Question

Suppose that an ideal transformer has 400 turns in its primary coil and 100 turns in its secondary coil. The primary coil is connected to a 120 V (rms) electric outlet and carries an rms current of 10 mA. What are the rms values of the voltage and current for the secondary?

Answer 1

To solve this problem we need to use the induced voltage ratio law with respect to the number of turns in a solenoid. So

$\frac{\epsilon_2}{\epsilon_1} = -\frac{N_2}{N_1}$

For the given values we have to

$N_1 = 400$

$N_2 = 100$

$\epsilon_2 = 120V$

Replacing we have that,

$\frac{\epsilon_2}{120} = -\frac{100}{400}$

$\epsilon = 30V$

Therefore the RMS value for secondary is 30V.

The current can be calculated at the same way, but here are inversely proportional then,

$\frac{I_2}{I_1} = -\frac{N_1}{N_2}$

Replacing we have

$\frac{I_2}{10mA} = -\frac{400}{100}$

$I_2 = 40mA$

Therefore the rms value of current for secondary is 40mA