Question

An ideal transformer has 60 turns on its primary coil and 300 turns on its secondary coil. If 120 V at 2.0 A is applied to the primary, what voltage and current are present in the secondary?

Answer 1

Explanation:

It is given that,

Number of turns in primary coil, $N_p=60$

Number of turns in secondary coil, $N_s=300$

Voltage in primary coil, $V_p=120\ V$

Current in primary coil, $I_p=2\ A$

We have to find the voltage and current in the secondary coli. Firstly calculating the voltage in secondary coil as :

$\dfrac{N_p}{N_s}=\dfrac{V_p}{V_s}$

$\dfrac{60}{300}=\dfrac{120}{V_s}$

$V_s=720\ Volts$

Now, calculating the current present in the secondary coil as :

$\dfrac{N_p}{N_s}=\dfrac{I_s}{I_p}$

$\dfrac{60}{300}=\dfrac{I_s}{2\ A}$

$I_s=0.4\ A$

Hence, this is the required solution.