Question

An ideal step-down transformer has a primary coil of 300 turns and a secondary coil of 18 turns. It is plugged into an outlet with 230.0 V (AC) and it draws a current of 0.31 A. Calculate the voltage in the secondary coil. (in V) 1.38×101 V You are correct. Your receipt no. is 154-2147 Help: Receipt Previous Tries Calculate the current in the secondary coil. (in A) 5.17 A You are correct. Your receipt no. is 154-6875 Help: Receipt Previous Tries Calculate the average power dissipated.

Answer 1

1) 13.8 V

We can use the transformer equation:

`(N_p)/(N_s)=(V_p)/(V_s)`

where we have

`N_p = 300` is the number of turns in the primary coil

`N_s=18` is the number of turns in the secondary coil

`V_p=230.0 V` is the voltage in the primary coil

`V_s = ?` is the voltage in the secondary coil

Solving for Vs, we find

`V_s = (N_s)/(N_p)V_p=(18)/(300)(230.0 V)=13.8 V`

2) 5.17 A

For an ideal transformer, the power in input is equal to the power in output, so we can write:

`P_(in) = P_(out)\\V_p I_p = V_s I_s`

where

`V_p=230.0 V` is the voltage in the primary coil

`V_s = 13.8 V` is the voltage in the secondary coil

`I_p=0.31 A` is the current in the primary coil

`I_s = ?` is the current in the secondary coil

Solving for Is, we find

`I_s = (I_p V_p)/(V_s)=((0.31 A)(230.0 V))/(13.8 V)=5.17 A`