Question

For the circuit shown in Figure 3.31, determine ${\mathit{V}}_{\mathbf{o}\mathbf{u}\mathbf{t}}\left(t\right)$.

Short answer

Answer

Therefore, the output voltage is ${\mathit{V}}_{\mathbf{o}\mathbf{u}\mathbf{t}}\left(t\right)\left(=3.42sin10tV\right)$.

Step-by-step solution

Determine the formulas of voltage divider rule, the impedance referred to the primary and transformer ratio.

The voltage across resistance using voltage divider rule is

${\mathit{V}}_{\mathbf{b}}\mathit{V}\left(\frac{R_b}{R_b+R_a}\right)$ ……. (1)

Here, $V_b$is a voltage across resistance$R_b$, $R_a$is a series resistance with $R_b$and V is a voltage across $R_b+R_a$.

Write the formula of transformer ratio.

$\frac{{\mathbf{E}}_{\mathbf{2}}}{{\mathbf{E}}_{\mathbf{1}}}\mathbf{=}\frac{{\mathbf{N}}_{\mathbf{2}}}{{\mathbf{N}}_{\mathbf{1}}}$ ……. (2)

Write a formula to find the secondary impedance referred to the primary.

${\mathit{Z}}_{\mathbf{2}}^{\mathbf{\text{'}}}\mathbf{=}{\mathit{Z}}_{\mathbf{2}}{\left(\frac{N_1}{N_2}\right)}^{\mathbf{2}}$ ……. (3)

Determine the primary and secondary voltage of first ideal transformer.

Determine the secondary impedance referred to the primary for second ideal transformer.

Substitute $4\Omega$for $Z_2,{Z_{22}}^{\text{'}}for{Z_2}^{\text{'}},1for{N}_{1}and2for{N}_2$, in equation (3).

$$\begin{gathered} {Z_{22}}^{\text{'}}=8\times {\left(\frac{1}{2}\right)}^2 \\ =2\Omega \end{gathered}$$

The secondary impedance of first ideal transformer is,

$$\begin{gathered} Z_{21\left(total\right)}=Z_{12}+Z_{22}\text{'} \\ =2\Omega +2\Omega \\ =4\Omega \end{gathered}$$.

Determine the secondary impedance referred to the primary for first ideal transformer.

Substitute $4\Omega$for $Z_{21\left(total\right)},2for{N}_{1}and1for{N}_2$, in equation (3).

$$\begin{gathered} Z_{21\left(total\right)}\text{'}=4\times {\left(\frac{2}{1}\right)}^2 \\ =16\Omega \end{gathered}$$

The equivalent resistance of $Z_{21\left(total\right)}\text{'}$and $R_b=36\Omega$is,

$$\begin{gathered} R_b\text{'}=36\parallel 16 \\ =11.076\Omega \end{gathered}$$

Determine the primary voltage $E_{11}$of first ideal transformer.

Substitute $18\sin 10tV$for $V,E_{11}$for $V_b,11.076\Omega$for $R_b$and $18\Omega$for $R_a$in equation (1).

$$\begin{gathered} E_{11}=18\sin 10tV\times \frac{11.076}{11.076+18} \\ =6.84\sin 10tV \end{gathered}$$

Determine the secondary voltage $E_{21}$of first ideal transformer.

Substitute $6.84\sin 10tV$for $E_{11},2for{N}_{1}and1for{N}_2$in equation (2).

$$\begin{gathered} \frac{E_{21}}{6.84\sin 10tV}=\frac{1}{2} \\ {E}_{21}=3.42\sin 10tV \end{gathered}$$

Determine the primary and secondary voltage of second ideal transformer.

Determine the primary voltage $E_{12}$of second ideal transformer.

Using equation (1), the primary voltage of second transformer is,

$$\begin{gathered} E_{12}=E_{21}\frac{Z_{22}\text{'}}{Z_{22}\text{'}+Z_{12}} \\ =\left(\frac{2}{2+2}\right)3.42\sin 10tV \\ =1.71\sin 10tV \end{gathered}$$

Determine the secondary voltage $E_{22}$of second ideal transformer.

Using equation (1), the primary voltage of second transformer is,

Substitute $1.71\sin 10tV$for $E_{12},1for{N}_{1}and2for{N}_2$in equation (2).

$$\begin{gathered} \frac{E_{21}}{1.7\sin 10tV}=\frac{2}{1} \\ {E}_{21}=3.42\sin 10tV \end{gathered}$$

Note: $E_{21}$stands for secondary winding voltage for first ideal transformer. This convention applied for other variables also.

Therefore, the output voltage is $V_{out}\left(t\right)=3.42\sin 10tV$.