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. D) The maximum flux in the core

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

${\frac {480\ volts}{120\ volts}}={\frac {300\ turns}{N_{2}}}$

To solve for the number of turns required for the secondary, the equation is rearranged solving for $N_{2}=$:

$N_{2}={\frac {300\cdot 120}{480}}\Rightarrow N_{2}=75\ turns$

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 in the loop ($i_{2}$ ).
$i_{2}={\frac {V_{2}}{R_{L}}}$
Where $i_{2}=$ Current through secondary, $V_{2}=$Voltage across secondary, $R_{L}=$ Load Resistor ($R_{L}=$ 100 Ω)

$i_{2}={\frac {120\ volts}{100\ \Omega }}\Rightarrow i_{2}=1.2\ A$

Part C:

The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:

${\frac {i_{1}}{i_{2}}}={\frac {N_{2}}{N_{1}}}$
Where $i_{1}=$Current in primary, $i_{2}=$Current in secondary, $N_{1}=$ Number of turns in primary, $N_{2}=$ Number of turns in secondary

${\frac {i_{1}}{1.2\ A}}={\frac {75\ turns}{300\ turns}}$

Rearranging to solve for $i_{1}$:

$\ i_{1}=i_{2}\cdot {\frac {N_{2}}{N_{1}}}\Rightarrow 1.2A\cdot {\frac {75\ turns}{300\ turns}}\Rightarrow i_{1}=.3A$

Part D:

The maximum flux in the core is given by: $\Phi _{m}=$