Ideal Transformer

The ratio between the primary voltage $V_1$ and the secondary voltage $V_2$ of a transformer is proportional to the ratio between the numbers of turns:

$$\frac{V_2}{V_1}=\frac{n_2}{n_1}$$

If there is no power loss by the transformer, then the transformer is ideal and we have

$$P_1=V_1I_1=P_2=V_2I_2,\;\;\;\;\mbox{i.e.}\;\;\;\; \frac{I_2}{I_1}=\frac{V_1}{V_2}=\frac{n_1}{n_2}$$

The ratio between the primary current $I_1$ and the secondary current $I_2$ of a transformer is inversely proportional to the ratio between the numbers of turns. Also note the reference directions of the currents $I_1$ and $I_2$ and the reference polarities of the voltages $V_1$ and $V_2$, reflecting the fact that the secondary current $I_2$ is caused by the induced voltage $V_2$ (consistent polarity), while $V_1$ is the induced voltage opposing the current $I_1$.

We can also find the ratio between the primary and secondary impedances based on the assumption that there is no power loss in the transformer, i.e.,

$$P_1=\frac{V_1^2}{Z_1}=P_2=\frac{V_2^2}{Z_2}, \;\;\;\;\; \frac{Z_1}{Z_2}=(\frac{V_1}{V_2})^2=(\frac{n_1}{n_2})^2$$

Example 1: Assume $R_0=1000\Omega$, $R_L=10\Omega$, and the voltage source is $V=12V$. Find the turn ratio of the transformer so that the load resistor will get maximum power from the voltage source.

We know that when the load resistor will get maximum power if its resistance is equal to the internal resistance of the voltage source.

• Convert load resistance on the secondary side $R_L=10\Omega$ to $R'_L=n^2\; R_L=10\; n^2$ on the primary side.
• Match $R'_L=10\;n^2$ to $R_0$, i.e., $10\; n^2=1000\Omega$
• Find turn ratio $n=N_1/N_2=10$.
• Find power consumption on $R'_L$:
  
  $$P=(\frac{V}{2})^2/R'_L=36/1000=0.036W$$
  
  

• The power consumed by real load $R_L$ is the same, due to ideal transformer.

Example 2: Assume $R_0=10\Omega$, $R_L=5\Omega$, and the turn ratio is $n=n_1/n_2=2$. Describe this circuit as a two-port network.

• Set up basic equations:
  
  $$\left\{ \begin{array}{l} V_1=10\;I_1+2\;V_2 I_2-V_2/5=-2\;I_1 \end{array} \right.$$
  
  

• Rearrange the equations in the form of a Z-model. The second equation is
  
  $$V_2=10\;I_1+5\;I_2$$
  
  
  Substituting into the first equation, we get
  
  $$V_1=30\;I_1+10\;I_2$$
  
  
  The Z-model is:
  
  $$\left[ \begin{array}{l} V_1 V_2 \end{array} \right]= \lef... ...right] \left[ \begin{array}{l} I_1 I_2 \end{array} \right]$$
  
  
  As $Z_{12}=Z_{21}=10$, this is a reciprocal network.

Alternatively, we can set up the equations in terms of the currents:

• 
  
  $$\left\{ \begin{array}{l} I_1=(V_1-2\;V_2)/10 I_2=-2\;I_1+V_2/5 \end{array} \right.$$
  
  

• Rearrange the equations in the form of a Y-model. The first equation is
  
  $$I_1=V_1/10-V_2/5$$
  
  

• Substituting into the second equation, we get
  
  $$I_2=-V_1/5+2V_2/5+V_2/5=-V_1/5+3V_2/5$$
  
  
  The Y-model is:
  
  $$\left[ \begin{array}{l} I_1 I_2 \end{array} \right]= \lef... ...right] \left[ \begin{array}{l} V_1 V_2 \end{array} \right]$$
  
  

Finally, we can verify that ${\bf Z}^{-1}={\bf Y}$

$${\bf Z}^{-1}=\left[ \begin{array}{rr} 30 & 10 10 & 5 \end{... ...y}{rr} 1/10 & -1/5 -1/5 & 3/5 \end{array} \right] ={\bf Y}$$