A

I don't know where is my mistake , Could someone explain to my what I have done wrong on this problem

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well,

I >= 0.4

i.e., V/Z >= 0.4

V = 20

Z =sqrt(R^2 + X^2), NOT (R+X)

Solve.

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You have added the reluctance and resistance, you can't do that. You have to calculate the impedance of the circuit.

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  • \$\begingroup\$ I added them because they have the same unit , why not ? \$\endgroup\$ – user34755 Nov 10 '17 at 16:10
  • \$\begingroup\$ @user34755 Same reason as 5+4i =! 9 \$\endgroup\$ – winny Nov 10 '17 at 16:19
  • \$\begingroup\$ They have the same units, but you know current and voltage have a 90 degree phase relationship in inductor, which is not there in resistor. That phase shift creates this squared relation , when calculating impedance. Refer to the vector diagram of series RL AC circuit. \$\endgroup\$ – Mitu Raj Nov 10 '17 at 16:27

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