Transformer example problem

Problem:

An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 Ω resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary.

Solution:

Part A:

The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:

$\frac{V_1}{V_2}=\frac{N_1}{N_2}$

Where $V_1=$Voltage across primary, $V_2=$Voltage across secondary, $N_1=$ Number of turns in primary, $N_2=$ Number of turns in secondary

$$\begin{gathered} \frac{480 \\ volts}{120 \\ volts}=\frac{300turns}{N_2} \end{gathered}$$

To solve for the number of turns required for the secondary, the equation is rearranged solving for N_2:

$$\begin{gathered} N_2=\frac{300\cdot 120}{480}\Rightarrow N_2=75 \\ turns \end{gathered}$$

Part B:

The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, $V=i\cdot R$, we can solve for the current.
$i_2=\frac{V_2}{R_L}$
Where $i_2=$ Current through secondary, $V_2=$Voltage across secondary, $R_L=$ Load Resistor (R_L = 100 Ω)

$$\begin{gathered} i_2=\frac{120 \\ volts}{100 \\ \mathrm{\Omega }}\Rightarrow i_2=1.2 \\ A \end{gathered}$$