# Electrical Engineering - Transformer

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

Well,

$$\I >= 0.4 \$$

i.e., $$\V/Z >= 0.4\$$

$$\V = 20\$$

$$\Z =\sqrt{R^2 + X^2}\$$, NOT $$\(R+X)\$$

Solve.

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

• I added them because they have the same unit , why not ? – user34755 Nov 10 '17 at 16:10
• @user34755 Same reason as 5+4i =! 9 – winny Nov 10 '17 at 16:19
• 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. – Mitu Raj Nov 10 '17 at 16:27