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

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

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 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)

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

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

$$\begin{gathered} \frac{i_1}{1.2 \\ A}=\frac{75 \\ turns}{300 \\ turns} \end{gathered}$$

Rearranging to solve for $i_1$:

$$\begin{gathered} {i}_1=i_2\cdot \frac{N_2}{N_1}\Rightarrow 1.2A\cdot \frac{75 \\ turns}{300 \\ turns}\Rightarrow i_1=.3A \end{gathered}$$

Part D:

The induced emf of the secondary can be calculated by: $$\begin{gathered} V_2=4.44 \\ {f}\cdot N_2\cdot {\mathrm{\Phi }}_m \\ \mathrm{\angle }{0}^{\circ } \end{gathered}$$<ref>Guru and Huseyin, Electric Machinery and Transformers, 3rd ed. (New York: Oxford University Press, 2001), 209.</ref> Solving for ${\mathrm{\Phi }}_m$ we can calculate the maximum flux in the core: ${\mathrm{\Phi }}_m=\frac{V_1}{4.44\cdot {f}\cdot N_1}$

References:

<references/>

Authors:

Tim Rasmussen